Integration of the rate law for a simple first-order reaction $(rate=k[A])$ results in an equation describing how the reactant concentration varies with time: $$ ln(\frac{[A]_0}{ [A]_t })=kt \label{eq1}\tag{1}$$

where *k* is the rate constant, the initial concentration is [*A*]_{0} and [*A*]_{t} is the concentration present after any given time *t*. This equation can be arranged to other formats for convenience:

$$ ln(\frac{[A]_t}{ [A]_0 })=-kt \label{eq2}\tag{2}$$

and a linearized version: $$ln[A]_t = -kt+ln[A]_0 \label{eq3}\tag{3}$$

### Example: The Integrated Rate Law for a First-Order Reaction

The rate constant for the first-order decomposition of cyclobutane, C_{4}H_{8} at 500 °C is 9.2 × 10^{−3} s^{−1}: $$ C_4 H_8 \rightarrow 2\; C_2 H_ 4 $$

How long will it take for 80.0% of a sample of C_{4}H_{8} to decompose?

Solution We use the integrated form of the rate law to answer questions regarding time: $ ln(\frac{[A]_0}{ [A]_t})=kt $

There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [*A*]_{0}, [*A*], and *k*, and need to find *t*.

The initial concentration of C_{4}H_{8}, [*A*]_{0}, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let *x* be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of *x* or 0.200*x.* Rearranging the rate law to isolate *t* and substituting the provided quantities yields: $$ t=ln\frac {[x]}{[0.200x]}\times \frac{1}{k}\\

=ln(5)\times \frac{1}{9.2\times 10^{-3} s^{-1}} \\

=1.609 \times \frac{1}{9.2\times 10^{-3} s^{-1}} \\

=1.7\times 10^2 s $$

### Check Your Learning

Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:

$$^{131}I \rightarrow \: ^{131}Xe\;+ \text{electron} $$The decay is first-order with a rate constant of 0.138 d^{−1}. Knowing that all radioactive decay is first order, how many days will it take for exactly 90% of the iodine−131 in a 0.500 *M* solution of this substance to decay to Xe-131?

**Answer: **

If 90% of the I-131 has decaued, then 10% is left. In other words, if we started with 100 M of I-131, after the end of this time period, we will have 10 M.

Using the first-order integrated rate law (equation $\ref{eq1}$ above), we can see that the units in the ratio $ \frac{[A]_0}{ [A]_t} $ will cancel – so what units of concentration we input don’t matter, as long as that ratio (100:10) is maintained. We do not need to know the actual starting concentration: we can assume 100 if we want:
$$ ln \left(\frac{[A]_0}{ [A]_t } \right)=kt \\
ln \left(\frac{100 \, M}{10 \, M} \right)=(0.138 \text{d}^{-1})\, t \\
t = 16.6_{854} \text{d} \\
t = 16.7\: \text{days}$$
* Note – since “90%” is a ratio, not a measured value, and is specified as “exact”, we can assume it to mean 90.0000000000… and it (and the numbers we derived from it – 100 and 10 M) will not affect the significant digits in the final answer. *

We can use integrated rate laws with experimental data that consist of time and concentration information to determine the order and rate constant of a reaction. The integrated rate law can be rearranged to a standard linear equation format: $$ln[A]_t =−kt+ln[A]_0 \label{eq4}\tag{4} \\ y=mx+b$$

A plot of ln[*A*] versus *t* for a first-order reaction is a straight line with a slope of −*k* and an intercept of ln[*A*]_{0}. If a set of rate data are plotted in this fashion but do *not* result in a straight line, the reaction is not first order in *A*.

### Determination of Reaction Order by Graphing

Show that the data in the table below can be represented by a first-order rate law by graphing ln[H_{2}O_{2}] versus time. Determine the rate constant for the rate of decomposition of H_{2}O_{2} from this data.

Trial | Time (h) | [H_{2}O_{2}] (M) | ln[H_{2}O_{2}] |
---|---|---|---|

1 | 0 | 1.000 | 0.0 |

2 | 6.00 | 0.500 | −0.693 |

3 | 12.00 | 0.250 | −1.386 |

4 | 18.00 | 0.125 | −2.079 |

5 | 24.00 | 0.0625 | −2.772 |

**Solution**: The data from the table above are plotted here:

The plot of ln[H_{2}O_{2}] versus time is linear, thus we have verified that the reaction may be described by a first-order rate law.

The rate constant for a first-order reaction is equal to the negative of the slope of the plot of ln[H_{2}O_{2}] versus time where: $$ slope=\frac{change\;in\;y}{change\;in\;x}=\frac{\Delta y}{\Delta x}=\frac{\Delta ln[H_2 O_2]}{\Delta t} $$

In order to determine the slope of the line, we need two values of ln[H_{2}O_{2}] at different values of *t* (one near each end of the line is preferable). For example, the value of ln[H_{2}O_{2}] when *t* is 6.00 h is −0.693; the value when *t* = 12.00 h is −1.386:
$$ \text{slope} =\frac{−1.386−(−0.693)}{12.00\;h−6.00\;h} \\
=\frac{−0.693}{6.00\;h} \\
=−1.155\times 10^{−1}\; h^{−1} \\
k=\, − \text{slope} =−(−1.155×10^{−1}h^{−1})=1.155\times 10^{−1}\, h^{−1} $$

### Check Your Learning

Graph the following data to determine whether the reaction $ A \rightarrow B+C$ is first order.

Trial | Time (s) | [A] |
---|---|---|

1 | 4.0 | 0.220 |

2 | 8.0 | 0.144 |

3 | 12.0 | 0.110 |

4 | 16.0 | 0.088 |

5 | 20.0 | 0.074 |

## Answer:

The plot of ln[A] vs *t* is not a straight line. Therefore this reaction is not a first-order reaction.