The rate law is a differential equation linking rate to concentration. By integrating it we obtain a direct relation between concentration and time, which is far more useful for calculations: it lets us find concentration at any time, the rate constant from data, and the half-life.
Zero-order reactions
For a zero-order reaction $R \rightarrow P$, the rate is independent of concentration: $-\frac{d[R]}{dt} = k[R]^0 = k$. Integrating from $[R]_0$ at $t=0$ to $[R]$ at time $t$ gives
$$[R] = [R]_0 - kt$$
A plot of $[R]$ against $t$ is a straight line of slope $-k$ and intercept $[R]_0$. The unit of $k$ is $\text{mol L}^{-1}\text{s}^{-1}$. Examples include the decomposition of $NH_3$ or $HI$ on a hot metal surface and some photochemical and enzyme-catalysed reactions where a surface or catalyst is saturated.
First-order reactions
For a first-order reaction $-\frac{d[R]}{dt} = k[R]$. Separating variables and integrating gives the natural-log form $\ln\frac{[R]_0}{[R]} = kt$, usually written in the base-10 form used in textbooks:
$$k = \frac{2.303}{t}\log\frac{[R]_0}{[R]}$$
Here $k$ has units of $\text{s}^{-1}$ and does not depend on the concentration units, which is why first-order kinetics is so convenient. A plot of $\log[R]$ versus $t$ is a straight line of slope $-\frac{k}{2.303}$. All radioactive decays and many decompositions (such as $N_2O_5$ and the gas-phase decomposition of $SO_2Cl_2$) are first order.
Half-life
The half-life $t_{1/2}$ is the time taken for the reactant concentration to fall to half its initial value. For a first-order reaction, set $[R] = \frac{[R]_0}{2}$:
$$t_{1/2} = \frac{2.303}{k}\log 2 = \frac{0.693}{k}$$
This is a striking result: the first-order half-life is independent of the initial concentration. For a zero-order reaction the half-life does depend on the starting amount, $t_{1/2} = \frac{[R]_0}{2k}$, so it shortens as the reaction proceeds.
Pseudo-first-order reactions
Some reactions are higher order on paper but behave as first order in practice. If one reactant is present in large excess (or is the solvent), its concentration hardly changes, so it can be absorbed into the rate constant. The hydrolysis of an ester, $CH_3COOC_2H_5 + H_2O \rightarrow CH_3COOH + C_2H_5OH$, is truly second order ($\text{Rate} = k[\text{ester}][H_2O]$), but in dilute aqueous solution $[H_2O]$ is effectively constant, so $\text{Rate} = k'[\text{ester}]$ with $k' = k[H_2O]$. Such reactions are called pseudo-first-order. The inversion of cane sugar and the acid hydrolysis of esters are classic examples.
A first-order reaction has a rate constant $k = 1.15 \times 10^{-3}\ \text{s}^{-1}$. How long will it take for $5\ \text{g}$ of the reactant to reduce to $3\ \text{g}$?
Solution- $t = \frac{2.303}{k}\log\frac{[R]_0}{[R]} = \frac{2.303}{1.15\times10^{-3}}\log\frac{5}{3}$.
- $\log\frac{5}{3} = \log 1.667 = 0.2218$.
- $t = \frac{2.303}{1.15\times10^{-3}} \times 0.2218 = 2002 \times 0.2218 \approx 444\ \text{s}$.
Answer: $t \approx 444\ \text{s}$ (about $7.4$ minutes).
The half-life of a first-order reaction is $60\ \text{minutes}$. Calculate its rate constant.
Solution- For first order, $t_{1/2} = \frac{0.693}{k}$, so $k = \frac{0.693}{t_{1/2}}$.
- $t_{1/2} = 60\ \text{min} = 60 \times 60 = 3600\ \text{s}$.
- $k = \frac{0.693}{3600} = 1.925 \times 10^{-4}\ \text{s}^{-1}$ (or $1.155\times10^{-2}\ \text{min}^{-1}$).
Answer: $k = 1.925 \times 10^{-4}\ \text{s}^{-1}$.
A first-order reaction is $25\%$ complete in $40\ \text{minutes}$. Calculate the rate constant and the time for $75\%$ completion.
Solution- At $25\%$ completion, $[R] = 75\%$ of $[R]_0$, so $\frac{[R]_0}{[R]} = \frac{100}{75} = 1.333$.
- $k = \frac{2.303}{40}\log 1.333 = \frac{2.303}{40}(0.1249) = 7.19\times10^{-3}\ \text{min}^{-1}$.
- For $75\%$ completion, $\frac{[R]_0}{[R]} = \frac{100}{25} = 4$: $t = \frac{2.303}{k}\log 4 = \frac{2.303}{7.19\times10^{-3}}(0.602) = 192.9\ \text{min}$.
Answer: $k = 7.19 \times 10^{-3}\ \text{min}^{-1}$; $75\%$ completion takes about $193\ \text{min}$.
For a zero-order reaction, $[R]_0 = 0.50\ \text{mol L}^{-1}$ and $k = 2.0 \times 10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}$. Find $[R]$ after $10\ \text{s}$ and the half-life.
Solution- $[R] = [R]_0 - kt = 0.50 - (2.0\times10^{-2})(10) = 0.50 - 0.20 = 0.30\ \text{mol L}^{-1}$.
- For zero order, $t_{1/2} = \frac{[R]_0}{2k} = \frac{0.50}{2(2.0\times10^{-2})} = \frac{0.50}{0.04} = 12.5\ \text{s}$.
Answer: $[R] = 0.30\ \text{mol L}^{-1}$ after $10\ \text{s}$; $t_{1/2} = 12.5\ \text{s}$.
A first-order reaction takes $69.3\ \text{minutes}$ for $50\%$ completion. How long will $90\%$ of the reaction take?
Solution- $50\%$ completion is the half-life, so $t_{1/2} = 69.3\ \text{min}$, giving $k = \frac{0.693}{69.3} = 0.0100\ \text{min}^{-1}$.
- For $90\%$ completion, $[R] = 10\%$ of $[R]_0$, so $\frac{[R]_0}{[R]} = 10$.
- $t = \frac{2.303}{k}\log 10 = \frac{2.303}{0.0100}(1) = 230.3\ \text{min}$.
Answer: $t \approx 230.3\ \text{min}$ for $90\%$ completion.
Explain, with the rate expression, why the acid hydrolysis of an ester in dilute aqueous solution is a pseudo-first-order reaction.
Solution- The true rate law is $\text{Rate} = k[\text{ester}][H_2O]$, so the reaction is second order overall.
- In dilute solution water is in large excess; $[H_2O]$ is essentially constant throughout the reaction.
- Combine the constant into the rate constant: $\text{Rate} = k'[\text{ester}]$ with $k' = k[H_2O]$, which is the form of a first-order rate law.
Answer: Because $[H_2O]$ stays effectively constant, the second-order reaction behaves as first order in ester, i.e. pseudo-first-order.