Integrated Rate Equations and Half-Life: Quick Revision

A brief note on Integrated Rate Equations and Half-Life. Read it before a test. Try each question at the end before you open its answer.

Zero order

• Rate law: rate $= k$. The unit of $k$ is $\text{mol L}^{-1}\,\text{time}^{-1}$.
• Integrated equation: $[R] = [R]_0 - kt$. A graph of $[R]$ against $t$ is a straight line with slope $-k$.
• Half-life: $t_{1/2} = \dfrac{[R]_0}{2k}$. It is proportional to the start concentration.
• End of the reaction: $t = \dfrac{[R]_0}{k}$, which is $2t_{1/2}$.
• Where it happens: when a surface, an enzyme or the light limits the speed.

First order

• Rate law: rate $= k[R]$. The unit of $k$ is $\text{time}^{-1}$.
• Integrated equation: $\ln[R] = \ln[R]_0 - kt$, or $[R] = [R]_0e^{-kt}$.
• Rate constant: $k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}$.
• Graph: $\ln[R]$ against $t$ is a straight line with slope $-k$. A $\log[R]$ graph has slope $-k/2.303$.
• Gas reaction, one gas giving two: $k = \dfrac{2.303}{t}\log\dfrac{p_i}{2p_i - p_t}$.

Half-life and fractions

• First order: $t_{1/2} = \dfrac{0.693}{k}$. It is the same at every stage and for every start concentration.
• Fraction left: $\left(\tfrac{1}{2}\right)^n$ after $n = t/t_{1/2}$ half-lives.
• Any fraction reacted, $f$: $t = \dfrac{2.303}{k}\log\dfrac{1}{1-f}$. About ten half-lives take a first order reaction to 99.9 %.
• Reading the order: the half-life is proportional to $[R]_0$ for zero order, constant for first order, and proportional to $1/[R]_0$ for second order.
• Radioactive decay follows the first order law.

Pseudo first order

• Meaning: one reactant is in large excess, so its concentration hardly changes.
• Constant: $k' = k[\text{B}]_0$, with the unit $\text{s}^{-1}$. The true rate constant is $k = k'/[\text{B}]_0$.
• Test: change the excess, and $k'$ changes in the same ratio.
• Examples: sucrose in dilute acid with water as the excess, and a dye fading in an excess of alkali.

A worked example

A first order reaction has $k = 0.0173\ \text{min}^{-1}$ and starts at $0.56\ \text{mol/L}$. Find the half-life, the concentration after $100\ \text{min}$ and the time for 75 % to react.

1. Half-life. $t_{1/2} = \dfrac{0.693}{0.0173\ \text{min}^{-1}} = 40.1\ \text{min}$.
2. After $100\ \text{min}$. $kt = 0.0173\ \text{min}^{-1} \times 100\ \text{min} = 1.73$. Then $[R] = 0.56\ \text{mol/L} \times e^{-1.73} = 0.0993\ \text{mol/L}$.
3. For 75 %. This is two half-lives, so $2 \times 40.1\ \text{min} = 80.2\ \text{min}$.

Practice with new numbers

A program to try

The program checks which graph is straight for a set of readings. A straight $\ln[R]$ graph means first order.

Mistakes to avoid

• Forgetting 2.303 when you use $\log$ instead of $\ln$.
• Turning $\dfrac{0.693}{k}$ upside down. A bigger $k$ gives a shorter half-life.
• Using the total pressure as the pressure of the reactant in a gas reaction.
• Using a zero order equation after the reactant is gone. It holds only up to $t = [R]_0/k$.
• Calling $k'$ the true rate constant of a pseudo first order reaction.

Check yourself

1. Which order has a half-life that does not depend on the start concentration?

2. About how many half-lives does a first order reaction need to be 99.9 % complete?

3. A zero order reaction starts at $0.40\ \text{mol/L}$ and has $k = 0.020\ \text{mol L}^{-1}\,\text{min}^{-1}$. When is the reactant all used up?

4. How do you find $k$ from a straight-line graph of $\ln[R]$ against $t$?