Unit 5 — Kinetics

Check — numeric entry

Dinitrogen pentoxide, N₂O₅, decomposes in a sealed flask. Its concentration falls from 0.50 M to 0.44 M over 30 s. What is the rate of consumption of N₂O₅? Give your answer to two significant figures.

Feedback when correct: Correct. The concentration of N₂O₅ fell by 0.06 M in 30 s, so the rate is 0.06 M ÷ 30 s = 0.0020 M/s.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Iodine, I₂, forms in a sealed flask as hydrogen iodide decomposes. The concentration of I₂ rises from 0 M to 0.030 M over 60 s. What is the rate of formation of I₂? Give your answer to two significant figures.

Feedback when correct: Correct. The concentration of I₂ rose by 0.030 M in 60 s, so the rate of formation is 0.030 M ÷ 60 s = 0.00050 M/s.

Anticipated wrong answers (feedback shown):

Check — multiple choice

In reaction A, the reactant's concentration falls by 0.20 M. In reaction B, the reactant's concentration falls by 0.05 M. Can you tell from these two changes alone which reaction had the higher rate?

Feedback if chosen: This option is wrong — you compared the concentration changes without their times. A rate is change per unit time: 0.05 M in one second beats 0.20 M spread over an hour.

Feedback if chosen: Correct. A rate is a concentration change per unit time. Neither change names its time, so the two rates cannot be compared.

Check — multiple choice

A reaction's change in concentration per unit time, with units such as M/s, is called what?

Feedback if chosen: Correct. The change in a reactant's or product's concentration per unit time is the reaction rate.

Feedback if chosen: This option is wrong — molarity is a concentration itself, the moles of solute per liter. The change per unit time is the reaction rate.

Feedback if chosen: This option is wrong — a mole ratio compares two species' coefficients and carries no time. The change per unit time is the reaction rate.

Check — multiple choice

Sulfur dioxide is consumed in the reaction 2 SO₂ + O₂ → 2 SO₃. While the SO₂ is being consumed, what is the sign of Δ[SO₂]/Δt?

Feedback if chosen: Correct. The reaction consumes the SO₂, so [SO₂] falls, and a falling concentration makes Δ[SO₂]/Δt negative.

Feedback if chosen: This option is wrong — you gave the sign of the reported rate. The reaction consumes the SO₂, so [SO₂] falls, and Δ[SO₂]/Δt is negative.

Check — multiple choice

Chlorine, Cl₂, is consumed as it reacts with hydrogen in a flask. Which expression gives chlorine's rate of consumption as a positive quantity?

Feedback if chosen: This option is wrong — you left the minus sign out. Δ[Cl₂]/Δt is negative for a consumed reactant, so this rate would come out negative.

Feedback if chosen: Correct. Δ[Cl₂]/Δt is negative because the Cl₂ is consumed; the built-in minus sign flips it into a positive rate.

Feedback if chosen: This option is wrong — you inverted the fraction. A rate is a concentration change divided by a time interval, with the minus sign in front: rate = −Δ[Cl₂]/Δt.

Check — multiple choice

In the reaction 2 SO₂ + O₂ → 2 SO₃, a student computes Δ[SO₂]/Δt = −0.0040 M/s and says the negative sign shows the reaction is going in reverse, turning SO₃ back into SO₂ and O₂. Is the student correct?

Feedback if chosen: This option is wrong — you read the minus sign as a direction. A forwards reaction consumes its reactant, so Δ[SO₂]/Δt is negative even while the reaction goes forwards.

Feedback if chosen: Correct. A forwards reaction consumes its reactant, so a reactant's Δ[X]/Δt is negative. The sign marks consumption, not direction.

Check — numeric entry

In the reaction 2 SO₂ + O₂ → 2 SO₃, [SO₂] falls from 0.060 M to 0.036 M over 120 s. What is Δ[SO₂]/Δt over this stretch, with its sign? Give your answer to two significant figures.

Feedback when correct: Correct. Δ[SO₂] = 0.036 M − 0.060 M = −0.024 M, over Δt = 120 s: −0.00020 M/s. The negative sign marks the falling concentration.

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Check — numeric entry

Nitrosyl chloride, NOCl, decomposes in a sealed flask. [NOCl] is 0.500 M at t = 0 and 0.420 M at t = 100 s. What is the average rate of consumption of NOCl between t = 0 and t = 100 s? Give your answer to two significant figures.

Feedback when correct: Correct. [NOCl] fell by 0.080 M over the 100 s stretch, so the average rate is 0.080 M ÷ 100 s = 0.00080 M/s.

Anticipated wrong answers (feedback shown):

Check — multiple choice

A lab report says a reaction's average rate was 0.0025 M/s. Which further detail must the report give for that number to carry its meaning?

Feedback if chosen: This option is wrong — concentration already accounts for volume: molarity is moles per liter. What the average still hides is its time stretch.

Feedback if chosen: This option is wrong — the rate tracks the concentration change, not the amount present. What the average still hides is its time stretch.

Feedback if chosen: Correct. An average rate always names its interval: 0.0025 M/s over one minute and 0.0025 M/s over an hour describe different experiments.

Check — multiple choice

A concentration change divided by the stretch of time over which the change happened is called what?

Feedback if chosen: This option is wrong — molarity is a single concentration, not a change over time. A change divided by its time stretch is the average rate.

Feedback if chosen: This option is wrong — a mole ratio compares coefficients and has no time in it. A change divided by its time stretch is the average rate.

Feedback if chosen: Correct. The concentration change divided by the finite time interval it happened over is the average rate over that interval.

Check — multiple choice

The graph shows [NO₂] falling as NO₂ reacts away in a sealed flask. Two straight lines, J and K, are drawn on the curve. Which line's slope gives the instantaneous rate at t = 40 s?

Feedback if chosen: Correct. Line J touches the curve at t = 40 s and matches its steepness there — the tangent. Its slope is the instantaneous rate at 40 s.

Feedback if chosen: This option is wrong — line K passes through the curve at 20 s and again at 80 s, so its slope averages the change over that stretch. It does not match the curve's steepness at 40 s.

Check — multiple choice

A student reads a concentration–time curve. At t = 90 s the curve sits at a height of 0.25 M. The student says the instantaneous rate at 90 s is therefore 0.25 M/s. Is the student correct?

Feedback if chosen: This option is wrong — you read the curve's height, which is a concentration in M. The instantaneous rate is the slope of the tangent at 90 s, in M/s.

Feedback if chosen: Correct. The curve's height at 90 s is a concentration, 0.25 M. The instantaneous rate at 90 s is the slope of the tangent there: how fast that concentration is changing.

Check — numeric entry

The graph shows [NO₂] against time, with the tangent at t = 50 s drawn and two points on the tangent marked. What is the instantaneous rate of consumption of NO₂ at t = 50 s? Give your answer to two significant figures. Scientific notation such as 1.2e-4 works.

Feedback when correct: Correct. The tangent drops from 0.040 M at 25 s to 0.020 M at 75 s: a rise of −0.020 M over a run of 50 s, so the rate is 4.0 × 10⁻⁴ M/s.

Anticipated wrong answers (feedback shown):

Check — multiple choice

The rate at one single moment, read as the slope of the tangent at that instant, is called what?

Feedback if chosen: This option is wrong — an average rate covers a stretch of time between two clock readings. The rate at one single moment is the instantaneous rate.

Feedback if chosen: Correct. The slope of the tangent at one instant is the instantaneous rate at that instant.

Feedback if chosen: This option is wrong — the curve's height at an instant is a concentration, in M. The rate at that instant is the slope of the tangent there, in M/s.

Check — multiple choice

A chemist watches one reaction for 90 s, then divides the whole concentration change over those 90 s by the 90 s. Which rate has the chemist computed? Choose the most exact name.

Feedback if chosen: Correct. A concentration change divided by a finite stretch of time, here the first 90 s, is the average rate over that stretch.

Feedback if chosen: This option is wrong — an instantaneous rate belongs to one single moment, read from a tangent. Dividing a change by a 90 s stretch averages over the whole stretch.

Feedback if chosen: This option is wrong — the initial rate is a tangent's slope at the single instant t = 0. Dividing a change by a 90 s stretch averages over the whole stretch.

Check — multiple choice

A chemist draws the tangent to a concentration–time curve at t = 45 s and computes its slope. Which rate has the chemist computed? Choose the most exact name.

Feedback if chosen: This option is wrong — no stretch of time was averaged. A tangent's slope belongs to one single instant: the instantaneous rate at 45 s.

Feedback if chosen: Correct. The slope of the tangent at t = 45 s is the instantaneous rate at that instant.

Feedback if chosen: This option is wrong — the initial rate sits at t = 0, the moment of mixing. This tangent sits at t = 45 s, so its slope is the instantaneous rate there.

Check — multiple choice

A chemist draws the tangent to a concentration–time curve at t = 0 and computes its slope. Which rate has the chemist computed? Choose the most exact name.

Feedback if chosen: This option is wrong — no stretch of time was averaged; a tangent's slope belongs to one instant. At t = 0, that slope has its own name: the initial rate.

Feedback if chosen: This option is wrong — the slope at t = 0 is indeed an instantaneous rate, but the instantaneous rate at t = 0 has a more exact name: the initial rate.

Feedback if chosen: Correct. The tangent's slope at t = 0 is the instantaneous rate at the moment the reaction begins — the initial rate.

Check — multiple choice

Chemists give the instantaneous rate at the moment a reaction begins, t = 0, its own name. What is that name?

Feedback if chosen: Correct. The instantaneous rate at t = 0 is called the initial rate.

Feedback if chosen: This option is wrong — an average rate covers a stretch of time. The rate at the single moment t = 0 is the initial rate.

Feedback if chosen: This option is wrong — a molarity is a concentration, not a rate. The rate at the moment of mixing is the initial rate.

Check — multiple choice

In the reaction N₂ + 3 H₂ → 2 NH₃, how many moles of H₂ are consumed for every 1 mol of N₂ consumed?

Feedback if chosen: This option is wrong — you matched the two reactants one-for-one. The coefficients say 3 H₂ for every 1 N₂.

Feedback if chosen: This option is wrong — 2 is the coefficient of the NH₃ formed. The H₂ : N₂ mole ratio is 3 : 1.

Feedback if chosen: Correct. The coefficients give the mole ratio: 3 mol of H₂ are consumed for every 1 mol of N₂.

Check — multiple choice

In N₂ + 3 H₂ → 2 NH₃, both reactants are consumed. Does [H₂] fall faster than [N₂] falls, slower, or at the same speed?

Feedback if chosen: Correct. Every 1 mol of N₂ consumed takes 3 mol of H₂ with it, so [H₂] falls three times as fast as [N₂].

Feedback if chosen: This option is wrong — you inverted the ratio. The coefficients say 3 H₂ per 1 N₂, so the H₂ is consumed three times as fast.

Feedback if chosen: This option is wrong — the same speed would need equal coefficients. The coefficients say 3 H₂ per 1 N₂, so [H₂] falls three times as fast.

Check — multiple choice

In N₂ + 3 H₂ → 2 NH₃, ammonia forms while hydrogen is consumed. Does [NH₃] rise faster than [H₂] falls, slower, or at the same speed?

Feedback if chosen: This option is wrong — faster would need NH₃'s coefficient to beat H₂'s. The coefficients say 2 NH₃ form for every 3 H₂ consumed.

Feedback if chosen: Correct. The coefficients say 2 mol of NH₃ form for every 3 mol of H₂ consumed, so [NH₃] rises at two-thirds of the speed [H₂] falls.

Feedback if chosen: This option is wrong — the same speed would need equal coefficients. The coefficients say 2 NH₃ per 3 H₂, so the NH₃ appears more slowly.

Check — multiple choice

In the reaction 2 CO + O₂ → 2 CO₂, the CO is consumed at 0.0080 M/s. At what rate is the O₂ consumed?

Feedback if chosen: Correct. The coefficients say 1 mol of O₂ per 2 mol of CO, so the O₂ is consumed at half the CO's rate: 0.0040 M/s.

Feedback if chosen: This option is wrong — you matched the two reactants one-for-one. The coefficients say 1 O₂ per 2 CO, so the O₂ rate is half the CO's.

Feedback if chosen: This option is wrong — you doubled instead of halving. The ratio of O₂ to CO is 1 : 2, so the O₂ is consumed at half the CO's rate.

Check — multiple choice

For the reaction 2 N₂O₅ → 4 NO₂ + O₂, which expression gives the rate of the reaction from each species?

Feedback if chosen: This option is wrong — you ignored the coefficients and the sign. The N₂O₅'s change must be divided by 2 and sign-flipped, and the NO₂'s divided by 4.

Feedback if chosen: Correct. Each species' change is divided by its own coefficient, and the consumed reactant's term carries the minus sign.

Feedback if chosen: This option is wrong — you multiplied by the coefficients instead of dividing. Dividing each change by its own coefficient is what makes every species give the same number.

Feedback if chosen: This option is wrong — you put a minus sign on the NO₂ term, but NO₂ is a formed product. Only the consumed reactant's term needs the minus sign.

Check — multiple choice

Two chemists watch the same flask of reacting N₂O₅: one measures the N₂O₅, the other measures the O₂. A student says the two chemists will compute the same reaction rate, because each will divide the species' Δ[X]/Δt by its own coefficient and fix the sign. Is the student correct?

Feedback if chosen: Correct. Dividing each species' Δ[X]/Δt by its own coefficient, with the sign fixed, turns both measurements into the same single number: the rate of the reaction.

Feedback if chosen: This option is wrong — you stopped at the raw Δ[X]/Δt values, which do differ between species. Dividing each by its own coefficient, sign fixed, gives both chemists the same rate.

Check — numeric entry

In 2 N₂O₅ → 4 NO₂ + O₂, a measurement gives Δ[NO₂]/Δt = +0.024 M/s. What is the rate of the reaction? Give your answer to two significant figures.

Feedback when correct: Correct. rate = (1/4)Δ[NO₂]/Δt = 0.024 ÷ 4 = 0.0060 M/s. Any species, divided by its own coefficient, gives this same number.

Anticipated wrong answers (feedback shown):

Check — numeric entry

In the reaction 2 NO + O₂ → 2 NO₂, a measurement gives Δ[NO]/Δt = −0.062 M/s. What is Δ[O₂]/Δt, with its sign? Give your answer to two significant figures.

Feedback when correct: Correct. The ratio of O₂ to NO is 1 : 2, so the size is 0.062 ÷ 2 = 0.031 M/s; the O₂ is consumed, so the sign is negative.

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Check — numeric entry

In the reaction 2 NO + O₂ → 2 NO₂, a measurement gives Δ[NO]/Δt = −0.062 M/s. What is Δ[NO₂]/Δt, with its sign? Give your answer to two significant figures.

Feedback when correct: Correct. The ratio of NO₂ to NO is 2 : 2, one for one, so the size stays 0.062 M/s; the NO₂ is a formed product, so the sign is positive.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Ozone decomposes by the reaction 2 O₃ → 3 O₂. A measurement gives Δ[O₃]/Δt = −0.040 M/s. What is Δ[O₂]/Δt, with its sign? Give your answer to two significant figures.

Feedback when correct: Correct. The ratio of O₂ to O₃ is 3 : 2, so the size is 0.040 × (3/2) = 0.060 M/s; the O₂ is a formed product, so the sign is positive.

Anticipated wrong answers (feedback shown):

Check — multiple choice

A calibration line is built from a set of standard solutions. What does the line let a chemist turn an absorbance reading into?

Feedback if chosen: This option is wrong — volumes come from glassware, not from the line. The calibration line maps an absorbance onto a concentration.

Feedback if chosen: This option is wrong — molar mass comes from the formula, not from the line. The calibration line maps an absorbance onto a concentration.

Feedback if chosen: Correct. The calibration line turns a measured absorbance into the colored species' concentration, read within the standards' range.

Check — multiple choice

Partway through a reaction, a solution's electrical conductivity is rising. What does the rise show is happening in the solution?

Feedback if chosen: Correct. A material carries current through charged particles that are free to move, so a rising conductivity means more mobile ions.

Feedback if chosen: This option is wrong — dissolved particles with no charge carry no current. Conductivity rises only when mobile charged particles become more plentiful.

Feedback if chosen: This option is wrong — electrons do not travel through the solution. The current in a solution is carried by ions, so a rising conductivity means more mobile ions.

Check — multiple choice

Heated in a sealed flask, solid calcium carbonate decomposes into solid calcium oxide and carbon dioxide gas. No species is colored, and no solution is present. Which instrument lets a chemist follow the reaction as it happens?

Feedback if chosen: This option is wrong — nothing in the flask is colored, so there is no absorbance change to read. The CO₂ gas raises the sealed flask's pressure; read that.

Feedback if chosen: Correct. The reaction releases CO₂ gas into a sealed flask, so the pressure increases as the CO₂ accumulates, and the gauge reads it continuously.

Feedback if chosen: This option is wrong — a conductivity probe needs a solution with mobile ions, and this flask holds two solids and a gas. Read the rising gas pressure instead.

Check — multiple choice

In a colorless solvent, orange-brown bromine reacts with excess cyclohexene, and the mixture's color fades. No gas forms and no ions form. Which instrument lets a chemist follow the reaction as it happens?

Feedback if chosen: Correct. The bromine's orange-brown color fades as the bromine is consumed, so the mixture's absorbance falls, and the spectrophotometer reads it continuously.

Feedback if chosen: This option is wrong — no gas forms, so a sealed flask's pressure would not change. The color is the property that changes: read the absorbance.

Feedback if chosen: This option is wrong — no ions form, so the conductivity does not change. The color is the property that changes: read the absorbance.

Check — multiple choice

Benzoyl chloride reacts with water, releasing H⁺(aq) and Cl⁻(aq) ions as it goes. The mixture stays colorless, and no gas leaves it. Which instrument lets a chemist follow the reaction as it happens?

Feedback if chosen: This option is wrong — the mixture stays colorless, so there is no absorbance change to read. Each new H⁺ and Cl⁻ pair raises the conductivity; read that.

Feedback if chosen: This option is wrong — no gas leaves the mixture, so there is no pressure change to read. Each new H⁺ and Cl⁻ pair raises the conductivity; read that.

Feedback if chosen: Correct. Every reacting molecule releases one H⁺ ion and one Cl⁻ ion, so the solution's conductivity climbs as the ions accumulate.

Check — multiple choice

In a sealed flask, gaseous cyclobutane decomposes, each molecule splitting into two ethene molecules, so the moles of gas increase. Every species is colorless, and no solution is present. Which instrument lets a chemist follow the reaction as it happens?

Feedback if chosen: This option is wrong — every species is colorless, so there is no absorbance change to read. The moles of gas increase in a sealed flask, so read the rising pressure.

Feedback if chosen: Correct. One molecule becomes two, so the moles of gas increase, and in a sealed flask the pressure increases with them.

Feedback if chosen: This option is wrong — a gas-filled flask holds no ionic solution to probe. The moles of gas increase, so read the rising pressure.

Check — numeric entry

The blue dye methylene blue loses its color as it reacts in a basic solution. The table gives its concentration at four times. What is Δ[methylene blue]/Δt between t = 0 and t = 180 s, with its sign? Give your answer to three significant figures. Scientific notation such as 1.23e-6 works.

Feedback when correct: Correct. Δ[methylene blue] = 4.85 × 10⁻⁶ − 8.00 × 10⁻⁶ = −3.15 × 10⁻⁶ M, over Δt = 180 s: −1.75 × 10⁻⁸ M/s.

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Check — numeric entry

The red food dye allura red loses its color as it reacts with bleach. The table gives its concentration at four times. What is the average rate of consumption of the dye between t = 50 s and t = 150 s? Give your answer to two significant figures. Scientific notation such as 1.2e-7 works.

Feedback when correct: Correct. The dye fell from 5.90 × 10⁻⁵ M at 50 s to 5.08 × 10⁻⁵ M at 150 s: 0.82 × 10⁻⁵ M ÷ 100 s = 8.2 × 10⁻⁸ M/s.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Sulfuryl chloride, SO₂Cl₂, decomposes in a sealed flask. The graph shows its concentration against time, with two points marked on gridline crossings. What is the average rate of consumption of SO₂Cl₂ between t = 100 s and t = 300 s? Give your answer to two significant figures. Scientific notation such as 1.2e-4 works.

Feedback when correct: Correct. The graph reads 0.08 M at 100 s and 0.04 M at 300 s: a change of 0.04 M over 200 s, so 2.0 × 10⁻⁴ M/s.

Anticipated wrong answers (feedback shown):

Check — multiple choice

A student computes a drawn tangent's slope from two points, using rise over run. Where must both points sit?

Feedback if chosen: Correct. The slope being computed is the tangent line's, so both read points must sit on the tangent line — far apart, on easy-to-read crossings.

Feedback if chosen: This option is wrong — the curve's steepness keeps changing, so two curve points give the average rate over the stretch between them, not the tangent's slope.

Feedback if chosen: This option is wrong — a mixed pair gives the slope of neither the tangent nor the curve. Both read points must sit on the tangent line itself.

Check — numeric entry

Cyclopropane gas slowly rearranges into propene in a sealed flask. The graph shows [cyclopropane] against time; the tangent at t = 100 s is drawn, with two points on the tangent marked on gridline crossings. What is Δ[cyclopropane]/Δt at t = 100 s, with its sign? Give your answer to two significant figures. Scientific notation such as 1.2e-4 works.

Feedback when correct: Correct. The tangent reads 0.45 M at 0 s and 0.15 M at 400 s: a rise of −0.30 M over a run of 400 s gives −7.5 × 10⁻⁴ M/s.

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Check — numeric entry

Azomethane gas decomposes in a sealed flask. The graph shows its concentration against time; the tangent at t = 0 is drawn, with two points on the tangent marked on gridline crossings. What is the initial rate of consumption of azomethane? Give your answer to two significant figures. Scientific notation such as 1.2e-4 works.

Feedback when correct: Correct. The tangent reads 0.032 M at 0 s and 0.008 M at 150 s: a rise of −0.024 M over a run of 150 s, so the initial rate is 1.6 × 10⁻⁴ M/s.

Anticipated wrong answers (feedback shown):

Check — multiple choice

Magnesium ribbon reacts with acetic acid in a flask. A chemist repeats the setup with a more concentrated acetic acid, with everything else held constant. What happens to the reaction's rate?

Feedback if chosen: Correct. A higher reactant concentration raises the rate, when all other conditions are held constant.

Feedback if chosen: This option is wrong — you moved the rate against the concentration. A higher reactant concentration raises the rate.

Feedback if chosen: This option is wrong — you treated the acid's concentration as irrelevant. A higher reactant concentration raises the rate.

Check — multiple choice

Sodium thiosulfate solution reacts with hydrochloric acid, and the mixture slowly turns cloudy. A chemist repeats the setup with twice the volume of each solution, at the same concentrations and temperature. What happens to the reaction's rate?

Feedback if chosen: This option is wrong — you read a bigger volume as a higher concentration. The volume and the particle count doubled together, so each liter holds the same number of particles.

Feedback if chosen: This option is wrong — you moved the rate down with no change to move it. The concentration is the same, so the rate is the same.

Feedback if chosen: Correct. The concentration is unchanged: each liter holds the same number of particles as before, so the rate is unchanged.

Check — multiple choice

Sodium thiosulfate solution reacts with hydrochloric acid, and the mixture slowly turns cloudy. A chemist dilutes the thiosulfate solution with water before adding the acid, with the temperature held constant. What happens to the rate of the clouding reaction?

Feedback if chosen: This option is wrong — you moved the rate up while the concentration went down. Diluting lowers the thiosulfate's concentration, and a lower reactant concentration lowers the rate.

Feedback if chosen: Correct. Diluting lowers the thiosulfate's concentration, and a lower reactant concentration lowers the rate.

Feedback if chosen: This option is wrong — you treated dilution as no change. Diluting lowers the thiosulfate's concentration, and a lower concentration lowers the rate.

Check — multiple choice

A 2 M solution and a 1 M solution of the same solute fill two identical flasks. Which solution holds more solute particles per liter?

Feedback if chosen: Correct. Molarity counts the moles of solute per liter, so a 2 M solution holds twice as many solute particles in each liter.

Feedback if chosen: This option is wrong — you put the larger count with the smaller molarity. Molarity counts the moles of solute per liter, so the 2 M solution holds more.

Feedback if chosen: This option is wrong — you read the two molarities as equal crowding. Molarity counts the moles of solute per liter, so the 2 M solution holds twice as many.

Check — multiple choice

Sodium thiosulfate solution reacts with hydrochloric acid, and the mixture slowly turns cloudy. A chemist doubles the thiosulfate's concentration, with everything else held constant. What happens to the reaction's rate?

Feedback if chosen: Correct. The doubled concentration puts more thiosulfate particles in each liter, so collisions between reactant particles occur more frequently.

Feedback if chosen: This option is wrong — you moved the rate against the concentration. More particles share each liter, so collisions occur more frequently and the rate increases.

Feedback if chosen: This option is wrong — you treated the doubled concentration as no change. More particles share each liter, so collisions occur more frequently and the rate increases.

Check — multiple choice

The doubled thiosulfate solution clouds faster than the original. Why does the higher concentration make the reaction faster?

Feedback if chosen: This option is wrong — you gave concentration temperature's job. A particle's speed follows the temperature, and the temperature did not change.

Feedback if chosen: Correct. A higher concentration puts more particles in a given volume, so collisions between reactant particles occur more frequently.

Feedback if chosen: This option is wrong — you gave concentration temperature's job. The energy a collision transfers follows the temperature, and the temperature did not change.

Check — multiple choice

A chemist dilutes acetic acid with water before lowering in a strip of magnesium. What happens to how often acid particles collide with the magnesium's surface?

Feedback if chosen: This option is wrong — you moved the collision frequency against the concentration. Diluting leaves fewer acid particles in each liter, so collisions with the magnesium occur less frequently.

Feedback if chosen: Correct. Diluting leaves fewer acid particles in each liter, so collisions with the magnesium's surface occur less frequently.

Feedback if chosen: This option is wrong — you treated dilution as no change. Diluting leaves fewer acid particles in each liter, so collisions occur less frequently.

Check — multiple choice

A glow stick makes its light with a chemical reaction. A student moves a glowing glow stick from a refrigerator into a cup of warm water. What happens to the rate of the light-making reaction?

Feedback if chosen: Correct. The warm water raises the glow stick's temperature, and raising the temperature raises the rate.

Feedback if chosen: This option is wrong — you moved the rate against the temperature. The warm water raises the temperature, and a higher temperature raises the rate.

Feedback if chosen: This option is wrong — you treated the warming as no change. The warm water raises the temperature, and a higher temperature raises the rate.

Check — multiple choice

Hydrogen and iodine vapor react inside a sealed glass tube held in a hot oven. A chemist moves the tube to a cooler oven. What happens to the reaction's rate?

Feedback if chosen: This option is wrong — you moved the rate up as the temperature went down. Cooling is the temperature rule read backwards: a lower temperature lowers the rate.

Feedback if chosen: Correct. Cooling is the temperature rule read backwards: a lower temperature lowers the rate.

Feedback if chosen: This option is wrong — you treated the cooling as no change. A lower temperature lowers the rate.

Check — multiple choice

A sealed tube of hydrogen and iodine vapor sits in a hot oven. A chemist moves the tube to a larger oven at the same temperature. What happens to the reaction's rate?

Feedback if chosen: This option is wrong — you read a bigger oven as a hotter oven. The temperature is unchanged, and nothing inside the sealed tube changed, so the rate is unchanged.

Feedback if chosen: This option is wrong — you moved the rate down with no change to move it. The temperature is unchanged, so the rate is unchanged.

Feedback if chosen: Correct. The oven's size is not a condition of the reaction. The temperature is unchanged, and nothing inside the sealed tube changed, so the rate is unchanged.

Check — multiple choice

Which quantity is proportional to the Kelvin temperature of a sample of particles?

Feedback if chosen: Correct. The Kelvin temperature of a sample is proportional to the average kinetic energy of its particles.

Feedback if chosen: This option is wrong — you tied the temperature to the particle count. The Kelvin temperature is proportional to the particles' average kinetic energy, not to their number.

Feedback if chosen: This option is wrong — you tied the temperature to each particle's mass. A particle's mass is fixed; the Kelvin temperature is proportional to the particles' average kinetic energy.

Check — multiple choice

A sealed tube of hydrogen and iodine vapor is moved from a 600 K oven to a 700 K oven. What happens to the reaction's rate?

Feedback if chosen: Correct. At 700 K the particles carry more average kinetic energy, so collisions occur more often and transfer more energy.

Feedback if chosen: This option is wrong — you moved the rate against the temperature. Hotter particles collide more often, and each collision transfers more energy.

Feedback if chosen: This option is wrong — you treated the 100 K rise as no change. Hotter particles collide more often, and each collision transfers more energy.

Check — multiple choice

The tube's reaction goes faster at 700 K than at 600 K. Why does the higher temperature make the reaction faster?

Feedback if chosen: This option is wrong — you gave the speed-up concentration's reason. The tube is sealed, so heating changes no particle count.

Feedback if chosen: This option is wrong — you reversed the energy change. Hotter particles carry more average kinetic energy, so each collision transfers more energy, not less.

Feedback if chosen: Correct. Kelvin temperature is proportional to the particles' average kinetic energy, so at 700 K collisions occur more often and transfer more energy.

Check — multiple choice

A student puts a glowing glow stick into iced water. What happens to how often the glow stick's reacting particles collide?

Feedback if chosen: This option is wrong — you moved the collisions against the temperature. The cold lowers the particles' average kinetic energy, so they move more slowly and collide less often.

Feedback if chosen: Correct. The cold lowers the particles' average kinetic energy, so they move more slowly and collide less often.

Feedback if chosen: This option is wrong — you treated the cooling as no change. The cold slows the particles, so they collide less often.

Check — multiple choice

An effervescent antacid tablet fizzes as it reacts with water. A student crushes an identical tablet into powder before dropping it into identical water. What happens to the reaction's rate?

Feedback if chosen: Correct. Crushing increases the solid's exposed surface area, and more exposed surface raises the rate.

Feedback if chosen: This option is wrong — you moved the rate against the exposed surface. Crushing exposes more surface, and more exposed surface raises the rate.

Feedback if chosen: This option is wrong — you treated crushing as no change. Crushing exposes more surface, and more exposed surface raises the rate.

Check — multiple choice

Loose sugar sprinkled through a burner flame burns in quick flashes. A cook presses the same mass of sugar into one hard cube and holds the cube in the same flame. What happens to the burning rate?

Feedback if chosen: This option is wrong — you moved the rate against the exposed surface. The cube buries most of the sugar inside, and less exposed surface lowers the rate.

Feedback if chosen: Correct. Pressing the grains into one cube buries most of the sugar inside, and less exposed surface lowers the rate.

Feedback if chosen: This option is wrong — you treated the pressing as no change. The cube exposes far less surface than the loose grains, and less exposed surface lowers the rate.

Check — multiple choice

Steel wool reacts with dilute acid in a beaker. A chemist pours the same acid into a wider beaker and lowers in the same steel wool. The acid still covers the steel wool. What happens to the reaction's rate?

Feedback if chosen: This option is wrong — you read the wider beaker as more exposed solid. The steel wool's exposed surface is unchanged, so the rate is unchanged.

Feedback if chosen: This option is wrong — you moved the rate down with no change to move it. The steel wool's exposed surface is unchanged, so the rate is unchanged.

Feedback if chosen: Correct. The beaker's shape is not a condition of the reaction. The steel wool's exposed surface and the acid's concentration are unchanged, so the rate is unchanged.

Check — multiple choice

Fine steel wool and a solid iron bar of equal mass are each held in a burner flame in air. Which iron reacts faster with the oxygen?

Feedback if chosen: Correct. The thin strands expose far more iron particles to the oxygen, and more exposed particles raise the rate.

Feedback if chosen: This option is wrong — you put the faster reaction with the smaller exposed surface. The bar buries most of its iron inside; the strands expose far more of theirs.

Feedback if chosen: This option is wrong — you matched the equal masses to equal rates. The masses are equal, but the strands expose far more iron particles to the oxygen.

Check — multiple choice

The steel wool reacts with the oxygen far faster than the iron bar does. Why does the steel wool react faster?

Feedback if chosen: This option is wrong — you overlooked the equal masses. The two samples hold the same number of iron particles; the strands simply expose more of them.

Feedback if chosen: This option is wrong — you gave the speed-up temperature's reason. Both samples sit in the same flame, so their particles' average kinetic energies match.

Feedback if chosen: Correct. More of the strands' iron particles are exposed and available to collide with oxygen particles, so collisions occur more often.

Check — multiple choice

A student presses loose antacid powder into one solid pellet before dropping it into water. What happens to the number of the solid's particles exposed to the water?

Feedback if chosen: Correct. Pressing the powder into one pellet buries most of the particles inside, so fewer particles are exposed to the water.

Feedback if chosen: This option is wrong — you moved the exposure against the pressing. The pellet buries most of the particles inside, so fewer are exposed.

Feedback if chosen: This option is wrong — you treated the pressing as no change. The pellet buries most of the particles inside, so fewer are exposed.

Check — multiple choice

A car's exhaust gases react as they pass over a platinum surface inside the catalytic converter. The reactions go far faster over the platinum, and the platinum is still there, unchanged, after years of driving. Is the platinum a catalyst for these reactions?

Feedback if chosen: Correct. The platinum increases the reactions' rates and is not consumed overall, so it is a catalyst.

Feedback if chosen: This option is wrong — you refused a substance that passes both tests. The platinum speeds the reactions and survives them unchanged, so it is a catalyst.

Check — multiple choice

Iron speeds the reaction N₂ + 3 H₂ → 2 NH₃ in an industrial reactor. The hydrogen in the reactor is consumed as the ammonia forms. Is the hydrogen a catalyst?

Feedback if chosen: This option is wrong — you overlooked the consumption. The reaction consumes the hydrogen, so the hydrogen is a reactant, not a catalyst.

Feedback if chosen: Correct. The reaction consumes the hydrogen, so the hydrogen is a reactant, not a catalyst.

Check — multiple choice

Argon gas is also present in the same ammonia reactor. The argon is all recovered unchanged, and the rate is the same with or without it. Is the argon a catalyst?

Feedback if chosen: This option is wrong — you stopped at the unchanged argon. A catalyst must also increase the rate, and the rate is the same with or without the argon.

Feedback if chosen: Correct. The argon survives unchanged but increases nothing, so it is simply present, not a catalyst.

Check — multiple choice

A substance that increases a reaction's rate without being consumed overall is called what?

Feedback if chosen: Correct. A substance that increases a reaction's rate without being consumed overall is a catalyst.

Feedback if chosen: This option is wrong — you named the substance a reaction consumes. A reactant's amount falls; this substance's amount stays the same.

Feedback if chosen: This option is wrong — you named a substance the reaction makes. A product forms during the reaction; this substance speeds the reaction and survives it.

Check — multiple choice

Iron is a suitable catalyst for the reaction N₂ + 3 H₂ → 2 NH₃. A chemist adds iron to a hot reactor holding the two gases. What happens to the reaction's rate?

Feedback if chosen: Correct. Adding a suitable catalyst increases the rate of the reaction.

Feedback if chosen: This option is wrong — you moved the rate down on a catalyst's arrival. Adding a suitable catalyst increases the rate.

Feedback if chosen: This option is wrong — you read the unconsumed iron as an idle iron. A catalyst is not consumed, yet it still increases the rate.

Check — multiple choice

A chemist drops powdered glass, which does not act on the reaction N₂ + 3 H₂ → 2 NH₃, into a hot reactor holding the two gases. What happens to the reaction's rate?

Feedback if chosen: This option is wrong — you treated any added solid as a catalyst. A substance that does not act on the reaction leaves its rate alone.

Feedback if chosen: This option is wrong — you moved the rate down with no change to move it. The glass does not act on the reaction, so the rate is unchanged.

Feedback if chosen: Correct. The glass does not act on this reaction, so it is not a suitable catalyst for it, and the rate is unchanged.

Check — multiple choice

An egg white sets as its proteins react in hot water. A cook moves the egg from water at 70 °C to water at 90 °C. What happens to the rate of the setting reactions?

Feedback if chosen: Correct. The hotter water raises the temperature, and raising the temperature raises the rate.

Feedback if chosen: This option is wrong — you moved the rate against the temperature. The 90 °C water is hotter, and a higher temperature raises the rate.

Feedback if chosen: This option is wrong — you treated the 20 °C rise as no change. A higher temperature raises the rate.

Check — multiple choice

Household bleach reacts with a food dye, and the dye's color fades. A chemist dilutes the bleach with water before adding the same amount of dye, with the temperature held constant. What happens to the rate of the fading reaction?

Feedback if chosen: This option is wrong — you moved the rate up while a reactant's concentration went down. Diluting lowers the bleach's concentration, and a lower concentration lowers the rate.

Feedback if chosen: Correct. Diluting lowers the bleach's concentration, so collisions between the reacting particles occur less frequently and the rate falls.

Feedback if chosen: This option is wrong — you treated dilution as no change. Diluting lowers the bleach's concentration, and a lower concentration lowers the rate.

Check — multiple choice

A camper's pile of thin kindling sticks burns briskly. The camper instead burns a single log of the same mass of the same wood, in the same fire pit. What happens to the burning rate?

Feedback if chosen: This option is wrong — you moved the rate against the exposed surface. The log buries most of its wood inside, and less exposed surface lowers the rate.

Feedback if chosen: Correct. The single log exposes far less wood surface to the oxygen than the pile of thin sticks, and less exposed surface lowers the rate.

Feedback if chosen: This option is wrong — you matched the equal masses to equal rates. The log exposes far less surface than the sticks, and less exposed surface lowers the rate.

Check — multiple choice

A reaction in a flask creeps along slowly. A chemist adds a spoonful of a solid that is a suitable catalyst for this reaction. What happens to the reaction's rate?

Feedback if chosen: Correct. Adding a suitable catalyst increases the rate of the reaction.

Feedback if chosen: This option is wrong — you moved the rate down on a catalyst's arrival. Adding a suitable catalyst increases the rate.

Feedback if chosen: This option is wrong — you read the unconsumed catalyst as an idle catalyst. A catalyst is not consumed, yet it still increases the rate.

Check — multiple choice

An egg sets faster in water at 90 °C than at 70 °C. Which particle-level reason explains the faster rate?

Feedback if chosen: This option is wrong — you gave the speed-up concentration's reason. Heating the water adds no protein particles to any liter.

Feedback if chosen: This option is wrong — you gave the speed-up surface area's reason. Heating the water exposes nothing new; the egg's surface is what it was.

Feedback if chosen: Correct. Kelvin temperature is proportional to the particles' average kinetic energy, so at 90 °C collisions occur more often and transfer more energy.

Check — multiple choice

A camper's thin kindling burns faster than a single log of the same mass. Which particle-level reason explains the faster rate?

Feedback if chosen: This option is wrong — you gave the speed-up concentration's reason. The same air surrounds both fires, so each liter of it holds the same number of oxygen particles.

Feedback if chosen: Correct. The thin sticks expose far more wood particles to the oxygen, so collisions with oxygen particles occur more often.

Feedback if chosen: This option is wrong — you gave the speed-up temperature's reason. Both fires start at the same temperature, so the particles' starting average kinetic energies match.

Check — multiple choice

In the heated flask, [NO]'s starting concentration is raised, with everything else held constant. What happens to the initial rate?

Feedback if chosen: Correct. A higher reactant concentration raises the rate for this reaction, when all other conditions are held constant.

Feedback if chosen: This option is wrong — you moved the rate against the concentration. A higher [NO] raises this reaction's rate.

Feedback if chosen: This option is wrong — you treated the concentration as irrelevant. A higher [NO] raises this reaction's rate.

Check — multiple choice

A chemist plans trials of the reaction 2 ClO₂ + 2 OH⁻ → ClO₃⁻ + ClO₂⁻ + H₂O, all at one temperature. Which pair of experiments isolates the effect of ClO₂ on the initial rate?

Feedback if chosen: Correct. From experiment 1 to experiment 2, only [ClO₂] changes, so any change in the initial rate belongs to ClO₂.

Feedback if chosen: This option is wrong — you picked a pair in which both concentrations change. A rate change between those trials could belong to either reactant.

Feedback if chosen: This option is wrong — you picked the pair in which only [OH⁻] changes. That pair isolates the effect of OH⁻, not of ClO₂.

Check — multiple choice

A chemist studies the reaction BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3 H₂O. To test how [H⁺] affects the initial rate, the chemist changes only [H⁺] from trial to trial, keeping [BrO₃⁻], [Br⁻] and the temperature fixed. Why must [BrO₃⁻] and [Br⁻] stay fixed?

Feedback if chosen: This option is wrong — you treated the fixed concentrations as irrelevant. They are fixed because they do affect the rate: fixing them keeps their effects out of the comparison.

Feedback if chosen: This option is wrong — you expected fixed concentrations to fix the rate. The rate still changes as [H⁺] changes; the fixing means only [H⁺] could have changed it.

Feedback if chosen: Correct. With [BrO₃⁻] and [Br⁻] fixed, only [H⁺] changes, so any change in the initial rate belongs to [H⁺].

Check — multiple choice

Here are four planned trials of the same bromate reaction, all at one temperature. Which pair of experiments isolates the effect of Br⁻ on the initial rate?

Feedback if chosen: This option is wrong — you picked the pair in which only [BrO₃⁻] changes. That pair isolates the effect of BrO₃⁻, not of Br⁻.

Feedback if chosen: Correct. From experiment 2 to experiment 3, only [Br⁻] changes, so any change in the initial rate belongs to Br⁻.

Feedback if chosen: This option is wrong — you picked a pair in which both [Br⁻] and [H⁺] change. A rate change could belong to either.

Feedback if chosen: This option is wrong — you picked the pair in which only [H⁺] changes. That pair isolates the effect of H⁺, not of Br⁻.

Check — multiple choice

Acetone reacts with Br₂ in acid: CH₃COCH₃ + Br₂ → CH₃COCH₂Br + HBr. Doubling [Br₂], with everything else held constant, leaves the initial rate unchanged. Does the rate depend on [Br₂]?

Feedback if chosen: This option is wrong — you read a doubled concentration as a changed rate. The measured initial rate did not change, so the rate does not depend on [Br₂].

Feedback if chosen: Correct. The unchanged initial rate means the rate does not depend on [Br₂].

Check — multiple choice

In the acetone reaction, CH₃COCH₃ + Br₂ → CH₃COCH₂Br + HBr, doubling [CH₃COCH₃] doubles the initial rate. Does the rate depend on [CH₃COCH₃]?

Feedback if chosen: Correct. The initial rate changed when only [CH₃COCH₃] changed, so the rate depends on [CH₃COCH₃].

Feedback if chosen: This option is wrong — you read a changed rate as no dependence. The rate doubled when [CH₃COCH₃] doubled, so the rate depends on it.

Check — multiple choice

CH₃Br reacts with OH⁻ in solution: CH₃Br + OH⁻ → CH₃OH + Br⁻. Doubling [OH⁻] doubles the initial rate. Does the rate depend on [OH⁻]?

Feedback if chosen: Correct. The initial rate changed when only [OH⁻] changed, so the rate depends on [OH⁻].

Feedback if chosen: This option is wrong — you read a changed rate as no dependence. The rate doubled when [OH⁻] doubled, so the rate depends on it.

Check — multiple choice

In the acetone reaction, CH₃COCH₃ + Br₂ → CH₃COCH₂Br + HBr, halving [Br₂], with everything else held constant, also leaves the initial rate unchanged. Does the rate depend on [Br₂]?

Feedback if chosen: This option is wrong — you read a halved concentration as a changed rate. The measured initial rate did not change, so the rate does not depend on [Br₂].

Feedback if chosen: Correct. The initial rate did not change when [Br₂] halved, so the rate does not depend on [Br₂].

Check — multiple choice

In the bromate reaction, BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3 H₂O, doubling [Br⁻] doubles the initial rate. Does the rate depend on [Br⁻]?

Feedback if chosen: Correct. The initial rate changed when only [Br⁻] changed, so the rate depends on [Br⁻].

Feedback if chosen: This option is wrong — you read a changed rate as no dependence. The rate doubled when [Br⁻] doubled, so the rate depends on it.

Check — multiple choice

Cyclopropane isomerizes to propene in the gas phase. Doubling [cyclopropane], with everything else held constant, doubles the initial rate. Is the rate directly proportional to [cyclopropane]?

Feedback if chosen: Correct. The rate changed by the same factor as the concentration, so the rate is directly proportional to [cyclopropane].

Feedback if chosen: This option is wrong — you refused a matched pair of factors. The rate doubled exactly as [cyclopropane] doubled, and that is direct proportionality.

Check — multiple choice

Acetone reacts with Br₂ in acid. Doubling [Br₂] leaves the initial rate unchanged. Is the rate directly proportional to [Br₂]?

Feedback if chosen: This option is wrong — you matched an unchanged rate to a doubled concentration. Direct proportionality would have doubled the rate; this rate does not depend on [Br₂] at all.

Feedback if chosen: Correct. A directly proportional rate would have doubled. This rate did not change, so it does not depend on [Br₂].

Check — multiple choice

A chemist halves [cyclopropane], with everything else held constant, and the initial rate halves. Is the rate directly proportional to [cyclopropane]?

Feedback if chosen: Correct. Halving [cyclopropane] halved the rate: the rate changes by the same factor as the concentration, which is direct proportionality.

Feedback if chosen: This option is wrong — you treated halving as outside the rule. Direct proportionality works in both directions: half the concentration, half the rate.

Check — multiple choice

In the reaction of CH₃Br with OH⁻, doubling [CH₃Br] doubles the initial rate. Is the rate directly proportional to [CH₃Br]?

Feedback if chosen: Correct. The rate changed by the same factor as the concentration, so the rate is directly proportional to [CH₃Br].

Feedback if chosen: This option is wrong — you refused a matched pair of factors. The rate doubled exactly as [CH₃Br] doubled, and that is direct proportionality.

Check — multiple choice

Butadiene dimerizes in the gas phase: 2 C₄H₆ → C₈H₁₂. Doubling [C₄H₆], with everything else held constant, quadruples the initial rate. Is the rate proportional to the square of [C₄H₆]?

Feedback if chosen: Correct. The doubled concentration multiplied the rate by two squared, four, so the rate is proportional to the square of [C₄H₆].

Feedback if chosen: This option is wrong — you refused the square response. A quadrupled rate from a doubled concentration is exactly two squared.

Check — multiple choice

CH₃Br reacts with OH⁻, and doubling [CH₃Br] doubles the initial rate. Is the rate proportional to the square of [CH₃Br]?

Feedback if chosen: This option is wrong — you matched a doubled rate to the square response. The square response would have quadrupled the rate; a doubled rate is direct proportionality.

Feedback if chosen: Correct. The square response would have quadrupled the rate. This rate only doubled, so it is directly proportional to [CH₃Br], not proportional to its square.

Check — multiple choice

Acetone reacts with Br₂ in acid, and doubling [H⁺] doubles the initial rate. What does the trial say about how the rate depends on [H⁺]?

Feedback if chosen: This option is wrong — you read a doubled rate as no effect. An unchanged rate would mean no dependence; this rate doubled.

Feedback if chosen: Correct. The rate changed by the same factor as the concentration, so the rate is directly proportional to [H⁺].

Feedback if chosen: This option is wrong — you matched a doubled rate to the square response. The square response would have quadrupled the rate.

Check — multiple choice

In the butadiene dimerization, halving [C₄H₆] divides the initial rate by four. What does the trial say about how the rate depends on [C₄H₆]?

Feedback if chosen: This option is wrong — you read a quartered rate as no effect. The rate changed, so it depends on [C₄H₆].

Feedback if chosen: This option is wrong — you matched a quartered rate to a halved concentration. Direct proportionality would have halved the rate, not quartered it.

Feedback if chosen: Correct. One half squared is one quarter, so the rate follows the square of [C₄H₆].

Check — multiple choice

In the acetone reaction, doubling [Br₂] leaves the initial rate unchanged. What does the trial say about how the rate depends on [Br₂]?

Feedback if chosen: Correct. The initial rate did not change when [Br₂] doubled, so the rate does not depend on [Br₂].

Feedback if chosen: This option is wrong — you matched an unchanged rate to the doubled response. Direct proportionality would have doubled the rate.

Feedback if chosen: This option is wrong — you matched an unchanged rate to the square response. The square response would have quadrupled the rate.

Check — multiple choice

In the bromate reaction, BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3 H₂O, doubling [H⁺] quadruples the initial rate. What does the trial say about how the rate depends on [H⁺]?

Feedback if chosen: This option is wrong — you read a quadrupled rate as no effect. The rate changed, so it depends on [H⁺].

Feedback if chosen: This option is wrong — you matched a quadrupled rate to the doubled response. Direct proportionality would have doubled the rate.

Feedback if chosen: Correct. The doubled concentration multiplied the rate by two squared, four, so the rate is proportional to the square of [H⁺].

Check — multiple choice

For the reaction 2 ClO₂ + 2 OH⁻ → ClO₃⁻ + ClO₂⁻ + H₂O, doubling [ClO₂] alone quadruples the initial rate, and doubling [OH⁻] alone doubles it. Which rate law records both responses?

Feedback if chosen: Correct. The quadrupled response puts a 2 on [ClO₂], and the doubled response leaves the unwritten 1 on [OH⁻].

Feedback if chosen: This option is wrong — you swapped the two responses. The quadrupled response belongs to ClO₂, so the 2 sits on [ClO₂].

Feedback if chosen: This option is wrong — you squared both factors. Only ClO₂ gave the quadrupled response; OH⁻ gave the doubled response, exponent 1.

Feedback if chosen: This option is wrong — you gave both reactants the doubled response. Doubling [ClO₂] quadrupled the rate, so [ClO₂] carries the exponent 2.

Check — multiple choice

A reaction of CH₃Br with OH⁻ has the rate law rate = k[CH₃Br][OH⁻]. A chemist doubles [CH₃Br] alone, holding everything else constant. What happens to the initial rate?

Feedback if chosen: This option is wrong — you read the exponent as 0. A reactant in the rate law with exponent 1 does change the rate: it doubles it.

Feedback if chosen: Correct. The unwritten exponent on [CH₃Br] is 1, which records the doubled response.

Feedback if chosen: This option is wrong — you read the exponent as 2. The exponent on [CH₃Br] is the unwritten 1, which records the doubled response.

Check — multiple choice

Which exponent in a rate law records the no-effect response?

Feedback if chosen: Correct. An exponent of 0 records the no-effect response: the rate does not depend on that concentration.

Feedback if chosen: This option is wrong — you picked the doubled response's exponent. The no-effect response is recorded by 0.

Feedback if chosen: This option is wrong — you picked the quadrupled response's exponent. The no-effect response is recorded by 0.

Check — multiple choice

An expression of the form rate = k[A]ᵐ[B]ⁿ, whose exponents record how the initial rate responds to each reactant's concentration, is called what?

Feedback if chosen: This option is wrong — you named the equation that counts the substances. A balanced equation records amounts, not rate responses.

Feedback if chosen: This option is wrong — you named a ratio of amounts. A mole ratio comes from coefficients, not from rate responses.

Feedback if chosen: Correct. The expression rate = k[A]ᵐ[B]ⁿ is called the rate law of the reaction.

Check — multiple choice

In the rate law rate = k[cyclopropane], what is k?

Feedback if chosen: This option is wrong — you read k as a concentration. The concentration sits inside the brackets; k is the constant that scales it.

Feedback if chosen: This option is wrong — you read k as the rate itself. The rate is the whole left side; k is the constant that scales the concentration factor.

Feedback if chosen: Correct. k is the rate law's proportionality constant, and each reaction carries its own value of it.

Check — multiple choice

Cyclopropane isomerizes to propene with rate = k[cyclopropane] and k = 1.2 × 10⁻⁴ s⁻¹. Sulfuryl chloride decomposes with rate = k[SO₂Cl₂] and k = 3.0 × 10⁻⁵ s⁻¹, measured at the same temperature. At equal starting concentrations, which reaction has the higher initial rate?

Feedback if chosen: This option is wrong — you put the faster reaction with the smaller k. At equal concentrations, the larger k gives the higher rate.

Feedback if chosen: Correct. The two rate laws have the same form and the concentrations match, so the larger k, 1.2 × 10⁻⁴, gives the higher rate.

Feedback if chosen: This option is wrong — you compared only the concentration factors. The two k values differ, and k scales the whole rate.

Check — multiple choice

The proportionality constant k in a rate law is called what?

Feedback if chosen: Correct. The proportionality constant k in the rate law is called the rate constant.

Feedback if chosen: This option is wrong — you named the whole left side of the rate law. The reaction rate changes with concentration; k is the constant that scales it.

Feedback if chosen: This option is wrong — you named the rate at t = 0. The initial rate is one measured rate; k is the constant in the rate law.

Check — multiple choice

How must the exponents in a rate law be determined?

Feedback if chosen: This option is wrong — you gave the balanced equation a job it cannot do. Coefficients record amounts, not rate responses.

Feedback if chosen: Correct. Only experiments record how the rate responds to each concentration, so only experiments can set the exponents.

Feedback if chosen: This option is wrong — you counted products. The exponents record how the rate responds to each concentration, measured in experiments.

Check — multiple choice

Acetone reacts with Br₂ in acid: CH₃COCH₃ + Br₂ → CH₃COCH₂Br + HBr. In initial-rate trials, doubling [Br₂], with everything else held constant, leaves the initial rate unchanged. Jonah copies the coefficient of Br₂ and writes the exponent on [Br₂] as 1. Is Jonah correct?

Feedback if chosen: This option is wrong — you accepted the copied coefficient. The unchanged rate is the no-effect response, so the experiment sets the exponent on [Br₂] at 0.

Feedback if chosen: Correct. The unchanged rate is the no-effect response, so the exponent on [Br₂] is 0. The exponent comes from the experiment, never from the coefficient.

Check — multiple choice

For the butadiene dimerization, 2 C₄H₆ → C₈H₁₂, a table shows that doubling [C₄H₆] quadruples the initial rate. Priya writes the exponent on [C₄H₆] as 2, because the quadrupled response is the square response. Is Priya correct?

Feedback if chosen: Correct. The quadrupled response sets the exponent at 2. The matching coefficient is a coincidence the experiment confirms.

Feedback if chosen: This option is wrong — you refused an exponent the experiment itself supports. Doubling quadrupled the rate, so the exponent on [C₄H₆] is 2.

Check — numeric entry

In the reaction 2 ClO₂ + 2 OH⁻ → ClO₃⁻ + ClO₂⁻ + H₂O, doubling [OH⁻] alone doubles the initial rate. The coefficient of OH⁻ is 2. Which exponent does OH⁻ take in the rate law? Give a whole number.

Feedback when correct: Correct. The doubled rate is the doubled response, so the exponent on [OH⁻] is 1. The coefficient 2 records amounts, not rate responses.

Anticipated wrong answers (feedback shown):

Check — numeric entry

The bromate reaction, BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3 H₂O, has the rate law rate = k[BrO₃⁻][Br⁻][H⁺]². What is the order of the reaction with respect to H⁺? Give a whole number.

Feedback when correct: Correct. The power on [H⁺] in the rate law is 2, so the reaction is second order in H⁺.

Anticipated wrong answers (feedback shown):

Check — numeric entry

For the same rate law, rate = k[BrO₃⁻][Br⁻][H⁺]², what is the order of the reaction with respect to Br⁻? Give a whole number.

Feedback when correct: Correct. The power on [Br⁻] is the unwritten 1, so the reaction is first order in Br⁻.

Anticipated wrong answers (feedback shown):

Check — multiple choice

The isomerization of cyclopropane to propene is first order in cyclopropane. A chemist doubles [cyclopropane], holding everything else constant. What happens to the initial rate?

Feedback if chosen: This option is wrong — you matched first order to the no-effect response. The power 0 records no effect; the power 1 doubles the rate.

Feedback if chosen: Correct. First order is the power 1, and the power 1 records the doubled response.

Feedback if chosen: This option is wrong — you matched first order to the square response. The power 1 records the doubled response.

Check — multiple choice

The power to which a reactant's concentration is raised in the rate law is called what?

Feedback if chosen: This option is wrong — you named the balanced equation's number. A coefficient records amounts; the power in the rate law is the order.

Feedback if chosen: Correct. That power is the order of the reaction with respect to that reactant.

Feedback if chosen: This option is wrong — you named the constant k. The rate constant scales the whole expression; the power on one concentration is the order.

Check — multiple choice

How is the overall order of a reaction found from its rate law?

Feedback if chosen: This option is wrong — you multiplied where the definition adds. The overall order is the sum of the powers.

Feedback if chosen: This option is wrong — you kept only one power. Every power enters the sum, including each unwritten 1.

Feedback if chosen: Correct. The overall order is the sum of the powers of the concentrations.

Check — numeric entry

The reaction of ClO₂ with OH⁻ has the rate law rate = k[ClO₂]²[OH⁻]. What is the overall order of the reaction? Give a whole number.

Feedback when correct: Correct. The powers are 2 and the unwritten 1, and 2 + 1 = 3: third order overall.

Anticipated wrong answers (feedback shown):

Check — numeric entry

The reaction of CH₃Br with OH⁻ has the rate law rate = k[CH₃Br][OH⁻]. What is the overall order of the reaction? Give a whole number.

Feedback when correct: Correct. Each factor carries the unwritten power 1, and 1 + 1 = 2: second order overall.

Anticipated wrong answers (feedback shown):

Check — multiple choice

The sum of the powers of the concentrations in the rate law is called what?

Feedback if chosen: This option is wrong — you named the constant k. The rate constant scales the rate law; the sum of the powers is the overall order.

Feedback if chosen: This option is wrong — you named a single power. One reactant's power is its own order; the sum of all the powers is the overall order.

Feedback if chosen: Correct. That sum is the overall order of the reaction.

Check — numeric entry

The bromate reaction has the rate law rate = k[BrO₃⁻][Br⁻][H⁺]². What is the order of the reaction with respect to BrO₃⁻? Give a whole number.

Feedback when correct: Correct. [BrO₃⁻] carries the unwritten power 1, so the reaction is first order in BrO₃⁻.

Anticipated wrong answers (feedback shown):

Check — numeric entry

For the same rate law, rate = k[BrO₃⁻][Br⁻][H⁺]², what is the overall order of the reaction? Give a whole number.

Feedback when correct: Correct. The powers are 1, 1 and 2, and 1 + 1 + 2 = 4: fourth order overall.

Anticipated wrong answers (feedback shown):

Check — numeric entry

The acid-catalyzed acetone bromination, CH₃COCH₃ + Br₂ → CH₃COCH₂Br + HBr, has the rate law rate = k[CH₃COCH₃][H⁺]. What is the order of the reaction with respect to H⁺? Give a whole number.

Feedback when correct: Correct. [H⁺] carries the unwritten power 1, so the reaction is first order in H⁺. The acid's H⁺ appears in the rate law even though the equation does not consume it; its power counts like any other.

Anticipated wrong answers (feedback shown):

Check — numeric entry

In that same acetone reaction, the rate law is rate = k[CH₃COCH₃][H⁺]. What is the order of the reaction with respect to Br₂? Give a whole number.

Feedback when correct: Correct. Br₂ does not appear in the rate law, so its power is 0: the rate does not depend on [Br₂].

Anticipated wrong answers (feedback shown):

Check — multiple choice

A chemist doubles the starting concentration of cyclopropane in its isomerization, at the same temperature. Does the value of k change?

Feedback if chosen: This option is wrong — you moved k with the concentration. The concentration factor carries the whole concentration effect; k does not change.

Feedback if chosen: Correct. k ignores concentration: the doubled starting concentration changes the rate, not k.

Check — multiple choice

Halfway through a sulfuryl chloride decomposition, with the temperature steady, half the SO₂Cl₂ has been consumed and the rate has fallen. Is the value of k smaller than it was at the start?

Feedback if chosen: This option is wrong — you moved k with the rate. The falling [SO₂Cl₂] explains the falling rate on its own; k is unchanged.

Feedback if chosen: Correct. The falling concentration factor explains the falling rate; the value of k is the same as at the start.

Check — multiple choice

A chemist makes up a fresh cyclopropane mixture at half the usual starting concentration, at the same temperature. Which quantity is lower for this mixture?

Feedback if chosen: This option is wrong — you moved k with the concentration. k is characteristic of the reaction and ignores concentration.

Feedback if chosen: Correct. The halved concentration factor halves the initial rate, while k stays what it was.

Feedback if chosen: This option is wrong — you moved both quantities. Only the rate falls; k ignores concentration.

Check — multiple choice

The temperature of a reaction mixture is raised, with everything else held constant. What happens to the reaction's rate?

Feedback if chosen: Correct. Raising the temperature increases the rate of a reaction.

Feedback if chosen: This option is wrong — you moved the rate against the temperature. Raising the temperature raises the rate.

Feedback if chosen: This option is wrong — you treated the warming as no change. Raising the temperature raises the rate.

Check — multiple choice

A chemist measures k for the isomerization of cyclopropane at 500 °C, and again at 550 °C. At which temperature is k larger?

Feedback if chosen: This option is wrong — you put the larger k at the lower temperature. k grows as the temperature rises.

Feedback if chosen: Correct. k for a reaction is larger at a higher temperature.

Feedback if chosen: This option is wrong — you treated k as temperature-proof. k ignores concentration, but it does grow with temperature.

Check — multiple choice

An oven holding a butadiene dimerization cools from 330 °C to 280 °C. What happens to the value of k?

Feedback if chosen: This option is wrong — you moved k against the temperature. Cooling lowers the temperature, and a lower temperature means a smaller k.

Feedback if chosen: Correct. k is larger at a higher temperature, so cooling makes k smaller.

Feedback if chosen: This option is wrong — you treated the cooling as no change. k follows the temperature: cooler means smaller.

Check — multiple choice

A chemist warms a mixture of CH₃Br and OH⁻, with both concentrations unchanged. Which quantities increase?

Feedback if chosen: This option is wrong — you kept k fixed through a temperature change. k carries the temperature effect, so it grows as the mixture warms.

Feedback if chosen: This option is wrong — you stopped at k. The rate equals k times the unchanged concentration factors, so the larger k raises the rate as well.

Feedback if chosen: Correct. Warming enlarges k, and with unchanged concentration factors the larger k raises the rate too.

Check — numeric entry

The acid-catalyzed acetone bromination has the rate law rate = k[CH₃COCH₃][H⁺]. What is the overall order of the reaction? Give a whole number.

Feedback when correct: Correct. The powers are 1 and 1, and 1 + 1 = 2: second order overall.

Anticipated wrong answers (feedback shown):

Check — multiple choice

A reaction is second order overall, so its units equation reads M/s = (units of k) × M². Which units must k carry?

Feedback if chosen: This option is wrong — you moved the M² onto k instead of cancelling it. M²·s⁻¹ × M² gives M⁴/s, not M/s.

Feedback if chosen: This option is wrong — you gave k the rate's own units, leaving the M² uncancelled. M·s⁻¹ × M² gives M³/s, not M/s.

Feedback if chosen: This option is wrong — you cancelled only one of the two M factors. s⁻¹ × M² gives M²/s, not M/s.

Feedback if chosen: Correct. (M/s) ÷ M² = M⁻¹·s⁻¹, and M⁻¹·s⁻¹ × M² gives back M/s.

Check — multiple choice

The cyclopropane isomerization is first order overall. Which units must its k carry so the calculated rate comes out in M/s?

Feedback if chosen: This option is wrong — you gave k the rate's own units instead of cancelling the M. M·s⁻¹ × M gives M²/s, not M/s.

Feedback if chosen: Correct. (M/s) ÷ M¹ = s⁻¹, and s⁻¹ × M gives back M/s.

Feedback if chosen: This option is wrong — you cancelled two M factors where the rate law has one. M⁻¹·s⁻¹ × M gives s⁻¹, not M/s.

Check — multiple choice

A reaction is zeroth order overall: its rate does not depend on any concentration. Which units must its k carry?

Feedback if chosen: Correct. With no concentration factor, the units equation reads M/s = units of k, so k itself carries M·s⁻¹.

Feedback if chosen: This option is wrong — you cancelled an M that is not there. With no concentration factor, k alone must supply the whole M/s.

Feedback if chosen: This option is wrong — you cancelled two M factors that are not there. With no concentration factor, k alone must supply the whole M/s.

Check — numeric entry

A rate constant for a gas-phase reaction is reported as k = 8.8 M⁻¹·s⁻¹. What is the overall order of the reaction? Give a whole number.

Feedback when correct: Correct. M⁻¹·s⁻¹ × M² = M/s, so the reaction is second order overall.

Anticipated wrong answers (feedback shown):

Check — numeric entry

A rate constant for a decomposition in solution is reported as k = 4.2 × 10⁻³ s⁻¹. What is the overall order of the reaction? Give a whole number.

Feedback when correct: Correct. s⁻¹ × M¹ = M/s, so the reaction is first order overall.

Anticipated wrong answers (feedback shown):

Check — numeric entry

A rate constant for a reaction on a metal surface is reported as k = 2.5 × 10⁻⁵ M·s⁻¹. What is the overall order of the reaction? Give a whole number.

Feedback when correct: Correct. M·s⁻¹ already is M/s, so no concentration factor is needed: zeroth order overall.

Anticipated wrong answers (feedback shown):

Check — numeric entry

A rate constant for a reaction between two dissolved ions is reported as k = 0.016 M⁻²·s⁻¹. What is the overall order of the reaction? Give a whole number.

Feedback when correct: Correct. M⁻²·s⁻¹ × M³ = M/s, so the reaction is third order overall.

Anticipated wrong answers (feedback shown):

Check — multiple choice

Doubling one reactant's concentration, with everything else held constant, quadruples the initial rate. What does the trial say about how the rate depends on that reactant's concentration?

Feedback if chosen: This option is wrong — you read a quadrupled rate as no effect. The rate changed, so it depends on that concentration.

Feedback if chosen: This option is wrong — you matched a quadrupled rate to the doubled response. Direct proportionality would only have doubled the rate.

Feedback if chosen: Correct. A quadrupled rate from a doubled concentration is two squared, the square response.

Check — numeric entry

Acetone reacts with Br₂ in acid: CH₃COCH₃ + Br₂ → CH₃COCH₂Br + HBr. The two trials below change only [H⁺]. What is the order of the reaction with respect to H⁺? Give the order as a whole number.

Feedback when correct: Correct. The rate ratio is 2 and the concentration ratio is 2, so 2ᵐ = 2 and m = 1.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Butadiene dimerizes: 2 C₄H₆ → C₈H₁₂. The two trials below change only [C₄H₆]. What is the order of the reaction with respect to C₄H₆? Give the order as a whole number.

Feedback when correct: Correct. The rate ratio is 4 and the concentration ratio is 2, so 2ᵐ = 4 and m = 2.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Cyclopropane isomerizes to propene. The two trials below triple [cyclopropane]. What is the order of the reaction with respect to cyclopropane? Give the order as a whole number.

Feedback when correct: Correct. The rate ratio is 3 and the concentration ratio is 3, so 3ᵐ = 3 and m = 1.

Anticipated wrong answers (feedback shown):

Check — numeric entry

In the acetone reaction, the two trials below change only [Br₂]. What is the order of the reaction with respect to Br₂? Give the order as a whole number.

Feedback when correct: Correct. The rate ratio is 1 while the concentration ratio is 2, so 2ᵐ = 1 and m = 0.

Anticipated wrong answers (feedback shown):

Check — multiple choice

The bromate reaction, BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3 H₂O, is studied in the four experiments below. Which pair of experiments isolates the effect of H⁺ on the initial rate?

Feedback if chosen: This option is wrong — you picked the pair in which only [BrO₃⁻] changes. That pair isolates BrO₃⁻, not H⁺.

Feedback if chosen: Correct. From experiment 1 to experiment 4, only [H⁺] changes, so any rate change belongs to H⁺.

Feedback if chosen: This option is wrong — you picked a pair in which both [BrO₃⁻] and [H⁺] change. A rate change could belong to either.

Feedback if chosen: This option is wrong — you picked a pair in which both [Br⁻] and [H⁺] change. A rate change could belong to either.

Check — numeric entry

Using the pair of experiments in the table that changes only [BrO₃⁻], what is the order of the reaction with respect to BrO₃⁻? Give the order as a whole number.

Feedback when correct: Correct. Experiments 1 and 2 change only [BrO₃⁻]: the rate ratio is 2 and the concentration ratio is 2, so 2ᵐ = 2 and m = 1.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Using the pair of experiments in the table that changes only [H⁺], what is the order of the reaction with respect to H⁺? Give the order as a whole number.

Feedback when correct: Correct. Experiments 1 and 4 change only [H⁺]: the rate ratio is 4 and the concentration ratio is 2, so 2ᵐ = 4 and m = 2.

Anticipated wrong answers (feedback shown):

Check — numeric entry

CH₃Br reacts with OH⁻: CH₃Br + OH⁻ → CH₃OH + Br⁻. Three experiments are listed below. What is the order of the reaction with respect to CH₃Br? Give the order as a whole number.

Feedback when correct: Correct. Experiments 1 and 2 change only [CH₃Br]: the rate ratio is 2 and the concentration ratio is 2, so 2ᵐ = 2 and m = 1.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Using the same three experiments on CH₃Br + OH⁻, what is the order of the reaction with respect to OH⁻? Give the order as a whole number.

Feedback when correct: Correct. Experiments 2 and 3 change only [OH⁻]: the rate ratio is 2 and the concentration ratio is 2, so 2ᵐ = 2 and m = 1.

Anticipated wrong answers (feedback shown):

Check — multiple choice

A reaction is third order overall. Which units must its rate constant carry so the calculated rate comes out in M/s?

Feedback if chosen: This option is wrong — you used the zeroth-order units. A third-order rate law needs three M factors cancelled: M⁻²·s⁻¹.

Feedback if chosen: Correct. (M/s) ÷ M³ = M⁻²·s⁻¹, and M⁻²·s⁻¹ × M³ gives back M/s.

Feedback if chosen: This option is wrong — you cancelled one M too many. (M/s) ÷ M³ leaves M⁻²·s⁻¹, not M⁻³·s⁻¹.

Check — numeric entry

The reaction 2 ClO₂ + 2 OH⁻ → ClO₃⁻ + ClO₂⁻ + H₂O has the determined rate law rate = k[ClO₂]²[OH⁻]. In one experiment, [ClO₂] = 0.010 M, [OH⁻] = 0.030 M, and the initial rate is 6.0 × 10⁻⁴ M/s. What is k, in M⁻²·s⁻¹? Give your answer to two significant figures. Scientific notation such as 1.5e3 works.

Feedback when correct: Correct. k = 6.0 × 10⁻⁴ ÷ ((0.010)² × 0.030) = 6.0 × 10⁻⁴ ÷ 3.0 × 10⁻⁶ = 2.0 × 10² M⁻²·s⁻¹.

Anticipated wrong answers (feedback shown):

Check — numeric entry

The reaction CH₃Br + OH⁻ → CH₃OH + Br⁻ has the determined rate law rate = k[CH₃Br][OH⁻]. In one experiment, [CH₃Br] = 0.10 M, [OH⁻] = 0.10 M, and the initial rate is 2.0 × 10⁻⁴ M/s. What is k, in M⁻¹·s⁻¹? Give your answer to two significant figures.

Feedback when correct: Correct. k = 2.0 × 10⁻⁴ ÷ (0.10 × 0.10) = 2.0 × 10⁻⁴ ÷ 0.010 = 0.020 M⁻¹·s⁻¹.

Anticipated wrong answers (feedback shown):

Check — numeric entry

The reaction 2 ClO₂ + 2 OH⁻ → ClO₃⁻ + ClO₂⁻ + H₂O is studied in the three experiments below. What is the order of the reaction with respect to ClO₂? Give the order as a whole number.

Feedback when correct: Correct. Experiments 1 and 2 change only [ClO₂]: the rate ratio is 4 and the concentration ratio is 2, so 2ᵐ = 4 and m = 2.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Using the same three ClO₂ experiments, what is the order of the reaction with respect to OH⁻? Give the order as a whole number.

Feedback when correct: Correct. Experiments 1 and 3 change only [OH⁻]: the rate ratio is 2 and the concentration ratio is 2, so 2ⁿ = 2 and n = 1.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Substituting experiment 1 into the determined rate law, rate = k[ClO₂]²[OH⁻], what is k, in M⁻²·s⁻¹? Give your answer to two significant figures. Scientific notation such as 1.5e3 works.

Feedback when correct: Correct. k = 4.0 × 10⁻⁴ ÷ ((0.010)² × 0.020) = 4.0 × 10⁻⁴ ÷ 2.0 × 10⁻⁶ = 2.0 × 10² M⁻²·s⁻¹.

Anticipated wrong answers (feedback shown):

Check — numeric entry

The bromate reaction, BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3 H₂O, is studied in the four experiments below. What is the order of the reaction with respect to BrO₃⁻? Give the order as a whole number.

Feedback when correct: Correct. Experiments 1 and 2 change only [BrO₃⁻]: the rate ratio is 2 and the concentration ratio is 2, so 2ᵐ = 2 and m = 1.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Using the same four bromate experiments, what is the order of the reaction with respect to Br⁻? Give the order as a whole number.

Feedback when correct: Correct. Experiments 2 and 3 change only [Br⁻]: the rate ratio is 2 and the concentration ratio is 2, so 2ᵐ = 2 and m = 1.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Using the same four bromate experiments, what is the order of the reaction with respect to H⁺? Give the order as a whole number.

Feedback when correct: Correct. Experiments 2 and 4 change only [H⁺]: the rate ratio is 4 and the concentration ratio is 2, so 2ᵐ = 4 and m = 2.

Anticipated wrong answers (feedback shown):

Check — multiple choice

Which rate law follows from the four bromate experiments?

Feedback if chosen: This option is wrong — you gave H⁺ the doubled response. Doubling [H⁺] quadrupled the rate, so [H⁺] carries the exponent 2.

Feedback if chosen: This option is wrong — you moved the square onto BrO₃⁻. Doubling [BrO₃⁻] only doubled the rate; the quadrupling belongs to H⁺.

Feedback if chosen: Correct. The trials set the orders at 1, 1 and 2, and only [H⁺] carries the 2.

Feedback if chosen: This option is wrong — you copied the balanced equation's coefficients into the exponents. The exponents come from the trials: 1, 1 and 2.

Check — numeric entry

The determined rate law for the bromate reaction is rate = k[BrO₃⁻][Br⁻][H⁺]². What is the overall order of the reaction? Give a whole number.

Feedback when correct: Correct. The powers are 1, 1 and 2, and 1 + 1 + 2 = 4: fourth order overall.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Substituting experiment 1 into the determined rate law, rate = k[BrO₃⁻][Br⁻][H⁺]², what is k, in M⁻³·s⁻¹? Give your answer to two significant figures.

Feedback when correct: Correct. k = 4.0 × 10⁻⁴ ÷ (0.050 × 0.10 × (0.10)²) = 4.0 × 10⁻⁴ ÷ 5.0 × 10⁻⁵ = 8.0 M⁻³·s⁻¹.

Anticipated wrong answers (feedback shown):

Check — numeric entry

Dinitrogen monoxide, N₂O, decomposes on a hot platinum surface, and the reaction is zeroth order in N₂O. The graph shows [N₂O] falling along a straight line, with two points marked. How much N₂O does the reaction remove in each 100 s stretch? Give your answer in M to two significant figures.

Feedback when correct: Correct. The line falls 0.40 M between the marked points, over 500 s. That span holds five equal 100 s stretches, so each stretch removes 0.40 M ÷ 5 = 0.080 M.

Anticipated wrong answers (feedback shown):

Check — multiple choice

A zeroth-order reactant's concentration–time graph is a straight line. Why?

Feedback if chosen: This option is wrong — you described a first-order reactant. A rate that falls with the concentration draws a curve that flattens, not a straight line.

Feedback if chosen: This option is wrong — you read the straight line as flat. The concentration still falls; it falls at a constant rate, so the line slopes down without bending.

Feedback if chosen: Correct. A zeroth-order rate ignores the concentration, so equal amounts are removed in equal stretches of time, and the line never bends.

Check — multiple choice

A student says: "A reaction can remove its reactant at a constant rate, never slowing, until the reactant is gone." Is the student correct?

Feedback if chosen: Correct. A zeroth-order rate ignores the concentration, so the removal never slows, and the concentration–time graph is a straight declining line.

Feedback if chosen: This option is wrong — you expected every reaction to fall along the familiar bending curve. A zeroth-order rate ignores the concentration and stays constant until the reactant is gone.

Check — multiple choice

Cyclopropane gas slowly rearranges into propene, and the reaction is first order in cyclopropane. The graph shows [cyclopropane] against time, with three points marked. Compare the time taken to fall from 0.60 M to 0.30 M with the time taken to fall from 0.30 M to 0.15 M.

Feedback if chosen: This option is wrong — you read the smaller fall as a faster one. The marked points show both halvings spanning 300 s each.

Feedback if chosen: Correct. The marked points show each halving taking 300 s: 0.60 M at 0 s, 0.30 M at 300 s, 0.15 M at 600 s. A first-order reactant halves in the same time wherever it starts.

Feedback if chosen: This option is wrong — you expected the flattening curve to stretch the second halving. The marked points show both halvings take the same 300 s.

Check — multiple choice

Azomethane decomposes in the gas phase, and the reaction is first order in azomethane. In each equal stretch of time, what happens to [azomethane]?

Feedback if chosen: Correct. A first-order reactant loses the same fraction of what remains in every equal stretch of time. That equal-fraction fall is what keeps the halving time constant.

Feedback if chosen: This option is wrong — you described a zeroth-order reactant. Equal amounts in equal stretches make a straight line; a first-order curve loses equal fractions instead.

Feedback if chosen: This option is wrong — you made the fall speed up. The fraction lost per stretch stays the same; only the amount of azomethane that fraction represents shrinks.

Check — multiple choice

Azomethane's concentration halves in 173 s. The reaction is first order in azomethane. A student says the next halving must take longer than 173 s, because less azomethane remains and the reaction has slowed. Is the student correct?

Feedback if chosen: This option is wrong — you kept only the slowing. The reaction also works on half as much azomethane: half the material at half the rate needs 173 s again.

Feedback if chosen: Correct. When the concentration halves, a first-order rate halves with it. Half the material, removed at half the speed, takes the same 173 s to halve again.

Check — multiple choice

Nitrosyl bromide, NOBr, decomposes in the gas phase, and the reaction is second order in NOBr. As the reaction proceeds, how does the time taken by each successive halving of [NOBr] compare with the one before?

Feedback if chosen: This option is wrong — you made the halvings speed up. The rate falls to one quarter at each halving, so each halving takes longer, not less time.

Feedback if chosen: This option is wrong — you gave the curve the first-order pattern. A constant halving time needs the rate to halve with the concentration; this rate falls to one quarter.

Feedback if chosen: Correct. Halving [NOBr] cuts a second-order rate to one quarter, so each halving works slower on less material and takes longer than the one before.

Check — numeric entry

Hydrogen iodide decomposes by the second-order reaction 2 HI → H₂ + I₂. The graph shows [HI] against time, with three points marked; the fall from 0.40 M to 0.20 M takes 500 s. Read from the graph: how long does the fall from 0.20 M to 0.10 M take? Give your answer in s.

Feedback when correct: Correct. The marked points put 0.20 M at 500 s and 0.10 M at 1500 s. The second halving spans 1500 s − 500 s = 1000 s, twice as long as the first.

Anticipated wrong answers (feedback shown):

Check — multiple choice

A student tracks the second-order decomposition of NOBr and says: "[NOBr] falls by the same fraction in every equal stretch of time." Is the student correct?

Feedback if chosen: This option is wrong — you gave the second-order curve the first-order pattern. Halving [NOBr] cuts the rate to one quarter, so the fraction lost per stretch keeps shrinking.

Feedback if chosen: Correct. Equal fractions in equal times belongs to first order. A second-order rate falls to one quarter when the concentration halves, so each halving takes longer and the fraction lost per stretch shrinks.

Check — multiple choice

Cyclopropane gas rearranges into propene in a sealed flask. The graph shows [cyclopropane] against time, with three points marked. From the graph, what is the order with respect to cyclopropane?

Feedback if chosen: This option is wrong — you matched a bending curve to the straight-line shape. The curve bends, so the rate changes as cyclopropane is consumed.

Feedback if chosen: Correct. The marked points show equal halving times: 0.48 M to 0.24 M in 300 s, then 0.24 M to 0.12 M in the next 300 s. A constant halving time is the first-order shape.

Feedback if chosen: This option is wrong — you expected the halvings to lengthen. The marked points show both halvings taking 300 s, and equal halving times are the first-order shape.

Check — multiple choice

Hydrogen iodide decomposes by the reaction 2 HI → H₂ + I₂. The graph shows [HI] against time, with its first two halvings bracketed. From the graph, what is the order with respect to HI?

Feedback if chosen: This option is wrong — you matched a bending curve to the straight-line shape. The brackets show the rate changing as HI is consumed.

Feedback if chosen: This option is wrong — you read the two halvings as equal. The brackets show 500 s, then 1000 s; lengthening halvings rule first order out.

Feedback if chosen: Correct. The first halving takes 500 s and the second takes 1000 s. Halvings that lengthen, with a long slowing tail, are the second-order shape.

Check — multiple choice

Dinitrogen monoxide, N₂O, decomposes on a hot platinum surface. The graph shows [N₂O] against time. From the graph, what is the order with respect to N₂O?

Feedback if chosen: Correct. The graph is a straight declining line: the reaction removes the same amount of N₂O in every equal stretch, so the rate ignores [N₂O].

Feedback if chosen: This option is wrong — you expected a curve with equal halving times. This graph never bends: equal amounts, not equal fractions, are removed in equal stretches.

Feedback if chosen: This option is wrong — you expected a steep drop with a slowing tail. This graph is a straight line; the rate never changes as N₂O is consumed.

Check — drawing task (pen and paper, self-marked)

Take a pen and paper and draw one pair of axes: concentration [A] on the vertical axis, time on the horizontal axis. Starting all three curves from the same point, [A]₀, sketch and label the three curves expected when the decomposition is zeroth order, first order, and second order in A. When you are done, select Check to compare your answer with the model answer.

Check — multiple choice

A classmate makes three attempts, J, K, and L, at the concentration–time sketch for a zeroth-order decomposition. Which attempt is correct?

Feedback if chosen: This option is wrong — you accepted a bending fall. A bend means the rate is changing; a zeroth-order rate stays constant, so the line must be straight.

Feedback if chosen: Correct. A straight declining line is the zeroth-order signature: a constant rate removes equal amounts in equal times.

Feedback if chosen: This option is wrong — you accepted a rising line. The reactant is consumed, so its concentration must fall with time.

Check — multiple choice

Which quantity is the rate constant k?

Feedback if chosen: Correct. The rate law reads rate = k[A]ⁿ, and k is its proportionality constant, fixed at a given temperature.

Feedback if chosen: This option is wrong — you merged k with the rate itself. The rate changes as concentrations change; k stays fixed at a given temperature.

Feedback if chosen: This option is wrong — you reached for a concentration. k is the constant that ties the rate to the concentrations in the rate law.

Check — multiple choice

For a zeroth-order reactant, the plot of [A] against time is a straight line. What does its slope equal?

Feedback if chosen: Correct. The concentration falls at the constant rate k, so the line's slope is −k; flipping the sign gives the rate constant.

Feedback if chosen: This option is wrong — you dropped the fall. [A] decreases with time, so the slope is negative: −k.

Feedback if chosen: This option is wrong — you inverted the constant. The slope is the constant rate of fall with a minus sign, −k, not a reciprocal.

Check — numeric entry

Dinitrogen monoxide, N₂O, decomposes on a hot platinum surface, and the reaction is zeroth order in N₂O. The graph plots [N₂O] against time, with two points marked. Calculate k. Give your answer in M/s to two significant figures.

Feedback when correct: Correct. The slope is the rise over the run: (0.40 M − 0.80 M) ÷ 500 s = −8.0 × 10⁻⁴ M/s. The line falls, so flip the sign: k = 8.0 × 10⁻⁴ M/s.

Anticipated wrong answers (feedback shown):

Check — multiple choice

A reaction is first order in its reactant A. Which plot against time comes out as a straight line?

Feedback if chosen: Correct. A first-order reactant falls by equal fractions in equal times, and ln turns those equal fractions into the equal steps of a straight line.

Feedback if chosen: This option is wrong — you gave first order the zeroth-order plot. [A] itself is straight only when the rate ignores the concentration; a first-order [A] curve bends.

Check — numeric entry

Azomethane decomposes in the gas phase, and the reaction is first order in azomethane. The graph plots ln[azomethane] against time, with two points marked. Calculate k. Give your answer in s⁻¹ to two significant figures.

Feedback when correct: Correct. The slope is the rise over the run: (−3.00 − (−1.00)) ÷ 500 s = −0.0040 s⁻¹. The ln plot falls, so flip the sign: k = 0.0040 s⁻¹.

Anticipated wrong answers (feedback shown):

Check — multiple choice

A student says: "The ln move straightens a first-order fall only; a falling curve without the equal-fraction pattern stays bent under ln." Is the student correct?

Feedback if chosen: Correct. ln turns equal fractions into equal downward steps, so only the equal-fraction fall of a first-order reactant comes out straight.

Feedback if chosen: This option is wrong — you expected ln to straighten every falling curve. The straightening needs equal fractions in equal times, and only a first-order fall has them.

Check — multiple choice

For a second-order reactant, the plot of 1/[A] against time is a straight line. What does its slope equal?

Feedback if chosen: This option is wrong — you carried the falling plots' minus sign over. The 1/[A] plot climbs, so its slope is already positive and equals +k.

Feedback if chosen: Correct. 1/[A] grows by k's worth every second, so the climbing line's slope equals +k directly.

Feedback if chosen: This option is wrong — you inverted the constant. The reciprocal sits on the axis, not in the slope; the slope equals k itself.

Check — numeric entry

Nitrosyl bromide, NOBr, decomposes in the gas phase, and the reaction is second order in NOBr. The graph plots 1/[NOBr] against time, with two points marked. Calculate k. Give your answer in M⁻¹·s⁻¹ to two significant figures.

Feedback when correct: Correct. The slope is the rise over the run: (10.0 − 2.5) M⁻¹ ÷ 300 s = 0.025 M⁻¹·s⁻¹. The 1/[NOBr] plot climbs, so k equals the slope directly: k = 0.025 M⁻¹·s⁻¹.

Anticipated wrong answers (feedback shown):

Check — multiple choice

A student reads a slope of +0.040 M⁻¹·s⁻¹ off a 1/[A] plot and flips the sign before reporting the rate constant, saying every slope needs its sign flipped. Is the student correct?

Feedback if chosen: This option is wrong — you carried the falling plots' rule onto the climbing one. Falling plots have slope −k; the climbing 1/[A] plot's slope equals +k.

Feedback if chosen: Correct. The sign flip belongs to the falling plots, [A] and ln[A], whose slope is −k. The 1/[A] plot climbs; its slope equals +k, with nothing to flip.

Check — drawing task (pen and paper, self-marked)

Cyclopropane rearranges into propene by a first-order reaction; call its concentration [A]. Take a pen and paper and draw axes labeled ln[A] on the vertical axis and time on the horizontal axis. Sketch the plot you expect, marking its start on the vertical axis as ln[A]₀. When you are done, select Check to compare your answer with the model answer.

Check — multiple choice