Double the concentration of one reactant and a reaction may go twice as fast, four times as fast, or not change at all. Which one happens cannot be read from the balanced equation; it has to be measured. The rate law is the equation that captures the result, and it is the key to predicting how fast a reaction will go under new conditions.

In this lesson you will write a rate law, find reaction orders and the rate constant from initial-rate data, and predict how the rate changes when a concentration changes (Figure 1).

1. Datainitial rates of several runs
2. $f^{m}$two runs, only one concentration changes
3. Ordersone per reactant, from the data
4. $m,\ n$from the data, never the coefficients
5. Rate laworders and k
6. $k$any run, or multiply by f to the power m
7. Answerk with units, or a new rate

Order: $f^{m} = \dfrac{\text{rate}_2}{\text{rate}_1}$$\text{rate} = k[\mathrm{A}]^{m}[\mathrm{B}]^{n}$$k = \dfrac{\text{rate}}{[\mathrm{A}]^{m}[\mathrm{B}]^{n}}$

Learning Objectives

After this lesson, you should be able to:

1. Determine reaction orders and the rate constant by the method of initial rates.
2. Predict how the rate changes when a concentration changes (zero, first, second order).

Key Ideas

1. The Rate Law

For a reaction $a\mathrm{A} + b\mathrm{B} \rightarrow \text{products}$, the rate usually depends on the reactant concentrations raised to some powers (Whitten Sec. 16-3):

The exponents $m$ and $n$ are the reaction orders with respect to A and B, and their sum is the overall order. The constant $k$ is the rate constant: it is fixed for a reaction at one temperature and does not change with concentration (Figure 2).

$\text{rate} = k[\mathrm{A}]^{m}[\mathrm{B}]^{n}$

$k$: the rate constant, fixed for one reaction at one temperature

$[\mathrm{A}]$, $[\mathrm{B}]$: reactant concentrations, in M

$m$, $n$: the orders in A and in B, found by experiment (usually 0, 1 or 2)

$m + n$: the overall order

Key idea: the orders $m$ and $n$ are found by experiment. They are usually 0, 1 or 2, and they are not the coefficients of the balanced equation.

2. Zero, First and Second Order

The order tells how strongly the rate responds to a concentration. If a concentration is multiplied by a factor $f$ and the others stay the same, the rate is multiplied by $f^{m}$:

• Zero order ($m = 0$): the rate does not change, because $f^{0} = 1$.
• First order ($m = 1$): the rate changes by the same factor; doubling the concentration doubles the rate.
• Second order ($m = 2$): the rate changes by the square of the factor; doubling the concentration makes the rate 4 times as fast, and tripling it makes the rate 9 times as fast.

Figure 3 compares the three orders for a doubling and a tripling.

Key idea: new rate = old rate $\times f^{m}$ for each concentration that changes by a factor $f$.

3. The Method of Initial Rates

To find the orders, run the reaction several times from different starting concentrations and measure each initial rate, the instantaneous rate at $t = 0$ from the previous lesson. Then compare two runs in which only one concentration changes. Everything else cancels in the ratio:

If doubling [A] doubles the rate, $2^{m} = 2$ and $m = 1$; if it quadruples the rate, $2^{m} = 4$ and $m = 2$; if the rate does not change, $m = 0$. Repeat with a pair of runs in which only [B] changes to find $n$ (Figure 4).

1. Two runs$\dfrac{[\mathrm{A}]_2}{[\mathrm{A}]_1} = f$only [A] changes
2. $\dfrac{\text{rate}_2}{\text{rate}_1}$Step 2: divide the rates
3. Rate ratio$f^{m} = \dfrac{\text{rate}_2}{\text{rate}_1}$from the data
4. $m$Step 3: solve for the power
5. Orderfor f of 2: ratio 1, 2, 4 gives 0, 1, 2

Key idea: pick two runs where only one concentration changes; the rate ratio equals the concentration ratio raised to the order.

Try it: find a rate law from initial rates

Choose a hidden rate law, set the starting concentrations for each run, and compare pairs of runs to find the orders. Then compute $k$ from any run. The worked-example buttons load each example’s runs.

4. The Rate Constant and Its Units

Once the orders are known, substitute any one run into the rate law and solve for $k$. The rate is always in M/s, so the units of $k$ depend on the overall order (Figure 5):

• overall order 0: $k$ in M/s;
• overall order 1: $k$ in $\text{s}^{-1}$;
• overall order 2: $k$ in $\text{M}^{-1}\,\text{s}^{-1}$.

Every run of the same experiment gives the same $k$, which is a useful check. A higher temperature gives a larger $k$, which is how temperature speeds up a reaction.

Overall order 0: $\text{rate} = k$; $k = \text{rate}$, in $\text{M/s}$

Overall order 1: $\text{rate} = k[\mathrm{A}]$; $k = \dfrac{\text{M/s}}{\text{M}}$, in $\text{s}^{-1}$

Overall order 2: $\text{rate} = k[\mathrm{A}]^{2}$ or $\text{rate} = k[\mathrm{A}][\mathrm{B}]$; $k = \dfrac{\text{M/s}}{\text{M}^{2}}$, in $\text{M}^{-1}\,\text{s}^{-1}$

Note: a rate law is written for the rate of reaction. If the data give the rate of formation of a product, divide by its coefficient first, as in the previous lesson.

Key idea: $k = \frac{\text{rate}}{[\mathrm{A}]^{m}[\mathrm{B}]^{n}}$ from any run; its units make the rate come out in M/s.

5. A Strategy for Rate-Law Problems

Follow the roadmap in Figure 1:

1. Find a pair of runs in which only [A] changes. Divide their rates and their [A] values; the order $m$ is the power that links the two ratios.
2. Repeat for B with a pair in which only [B] changes.
3. Write the rate law with the orders you found, never with the coefficients.
4. Find k from one run, with units for the overall order; or predict a new rate by multiplying by $f^{m}$ for each change.

Exam tip: check $k$ with a second run. If two runs give different values of $k$, an order is wrong.

Worked Example 1: Reaction Orders from Initial Rates

Problem: For the hypothetical reaction $2\mathrm{A} + \mathrm{B} \rightarrow \mathrm{C}$, the initial rate was measured in three runs at the same temperature (illustrative data). Find the order with respect to each reactant, write the rate law, and find $k$.

Step 1. Order in A: compare runs 1 and 2, where only [A] changes (it doubles).

Step 2. Order in B: compare runs 1 and 3, where only [B] changes (it triples).

Step 3. Write the rate law and find k from run 1. The overall order is 2, so $k$ is in $\text{M}^{-1}\,\text{s}^{-1}$.

Final Answer: $\text{rate} = k[\mathrm{A}][\mathrm{B}]$, first order in A and in B, second order overall; $k = 0.0200\,\text{M}^{-1}\,\text{s}^{-1}$

The coefficient of A is 2, yet the reaction is first order in A: the orders come from the data. Check with run 2: $\frac{4.00 \times 10^{-4}}{(0.200)(0.100)}$ also gives 0.0200.

1. Order in A$\dfrac{4.00 \times 10^{-4}}{2.00 \times 10^{-4}} = 2 = 2^{m}$, $m = 1$runs 1 and 2
2. $\dfrac{6.00 \times 10^{-4}}{2.00 \times 10^{-4}} = 3 = 3^{n}$Step 2: runs 1 and 3
3. Order in B$n = 1$B tripled
4. $\text{rate} = k[\mathrm{A}][\mathrm{B}]$Step 3: rate law
5. Rate constant$k = \dfrac{2.00 \times 10^{-4}\,\text{M/s}}{0.0100\,\text{M}^{2}}$, $k = 0.0200\,\text{M}^{-1}\,\text{s}^{-1}$from run 1

Your turn: find a reaction order

Same steps as Worked Example 1: find two runs where only one concentration changes, divide the rates, and find the power that matches the concentration ratio.

Worked Example 2: A Zero Order and the Rate Constant

Problem: For the hypothetical reaction $\mathrm{A} + 2\mathrm{B} \rightarrow \mathrm{C}$, three runs at one temperature gave the initial rates below (illustrative data). Find the rate law and the rate constant.

Step 1. Order in A (runs 1 and 2): [A] doubles and the rate is multiplied by 4.

Step 2. Order in B (runs 1 and 3): [B] triples but the rate does not change, so $3^{n} = 1$ and $n = 0$. B does not appear in the rate law.

Step 3. Find k from run 2. The overall order is 2.

Final Answer: $\text{rate} = k[\mathrm{A}]^{2}$ with $k = 0.00300\,\text{M}^{-1}\,\text{s}^{-1}$

The coefficient of B is 2, yet the reaction is zero order in B. Run 1 gives the same $k$: $\frac{3.00 \times 10^{-5}}{(0.100)^{2}}$ is also 0.00300.

Your turn: find the rate constant

Same idea as Step 3 of Worked Example 2: divide the rate by the concentrations raised to their orders, and give $k$ the units that make the rate come out in M/s.

Worked Example 3: Predicting a New Rate

Problem: For the hypothetical reaction $\mathrm{A} + \mathrm{B} \rightarrow 2\mathrm{C}$, the rate law is $\text{rate} = k[\mathrm{A}]$. At the start of one run, with [A] = 0.100 M, C forms at $4.00 \times 10^{-4}\,\text{M/s}$. Find $k$, then the rate at which C forms at the start of a run with [A] tripled to 0.300 M.

Step 1. Convert the rate of formation of C to the rate of reaction: divide by the coefficient of C, 2.

Step 2. Find k. The reaction is first order overall, so $k$ is in $\text{s}^{-1}$.

Step 3. Predict the new rate. [A] is multiplied by 3 and the order is 1, so the rate, and the rate of formation of C, are multiplied by $3^{1} = 3$.

Final Answer: $k = 0.00200\,\text{s}^{-1}$; C then forms at $1.20 \times 10^{-3}\,\text{M/s}$

Check with the rate law: $k$ times 0.300 M gives a rate of reaction three times larger, and C forms at twice that rate. B is zero order, so its concentration does not matter.

Your turn: a new initial rate

Same idea as Step 3 of Worked Example 3: multiply the old rate by $f^{m}$ for each concentration that changes.

Common Mistakes

Common mistake: Taking the orders from the coefficients. In Worked Example 2, the coefficients (1 for A, 2 for B) would give $\text{rate} = k[\mathrm{A}][\mathrm{B}]^{2}$ and predict that tripling [B] makes the rate 9 times as fast, yet the data show no change; $k$ from run 2 would come out as 0.0600, which no other run confirms. Orders come only from experiment.

Common mistake: Using the rate ratio as the order. In Worked Example 2, the rate ratio is 4, but the order is 2, because $2^{2} = 4$. The order is the power that turns the concentration ratio into the rate ratio.

Common mistake: Comparing runs where two concentrations change. Runs 2 and 3 of Worked Example 1 differ in both [A] and [B], so their rate ratio mixes the two orders. Pick a pair of runs in which only one concentration changes.

Common mistake: Setting up the k formula wrongly. In Worked Example 2, $k = \frac{\text{rate}}{[\mathrm{A}]^{2}}$ gives 0.00300. Dividing the rate by [A] instead of $[\mathrm{A}]^{2}$ gives $6.00 \times 10^{-4}$, and turning the formula upside down, $\frac{[\mathrm{A}]^{2}}{\text{rate}}$, gives about 333. Raise each concentration to its order and keep the rate on top.

Common mistake: Using a species rate as the rate of reaction. In Worked Example 3, using the rate of formation of C, $4.00 \times 10^{-4}\,\text{M/s}$, directly gives $k = 0.00400\,\text{s}^{-1}$, twice too large. Divide by the coefficient of C first.

Common mistake: Giving k the wrong units. The units of $k$ follow the overall order: M/s for zero order, $\text{s}^{-1}$ for first order, $\text{M}^{-1}\,\text{s}^{-1}$ for second order.

Key Takeaway

A rate law, $\text{rate} = k[\mathrm{A}]^{m}[\mathrm{B}]^{n}$, states how the rate depends on each concentration. The orders $m$ and $n$ come from experiment, usually by comparing initial rates of runs in which only one concentration changes; they are not the coefficients. Multiplying a concentration by $f$ multiplies the rate by $f^{m}$, and $k$, found from any run, has units that depend on the overall order.

Practice: randomized problems

Each problem below uses fresh numbers every time and mixes every kind of problem in this lesson: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures and simulator: original work by Dr. Ray’s Learning Center, licensed CC BY 4.0. Method: Whitten, Davis, Peck and Stanley, Chemistry, 9th ed. (Brooks/Cole, Cengage Learning, 2010), Section 16-3 (pp. 614 to 622) for the rate-law expression, reaction orders found by experiment, the method of initial rates and the units of the rate constant. Data: hypothetical reactions with illustrative initial rates generated from a stated rate law and checked with exact fractions in our own Python; they are not measured values. The simulator and the practice problems are audited; see the lesson audit log.