Topic 5.2: Introduction to Rate Law Quiz

Master AP Chemistry Topic 5.2 with this free practice quiz. Test your skills on method of initial rates, reaction orders, rate constant, and units of k to prepare for your AP exam.

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Topic 5.2: Introduction to Rate Law Quiz

18 MCQs

Master the rate laws with these 18 MCQs on AP Chemistry Topic 5.2: Introduction to Rate Law.

Covered Subtopics: Rate laws, reaction orders, rate constants

1 / 18

1. A reaction follows the rate law rate = k[A]^2 with k = 0.050 M^-1 s^-1. If [A] = 0.40 M, what is the initial rate?

0.020 M/s

0.0020 M/s

0.0080 M/s

0.080 M/s

Substituting into rate = k[A]^2 gives 0.050 times 0.40 squared, which equals 0.050 times 0.16, or 0.0080 M/s.

Substituting into rate = k[A]^2 gives 0.050 times 0.40 squared, which equals 0.050 times 0.16, or 0.0080 M/s.

2 / 18

2. For a reaction with the rate law rate = k[X]^2, [X] is tripled. By what factor does the rate increase?

27

6

9

3

The rate depends on [X] squared, so tripling [X] multiplies the rate by three squared, which is a factor of 9.

The rate depends on [X] squared, so tripling [X] multiplies the rate by three squared, which is a factor of 9.

3 / 18

3. A reaction follows the rate law rate = k[NO]^2[O2]. In an experiment with [NO] = 0.10 M and [O2] = 0.10 M, the initial rate is 2.0 x 10^-4 M/s. What is the value of k?

2.0 M^-2 s^-1

0.020 M^-2 s^-1

0.20 M^-2 s^-1

20 M^-2 s^-1

Substituting the data into rate = k[NO]^2[O2] gives k = 2.0 x 10^-4 divided by (0.10)^2(0.10), which equals 0.20 M^-2 s^-1.

Substituting the data into rate = k[NO]^2[O2] gives k = 2.0 x 10^-4 divided by (0.10)^2(0.10), which equals 0.20 M^-2 s^-1.

4 / 18

4. Which change will increase the value of the rate constant k for a reaction?

Increasing reactant concentrations

Increasing the pressure

Adding more reactant particles

Increasing the temperature

The rate constant depends on temperature and the activation energy, so raising the temperature increases k, while concentration and pressure changes do not.

The rate constant depends on temperature and the activation energy, so raising the temperature increases k, while concentration and pressure changes do not.

5 / 18

5. The rate law for a reaction is rate = k[A]^0[B]. Which statement about this reaction is correct?

The rate depends on [A] and [B] equally

The rate is constant under all conditions

The rate depends only on [A]

The rate depends only on [B]

Since [A]^0 equals 1, the [A] term drops out of the rate law, leaving a rate that depends only on the concentration of B.

Since [A]^0 equals 1, the [A] term drops out of the rate law, leaving a rate that depends only on the concentration of B.

6 / 18

6. For a reaction with the rate law rate = k[A]^2[B], what happens to the rate if [A] is doubled while [B] is held constant?

It stays the same

It quadruples

It doubles

It triples

The rate depends on the square of [A], so doubling [A] multiplies the rate by two squared, which quadruples the rate.

The rate depends on the square of [A], so doubling [A] multiplies the rate by two squared, which quadruples the rate.

7 / 18

7. The initial rate of a reaction between A and B was measured in three experiments: [table data]. What is the order with respect to A?

1

2

3

0

Comparing experiments 1 and 2, doubling [A] while [B] is constant doubles the rate, so the reaction is first order in A.

Comparing experiments 1 and 2, doubling [A] while [B] is constant doubles the rate, so the reaction is first order in A.

8 / 18

8. For a reaction with the rate law rate = k[A][B]^2, what are the units of the rate constant k?

M s^-1

s^-1

M^-2 s^-1

M^-1 s^-1

Since rate has units of M/s and [A][B]^2 has units of M^3, dividing gives M/s divided by M^3, which is M^-2 s^-1.

Since rate has units of M/s and [A][B]^2 has units of M^3, dividing gives M/s divided by M^3, which is M^-2 s^-1.

9 / 18

9. Which statement about reaction orders is correct?

They are always whole numbers like 1 or 2

They are found from experimental rate data

They equal the molecularity of the reaction

They always match the balanced equation coefficients

Reaction orders describe how the measured rate responds to concentration changes, so they must be determined experimentally and cannot be read from the balanced equation.

Reaction orders describe how the measured rate responds to concentration changes, so they must be determined experimentally and cannot be read from the balanced equation.

10 / 18

10. For a reaction that is second order overall, what are the units of the rate constant k?

M^-1 s^-1

s^-1

M s^-1

M^-2 s^-1

For an overall second-order reaction, rate = k times M^2, so k must carry units of M^-1 s^-1 to give a rate in M/s.

For an overall second-order reaction, rate = k times M^2, so k must carry units of M^-1 s^-1 to give a rate in M/s.

11 / 18

11. Using the same data table as Question 2, what is the overall order of the reaction?

3

4

2

5

The reaction is first order in A and second order in B, and the overall order is the sum of the individual orders, which is 3.

The reaction is first order in A and second order in B, and the overall order is the sum of the individual orders, which is 3.

12 / 18

12. Using the same data table as Question 2, what is the order of the reaction with respect to B?

2

0

1

3

Comparing experiments 1 and 3, doubling [B] while [A] is constant quadruples the rate, so the reaction is second order in B.

Comparing experiments 1 and 3, doubling [B] while [A] is constant quadruples the rate, so the reaction is second order in B.

13 / 18

13. In a series of initial-rate experiments, [A] is tripled while all other concentrations are held constant, and the initial rate does not change. What is the order of the reaction with respect to A?

1

2

0

3

Because changing [A] has no effect on the rate, the rate is independent of [A], which means the reaction is zero order in A.

Because changing [A] has no effect on the rate, the rate is independent of [A], which means the reaction is zero order in A.

14 / 18

14. The initial rate of a reaction between P and Q was measured in three experiments: [table]. What is the overall order?

5

3

4

2

Tripling [P] multiplies the rate by 9, so the order in P is 2, and tripling [Q] also multiplies the rate by 9, so the order in Q is 2, giving an overall order of 4.

Tripling [P] multiplies the rate by 9, so the order in P is 2, and tripling [Q] also multiplies the rate by 9, so the order in Q is 2, giving an overall order of 4.

15 / 18

15. For a first-order reaction, if the concentration of the reactant is halved, the reaction rate

doubles

is unchanged

is halved

drops to one fourth

In a first-order reaction the rate is directly proportional to the reactant concentration, so halving the concentration halves the rate.

In a first-order reaction the rate is directly proportional to the reactant concentration, so halving the concentration halves the rate.

16 / 18

16. The initial rate of a reaction was measured at two concentrations of C: [table]. What is the order with respect to C?

0

2

3

1

Doubling [C] leaves the rate unchanged, so the rate does not depend on [C], which means the reaction is zero order in C.

Doubling [C] leaves the rate unchanged, so the rate does not depend on [C], which means the reaction is zero order in C.

17 / 18

17. For a reaction with the rate law rate = k[A], what are the units of k?

unitless

1/s

M^-1 s^-1

M/s

Since the rate has units of M/s and [A] has units of M, k must have units of 1/s so that the product comes out in M/s.

Since the rate has units of M/s and [A] has units of M, k must have units of 1/s so that the product comes out in M/s.

18 / 18

18. Using the same data table as Question 2, what is the value of the rate constant k (in M^-2 s^-1)?

20

0.20

2.0

200

Substituting experiment 1 into rate = k[A][B]^2 gives k = 2.0 x 10^-3 divided by (0.10)(0.10)^2, which equals 2.0 M^-2 s^-1.

Substituting experiment 1 into rate = k[A][B]^2 gives k = 2.0 x 10^-3 divided by (0.10)(0.10)^2, which equals 2.0 M^-2 s^-1.

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The Rate Law and What the Orders Mean

For most reactions the rate depends on reactant concentrations through a rate law:

rate = k[A]ᵐ[B]ⁿ

The exponents m and n are the orders with respect to A and B. The overall order is their sum. Orders are found from experiment, never from the balanced equation: the coefficients of an overall reaction say nothing about its orders, which can be zero, fractions, or whole numbers unrelated to the stoichiometry.

The order tells you exactly how the rate responds to a concentration change. First order: doubling [A] doubles the rate. Second order: doubling [A] quadruples the rate, because the factor is 2². Third order: doubling multiplies the rate by 8. Tripling a second-order reactant multiplies the rate by 9.

Zero order is a real case, not an absence of data. If the order in A is zero, [A]⁰ = 1 and A drops out of the rate law entirely: doubling or tripling [A] leaves the rate unchanged.

Finding Orders: The Method of Initial Rates

Measure the initial rate in several experiments with different starting concentrations. To find the order in one reactant, compare two experiments in which only that reactant’s concentration changes and everything else is held constant.

Take the ratio of the rates and the ratio of the concentrations. If the concentration doubles and the rate doubles, the order is 1. If the concentration doubles and the rate quadruples, the order is 2. If the concentration doubles and the rate is unchanged, the order is 0. Repeat the comparison for each reactant, then add the orders for the overall order.

Initial rates are used because concentrations are known exactly at the start, before the rate drifts as reactants are consumed.

The Rate Constant k: Value and Units

The rate constant k folds in everything about the reaction except concentration: temperature, activation energy, and the nature of the reactants. Changing a concentration changes the rate but not k. Raising the temperature increases k.

Once the orders are known, k comes from any one experiment: k = rate / ([A]ᵐ[B]ⁿ).

The units of k are set by the overall order, because rate must always come out in M/s. Overall order 0: k has units M s⁻¹. Order 1: s⁻¹. Order 2: M⁻¹ s⁻¹. Order 3: M⁻² s⁻¹. The pattern is M^(1 − overall order) s⁻¹, and exam questions often give the units of k as a route to the overall order: s⁻¹ can only mean first order overall.

Worked Examples

Example 1: orders and k from initial rates

Three experiments for a reaction between A and B:

Experiment 1: [A] = 0.10 M, [B] = 0.10 M, rate = 2.0 × 10⁻³ M/s

Experiment 2: [A] = 0.20 M, [B] = 0.10 M, rate = 4.0 × 10⁻³ M/s

Experiment 3: [A] = 0.10 M, [B] = 0.20 M, rate = 8.0 × 10⁻³ M/s

Experiments 1 to 2: [A] doubles, [B] fixed, rate doubles from 2.0 × 10⁻³ to 4.0 × 10⁻³. Order in A is 1.

Experiments 1 to 3: [B] doubles, [A] fixed, rate quadruples from 2.0 × 10⁻³ to 8.0 × 10⁻³. Order in B is 2. Overall order is 3.

k from experiment 1: k = 2.0 × 10⁻³ / (0.10 × 0.10²) = 2.0 × 10⁻³ / 1.0 × 10⁻³ = 2.0 M⁻² s⁻¹. The units M⁻² s⁻¹ confirm a third-order reaction.

Example 2: using the law to predict a rate

With rate = k[A][B]² and k = 2.0 M⁻² s⁻¹, find the rate at [A] = 0.30 M and [B] = 0.30 M.

rate = 2.0 × 0.30 × 0.30² = 2.0 × 0.30 × 0.090 = 0.054 M/s.

Traps Specific to This Topic

Reading orders off the balanced equation is the central error this topic tests. For an overall reaction the equation and the rate law are connected only through experiment. Only an elementary step, covered in Topic 5.4, lets you write orders from coefficients.

Comparing the wrong pair of experiments gives a meaningless ratio. If two experiments differ in both [A] and [B], the rate change mixes both effects and no single order can be extracted. Always find the pair where only one concentration moves.

A reactant can be present in the reaction and absent from the rate law. Zero order means the rate genuinely does not respond to that concentration, and including [A] with exponent 1 out of habit invents a dependence that the data deny.

Units of k are not decoration. Work out the units from the overall order every time: M^(1 − n) s⁻¹. Reporting s⁻¹ for a third-order reaction means the division was done with the wrong powers.

Changing concentration never changes k; only the rate changes. Temperature is the variable that moves k.