Rate law and the order of a reaction

Why concentration matters

Picture a school corridor at break time. With ten students, bumps are rare. With a hundred students in the same corridor, bumps happen all the time. In a solution, molecules must meet before they can react. More molecules in each litre mean more meetings every second.

So the rate depends on the concentrations of the reactants. The question is how. Does the rate double when a concentration doubles? Or does it go up four times? Only experiment can tell.

The rate law

The equation that links the rate to the concentrations is the rate law, also called the rate equation. For a reaction of A and B it has the form:

• $k$ is the rate constant. It is the rate when every concentration is $1\ \text{mol/L}$. It does not change with the concentrations, but it changes with temperature and with a catalyst.
• $x$ is the order in A, and $y$ is the order in B.
• The overall order of the reaction is $x + y$.

Here are four real reactions with their rate laws, all found by experiment:

• $\text{H}_2 + \text{I}_2 \rightarrow 2\text{HI}$: rate $= k[\text{H}_2][\text{I}_2]$. The order is 2.
• $2\text{NO}_2 \rightarrow 2\text{NO} + \text{O}_2$: rate $= k[\text{NO}_2]^2$. The order is 2.
• $\text{CH}_3\text{CHO} \rightarrow \text{CH}_4 + \text{CO}$: rate $= k[\text{CH}_3\text{CHO}]^{3/2}$. The order is $\tfrac{3}{2}$.
• $\text{CH}_3\text{COCH}_3 + \text{I}_2 \rightarrow \text{CH}_3\text{COCH}_2\text{I} + \text{HI}$, in acid: rate $= k[\text{CH}_3\text{COCH}_3][\text{H}^+]$. The order is 2, and iodine does not appear in the law at all.

Look at these laws one at a time. In the second, the power equals the coefficient. In the third, the power is a fraction, though the coefficient is 1. In the fourth, the iodine is a reactant, but its power is zero. The acid is not even in the equation, but it appears in the law.

The orders cannot be read from the balanced equation. They come from experiment. An order can be zero, a whole number or a fraction.

The graphs show what an order means. For a reaction of order zero, the rate does not depend on the concentration. For order one, the rate doubles when the concentration doubles. For order two, the rate becomes four times as large.

The unit of k

The unit of $k$ is not the same for every reaction. It depends on the overall order $n$. From $\text{rate} = k[\text{A}]^n$ we get $k = \dfrac{\text{rate}}{[\text{A}]^n}$. The unit of $k$ is the unit of rate divided by $(\text{mol/L})^n$:

• Order 0: $\text{mol L}^{-1}\ \text{s}^{-1}$.
• Order 1: $\text{s}^{-1}$.
• Order 2: $\text{L mol}^{-1}\ \text{s}^{-1}$.
• Order 3: $\text{L}^2\ \text{mol}^{-2}\ \text{s}^{-1}$.
• Any order $n$: $\text{L}^{n-1}\ \text{mol}^{1-n}\ \text{s}^{-1}$.

So the unit of $k$ tells you the order. A rate constant in $\text{s}^{-1}$ belongs to a first order reaction. A rate constant in $\text{L mol}^{-1}\ \text{s}^{-1}$ belongs to a second order reaction.

Finding the order: the method of initial rates

How do we find the orders of a new reaction? We run it several times. Each time we change the starting concentrations and measure the initial rate, which is the slope of the concentration graph at $t = 0$. At that moment we know the concentrations exactly.

The rule is to change one concentration at a time. Compare two runs where only $[\text{A}]$ differs:

• Doubling $[\text{A}]$ leaves the rate the same: $x = 0$.
• Doubling $[\text{A}]$ doubles the rate: $x = 1$.
• Doubling $[\text{A}]$ makes the rate four times as large: $x = 2$.
• Doubling $[\text{A}]$ makes the rate eight times as large: $x = 3$.

For other factors, take logarithms: $x = \dfrac{\log(\text{rate}_2/\text{rate}_1)}{\log([\text{A}]_2/[\text{A}]_1)}$.

Example 1: persulphate and iodide

Persulphate ions oxidise iodide ions: $\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2$. The numbers below are chosen for practice. The rate law has the form $\text{rate} = k[\text{S}_2\text{O}_8^{2-}]^x[\text{I}^-]^y$.

• Run 1: $[\text{S}_2\text{O}_8^{2-}] = 0.040\ \text{mol/L}$, $[\text{I}^-] = 0.040\ \text{mol/L}$, rate $2.40 \times 10^{-5}\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• Run 2: $[\text{S}_2\text{O}_8^{2-}] = 0.080\ \text{mol/L}$, $[\text{I}^-] = 0.040\ \text{mol/L}$, rate $4.80 \times 10^{-5}\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• Run 3: $[\text{S}_2\text{O}_8^{2-}] = 0.040\ \text{mol/L}$, $[\text{I}^-] = 0.080\ \text{mol/L}$, rate $4.80 \times 10^{-5}\ \text{mol L}^{-1}\ \text{s}^{-1}$.

Step 1. The order in persulphate. Runs 1 and 2 differ only in persulphate, which is doubled. The rate doubles, so $x = 1$.

Step 2. The order in iodide. Runs 1 and 3 differ only in iodide, which is doubled. The rate doubles, so $y = 1$.

Step 3. The rate law. $\text{rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]$. The overall order is 2.

Step 4. The rate constant. We want $k$, so $k = \dfrac{\text{rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]}$. With run 1, $k = \dfrac{2.40 \times 10^{-5}\ \text{mol L}^{-1}\ \text{s}^{-1}}{0.040\ \text{mol/L} \times 0.040\ \text{mol/L}} = 0.0150\ \text{L mol}^{-1}\ \text{s}^{-1}$. The unit is the one for order 2.

Step 5. A check. A fourth run has both concentrations at $0.120\ \text{mol/L}$. The law predicts $0.0150\ \text{L mol}^{-1}\ \text{s}^{-1} \times 0.120\ \text{mol/L} \times 0.120\ \text{mol/L} = 2.16 \times 10^{-4}\ \text{mol L}^{-1}\ \text{s}^{-1}$. A measured rate of this size confirms the law.

Example 2: propanone and iodine in acid

Propanone (acetone) reacts with iodine in acid. The numbers are again chosen for practice.

• Run 1: $[\text{propanone}] = 0.80\ \text{mol/L}$, $[\text{H}^+] = 0.20\ \text{mol/L}$, $[\text{I}_2] = 0.0050\ \text{mol/L}$, rate $4.32 \times 10^{-6}\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• Run 2: propanone $1.60\ \text{mol/L}$, the rest as in run 1. The rate is $8.64 \times 10^{-6}\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• Run 3: $[\text{H}^+] = 0.40\ \text{mol/L}$, the rest as in run 1. The rate is $8.64 \times 10^{-6}\ \text{mol L}^{-1}\ \text{s}^{-1}$.
• Run 4: $[\text{I}_2] = 0.0100\ \text{mol/L}$, the rest as in run 1. The rate is $4.32 \times 10^{-6}\ \text{mol L}^{-1}\ \text{s}^{-1}$.

Doubling propanone doubles the rate, so the order in propanone is 1. Doubling the acid doubles the rate, so the order in $\text{H}^+$ is 1. Doubling the iodine changes nothing, so the order in iodine is 0.

The rate law is $\text{rate} = k[\text{propanone}][\text{H}^+]$, and $k = \dfrac{4.32 \times 10^{-6}\ \text{mol L}^{-1}\ \text{s}^{-1}}{0.80\ \text{mol/L} \times 0.20\ \text{mol/L}} = 2.7 \times 10^{-5}\ \text{L mol}^{-1}\ \text{s}^{-1}$.

The acid is a catalyst. It is not used up, yet it sets the speed. This shows again that the rate law and the balanced equation are different things.

Two more ways to find an order. Plot $\log(\text{rate})$ against $\log[\text{A}]$ for several runs: the points fall on a straight line, and its slope is the order. Or follow one run and test the shape of its concentration graph. That method comes with the integrated rate equations.

Practise with new numbers

Try it in Python

The program uses the persulphate and iodide runs. It finds each order from a pair of runs, and then the rate constant from every run.

Common mistakes

• Reading the orders from the balanced equation. The powers come from experiment. They may be zero, a fraction or different from the coefficients.
• Changing two concentrations at once. Then you cannot tell which one changed the rate.
• Calling $k$ the rate. The rate depends on the concentrations. The rate constant does not.
• Forgetting that $k$ depends on temperature. It is constant only at a fixed temperature.
• Writing the wrong unit for $k$. Work it out from the order: $\text{L}^{n-1}\ \text{mol}^{1-n}\ \text{s}^{-1}$.
• Adding the orders wrongly. The overall order is $x + y$, the sum of the powers in the rate law, not the sum of the coefficients.

Remember these

• Rate law: $\text{rate} = k[\text{A}]^x[\text{B}]^y$. The overall order is $x + y$.
• The orders come from experiment. They may be zero, whole numbers or fractions.
• Doubling a concentration multiplies the rate by $2^x$.
• The unit of $k$ is $\text{L}^{n-1}\ \text{mol}^{1-n}\ \text{s}^{-1}$ for order $n$.
• Method of initial rates: change one concentration at a time and compare the initial rates.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A reaction has the rate law $\text{rate} = k[\text{A}]^2[\text{B}]$. By what factor does the rate change when $[\text{A}]$ is tripled and $[\text{B}]$ is halved?

2. The rate constant of a second order reaction is measured with time in minutes. What is its unit?

3. A rate constant is given as $3.2 \times 10^{-4}\ \text{s}^{-1}$. What is the overall order of the reaction?

4. When $[\text{A}]$ is doubled, the initial rate rises eight times. What is the order in A?

5. A reactant appears in the balanced equation, but doubling its concentration does not change the rate. What is its order, and what does this tell you?