Integrated Rate Equation — First Order & Pseudo First Order

First Order Reactions

A first-order reaction has a rate proportional to the first power of the reactant concentration. For $R \rightarrow P$:

$$\text{Rate} = -\frac{d[R]}{dt} = k[R]$$

Integrating gives the natural-log form $\ln\dfrac{[R]_0}{[R]} = kt$, or in the base-10 form used in most numericals:

$$\boxed{k = \frac{2.303}{t}\log\frac{[R]_0}{[R]}}$$

Equivalently, $[R] = [R]_0 e^{-kt}$. The units of $k$ are s$^{-1}$ (or min$^{-1}$) — independent of concentration, a hallmark of first order.

Graph of a First-Order Reaction

Rearranging the integrated law:

$$\log[R] = \log[R]_0 - \frac{k}{2.303}t$$

So a plot of $\log[R]$ against $t$ is a straight line with slope $= -\dfrac{k}{2.303}$ and intercept $\log[R]_0$. (Equivalently, $\ln[R]$ vs $t$ has slope $-k$.)

Examples of first-order reactions: radioactive decay, decomposition of $\text{N}_2\text{O}_5$, decomposition of $\text{SO}_2\text{Cl}_2$, and several unimolecular decompositions studied in kinetics.

Pseudo First Order Reactions

Some reactions that are truly second order behave as first order because one reactant is present in such large excess that its concentration hardly changes. These are pseudo-first-order reactions.

Two classic examples:

1. Acid hydrolysis of an ester: $$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$$ True rate $= k[\text{ester}][\text{H}_2\text{O}]$, but water is in vast excess, so $[\text{H}_2\text{O}]$ is essentially constant. The rate then appears as $\text{Rate} = k'[\text{ester}]$ with $k' = k[\text{H}_2\text{O}]$ — pseudo first order.

2. Inversion of cane sugar: $$\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{glucose} + \text{fructose}$$ Again water is in large excess, so it is pseudo first order.

[JEE Tip] A reaction's true order can be higher than its observed order when a reactant is in large excess. Spotting "water in excess" or "one reactant in large excess" is the cue for pseudo first order.

Solved Examples

Example 1: First-order rate constant

A first-order reaction is 50% complete in 20 minutes. Find $k$.

Solution: 50% complete means $[R] = [R]_0/2$, so $\dfrac{[R]_0}{[R]} = 2$. $k = \dfrac{2.303}{t}\log 2 = \dfrac{2.303}{20}\times0.301 = 0.0347$ min$^{-1}$.

Example 2: Concentration after time

A first-order reaction has $k = 2.303\times10^{-3}$ s$^{-1}$. What fraction of reactant remains after 1000 s?

Solution: $\log\dfrac{[R]_0}{[R]} = \dfrac{kt}{2.303} = \dfrac{(2.303\times10^{-3})(1000)}{2.303} = 1$. So $\dfrac{[R]_0}{[R]} = 10$, i.e. $\dfrac{[R]}{[R]_0} = 0.1$. 10% remains.

Example 3: Time for a given completion

A first-order reaction has $k = 1.15\times10^{-3}$ s$^{-1}$. How long for the reactant to fall to one-fourth of its initial value?

Solution: $\dfrac{[R]_0}{[R]} = 4$. $t = \dfrac{2.303}{k}\log4 = \dfrac{2.303}{1.15\times10^{-3}}\times0.602 = 2003.5\times0.602 = 1206$ s.

Example 4: Percentage decomposition and time

A first-order reaction takes 40 min for 30% decomposition. Find $k$.

Solution: 30% decomposed means $[R] = 70\%$ of $[R]_0$, so $\dfrac{[R]_0}{[R]} = \dfrac{100}{70} = 1.4286$. $k = \dfrac{2.303}{40}\log1.4286 = \dfrac{2.303}{40}\times0.1549 = 8.92\times10^{-3}$ min$^{-1}$.

Example 5: Pseudo first order identification

Why is the acid hydrolysis of ethyl acetate called a pseudo-first-order reaction?

Solution: The true rate law is $\text{Rate} = k[\text{ester}][\text{H}_2\text{O}]$ (second order). But water is in large excess, so $[\text{H}_2\text{O}]$ is effectively constant. The rate reduces to $\text{Rate} = k'[\text{ester}]$, appearing first order — hence pseudo first order.

Example 6: Fraction remaining after n half-lives

For a first-order reaction, what fraction of reactant remains after 3 half-lives?

Solution: After each half-life the amount halves: after 3, fraction $= \left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8} = 0.125$. 12.5% remains.

Example 7: Rate constant from concentrations

For a first-order reaction, $[R]$ falls from 0.80 M to 0.20 M in 60 s. Find $k$.

Solution: $k = \dfrac{2.303}{60}\log\dfrac{0.80}{0.20} = \dfrac{2.303}{60}\log4 = \dfrac{2.303}{60}\times0.602 = 0.0231$ s$^{-1}$.

Example 8: Time for 99% completion

Show that for a first-order reaction, the time for 99% completion is twice the time for 90% completion.

Solution: $t_{99\%} = \dfrac{2.303}{k}\log\dfrac{100}{1} = \dfrac{2.303}{k}\times2$; $t_{90\%} = \dfrac{2.303}{k}\log\dfrac{100}{10} = \dfrac{2.303}{k}\times1$. So $t_{99\%} = 2\,t_{90\%}$.

Example 9: Find concentration after given time

A first-order reaction with $k = 0.0693$ min$^{-1}$ starts at $[R]_0 = 1.0$ M. Find $[R]$ after 10 min.

Solution: $\log\dfrac{[R]_0}{[R]} = \dfrac{kt}{2.303} = \dfrac{0.0693\times10}{2.303} = 0.301$. So $\dfrac{[R]_0}{[R]} = 10^{0.301} = 2.0$, giving $[R] = 0.50$ M.

Example 10: Order from constant units

The rate constant of a reaction is $3.0\times10^{-4}$ s$^{-1}$. What is the order?

Solution: Units s$^{-1}$ correspond to first order (k independent of concentration).