AP Chemistry 5.2: Rate Laws and Concentration Changes Over Time

A rate law is an equation that shows how the reaction rate depends on reactant concentrations.

A general rate law looks like:

• Rate = k[A]ᵐ[B]ⁿ
• where:
• Rate = reaction rate
• k = rate constant
• [A] and [B] = reactant concentrations
• m and n = reaction orders
• The values of the exponents are extremely important because they tell us how strongly the reaction rate depends on each reactant concentration.

For most overall reactions, the rate law must be determined experimentally. You cannot simply look at the balanced chemical equation and assume the coefficients become the exponents.

• That shortcut works only for certain elementary reactions, which are discussed later.

Suppose:

• Rate = k[A]²[B]
• The reaction is:
• second order in A
• because A has exponent 2.
• It is:
• first order in B
• because B has exponent 1.

The overall reaction order is the sum:

• 2 + 1 = 3
• So the reaction is third order overall.

If a reactant does not affect the rate, it has exponent 0:

• [A]⁰ = 1
• That reactant is said to be zero order.

Reaction order tells you what happens to the rate when concentration changes.

If:

• Rate = k[A]⁰
• then:
• Rate = k
• Changing [A] does not change the rate.
• For example, doubling [A] gives:
• 2⁰ = 1
• so the rate stays the same.

If:

• Rate = k[A]
• then rate is directly proportional to [A].

Doubling [A]:

• rate × 2

Tripling [A]:

• rate × 3

If:

• Rate = k[A]²
• then rate depends on the square of [A].

Doubling [A]:

• 2² = 4
• so the rate becomes four times larger.

Tripling [A]:

• 3² = 9
• so the rate becomes nine times larger.

For:

• Rate = k[A]ᵐ[B]ⁿ
• if concentrations change, compare rates using:
• Rate₂ / Rate₁ = ([A]₂/[A]₁)ᵐ([B]₂/[B]₁)ⁿ
• For example:
• Rate = k[A]²[B]
• If A doubles and B triples:
• effect from A:
• 2² = 4
• effect from B:
• 3¹ = 3
• overall rate change:
• 4 × 3 = 12
• The new rate is 12 times greater.

AP Chemistry often gives a table of experiments showing initial concentrations and initial rates.

• Example:

Assume:

• Rate = k[A]ᵐ[B]ⁿ

Compare Experiments 1 and 2 because B stays constant.

• [A] doubles:
• 0.10 → 0.20
• Rate increases:
• 0.020 → 0.080
• The rate became 4 times larger.
• So:
• 2ᵐ = 4
• Therefore:
• m = 2
• The reaction is second order in A.

Compare Experiments 2 and 3 because A stays constant.

• [B] doubles:
• 0.10 → 0.20
• Rate doubles:
• 0.080 → 0.160
• So:
• 2ⁿ = 2
• Therefore:
• n = 1
• The rate law is:
• Rate = k[A]²[B]

Suppose [A] triples while the rate increases by a factor of 9.

• Then:
• 3ᵐ = 9
• so:
• m = 2
• If the numbers are not obvious, you can use logarithms, but many AP problems use relationships that can be recognized directly.

Once you know the rate law, substitute data from any experiment.

• Using:
• Rate = k[A]²[B]
• suppose:
• Rate = 0.020 M/s [A] = 0.10 M [B] = 0.10 M
• Then:
• 0.020 = k(0.10)²(0.10)
• 0.020 = k(0.001)
• k = 20
• But k must also include units.

The units of k depend on overall reaction order.

• Because:
• Rate = k(concentration terms)
• k must have whatever units are needed to make the final rate units equal to M/s.
• For common cases:

For the previous third-order reaction:

• Rate = k[A]²[B]
• the units are:
• M⁻²s⁻¹

A useful general pattern is:

• units of k = M^(1 − overall order) · s⁻¹

Suppose:

• 2NO₂ + F₂ → 2NO₂F
• You should not automatically write:
• Rate = k[NO₂]²[F₂]
• unless the reaction is known to occur in one elementary step.
• For a normal overall reaction, the exponents must come from experimental data.
• This is one of the most important rules in kinetics.

For a particular reaction at a specific temperature, k has a particular value.

Increasing temperature usually increases k, which increases reaction rate.

Changing concentration does not normally change k. Instead, concentration changes the concentration terms in the rate law.

A catalyst can also change the effective pathway and therefore the associated rate constant.

Do not automatically use balanced-equation coefficients as rate-law exponents.

Do not confuse the reaction order in one reactant with the overall order.

Do not forget the exponent when predicting the effect of a concentration change.

Do not assume k has the same units for every reaction.

A rate law answers:

• “How does changing concentration change reaction speed?”

For:

• Rate = k[A]²[B]
• doubling A makes the reaction 4× faster, while doubling B makes it 2× faster.

The rate law tells us how rate depends on concentration. An integrated rate law goes one step further by connecting the concentration of a reactant to time.

AP Chemistry mainly focuses on zero-order, first-order, and second-order reactions.

These equations can help you determine:

• concentration after a certain time
• time needed to reach a concentration
• rate constant k
• reaction order from experimental graphs
• half-life

The three major integrated rate laws are:

[A]ₜ = [A]₀ − kt

ln[A]ₜ = ln[A]₀ − kt

• or:
• ln([A]ₜ/[A]₀) = −kt

1/[A]ₜ = 1/[A]₀ + kt

• Here:
• [A]₀ = initial concentration [A]ₜ = concentration at time t k = rate constant t = time

For a zero-order reaction:

• Rate = k
• The reaction rate does not depend on [A].

The integrated rate law is:

• [A]ₜ = [A]₀ − kt

Compare this with:

• y = mx + b

If you graph:

• [A] vs. time
• you get a straight line.
• The slope is:
• −k
• and the y-intercept is:
• [A]₀
• So if [A] vs. time is linear, the reaction is zero order in A.

Suppose:

• [A]₀ = 0.800 M k = 0.0500 M/s t = 6.00 s
• Use:
• [A]ₜ = [A]₀ − kt
• [A]ₜ = 0.800 − (0.0500)(6.00)
• [A]ₜ = 0.500 M

For a first-order reaction:

• Rate = k[A]

The integrated rate law is:

• ln[A]ₜ = ln[A]₀ − kt

If you graph:

• ln[A] vs. time
• you get a straight line.
• Slope:
• −k
• Intercept:
• ln[A]₀
• Therefore, if ln[A] vs. time is linear, the reaction is first order.

Suppose:

• [A]₀ = 1.00 M k = 0.200 s⁻¹ t = 5.00 s
• Use:
• ln([A]ₜ/[A]₀) = −kt
• ln([A]ₜ/1.00) = −(0.200)(5.00)
• ln[A]ₜ = −1.00

Take e to both sides:

• [A]ₜ = e⁻¹
• [A]ₜ ≈ 0.368 M

The half-life is the amount of time required for the concentration of a reactant to decrease to half its current value.

For a first-order reaction:

• t½ = 0.693/k

A special feature of first-order kinetics is that the half-life is constant and does not depend on initial concentration.

• Suppose:
• k = 0.231 s⁻¹
• Then:
• t½ = 0.693 / 0.231
• t½ = 3.00 s
• That means:
• 1.00 M → 0.500 M takes 3.00 s
• 0.500 M → 0.250 M also takes 3.00 s
• 0.250 M → 0.125 M also takes 3.00 s
• Each halving requires the same amount of time.

After one half-life:

• 1/2 remains

After two:

• 1/4 remains

After three:

• 1/8 remains

After n half-lives:

• fraction remaining = (1/2)ⁿ
• For example, after four half-lives:
• (1/2)⁴ = 1/16
• So 6.25% remains.

Radioactive decay follows first-order kinetics.

• This is why radioactive isotopes have constant half-lives.
• Suppose a radioactive isotope has a half-life of 10 years. A 100 g sample would decrease approximately:
• 100 g → 50 g → 25 g → 12.5 g
• at 10-year intervals.

The decay rate becomes slower in absolute amount because fewer radioactive nuclei remain, but the fraction that decays per unit time stays consistent with first-order behavior.

For a second-order reaction in A:

• Rate = k[A]²

The integrated rate law is:

• 1/[A]ₜ = 1/[A]₀ + kt

If you graph:

• 1/[A] vs. time
• you get a straight line.
• Slope:
• +k
• Intercept:
• 1/[A]₀
• Notice that unlike zero- and first-order graphs, the second-order linear plot has a positive slope.

Suppose:

• [A]₀ = 0.500 M k = 0.400 M⁻¹s⁻¹ t = 5.00 s
• Use:
• 1/[A]ₜ = 1/[A]₀ + kt
• 1/[A]ₜ = 1/0.500 + (0.400)(5.00)
• 1/[A]ₜ = 2.00 + 2.00
• 1/[A]ₜ = 4.00
• Therefore:
• [A]ₜ = 0.250 M

This is one of the most important Unit 5 skills.

You may be given several graphs and asked which one is linear.

The linear graph identifies the reaction order.

Then the slope gives k.

Suppose experimental plots show:

• [A] vs. t → curved ln[A] vs. t → straight line 1/[A] vs. t → curved
• Then the reaction is:
• first order
• If the slope of ln[A] vs. t is:
• −0.0350 s⁻¹
• then:
• k = 0.0350 s⁻¹

The units of k can also provide a clue.

• M/s → zero order
• s⁻¹ → first order
• M⁻¹s⁻¹ → second order

The simple equation:

• t½ = 0.693/k
• is specifically for first-order reactions.

Do not automatically use it for zero- or second-order reactions.

Do not use the wrong integrated rate law.

Do not forget that ln[A], not [A], is linear for first order.

Do not forget that 1/[A] is linear for second order.

Do not use the first-order half-life equation unless the process is first order.

Remember that slope is −k for zero and first order, but +k for second order.

Memorize the linear-plot pattern:

• Zero → [A]
• First → ln[A]
• Second → 1/[A]
• Then use the slope to find k.

This table is one of the most useful things to know for Unit 5.