# Integrated Rate Law

Integrated rate laws are derived from rate laws. These equations relate reactant concentration with time.

## Zeroth-order Integrated Rate Law

$[\mathrm{A}]_t = -kt + [\mathrm{A}]_0$

## First-order Integrated Rate Law

$\ln[\mathrm{A}]_{t} = -kt + \ln\mathrm[\mathrm{A}]_0$

Derivation from a first order rate law

\begin{align*} \mathrm{rate~of~reaction} &= k[\mathrm{A}] \\[2ex] -\dfrac{d[\mathrm{A}]}{dt} &= k[\mathrm{A}] \\[2ex] \dfrac{d[\mathrm{A}]}{dt} &= -k~dt \\[2ex] \int \dfrac{d[\mathrm{A}]}{dt} &= -k \int dt\\[2ex] \int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t} \dfrac{d[\mathrm{A}]}{dt} &= -k \int_0^t dt\\[2ex] \ln[\mathrm{A}]_t -\ln[\mathrm{A}]_0 = -kt \\[2ex] \ln[\mathrm{A}]_t = -kt + \ln[\mathrm{A}]_0 \end{align*}

## Second-order Integrated Rate Law

$\dfrac{1}{[\mathrm{A}]_t} = kt + \dfrac{1}{[\mathrm{A}]_0}$

See handout on how to transform and plot rate data (for a 0th, 1st, and 2nd order reaction).

## Practice Problems

### Problem 1

The rate constant for first-order decomposition of cyclobutane (C4H8) at 500 °C is 9.2 × 10–3 s–1.
How long (in s) will it take for 80.0% of the reactant to decompose?

C4H8 ⟶ 2C2H4

Hint: Assume initial 1 L solution and 1 M concentration. 80% decomposed means new concentration is 0.2 M.

Solution

\begin{align*} \ln [\mathrm{A}]_t &= -kt + \ln [\mathrm{A}]_0\\[2ex] t &= \dfrac{\ln [\mathrm{A}]_t - \ln [\mathrm{A}]_0}{-k} \\ &= \dfrac{ \ln (0.20) - \ln (1.00) }{-9.2\times 10^{-3}~\mathrm{s^{-1}}} \\[1ex] &= 175~\mathrm{s} \end{align*}

### Problem 2

The following reaction is 2nd-order with k = 5.76 × 10–2 M–1 min–1 under certain conditions. If the initial concentration of C4H6 is 0.200 M, what is the concentration (in M) after 10 minutes?

2C4H6 (g) ⟶ C8H12 (g)

Solution

\begin{align*} \dfrac{1}{[\mathrm{A}]_t} &= kt + \dfrac{1}{[\mathrm{A}]_0} \\ [\mathrm{A}]_t &= \dfrac{1}{kt+\dfrac{1}{[\mathrm{A}]_0}} \\ &= \dfrac{1}{\left ( 5.76\times 10^{-2} ~M^{-1}~\mathrm{min}^{-1} \right ) \left ( 10.0~\mathrm{min}\right ) + \left ( \dfrac{1}{0.200~M} \right ) } \\ &= 0.179~M \end{align*}

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