### Second-Order Reactions

The equations that relate the concentrations of reactants and the rate constant of second-order reactions are fairly complicated. We will limit ourselves to the simplest second-order reactions, namely, those with rates that are dependent upon just one reactant’s concentration and described by the differential rate law: $$rate=k[A]^2 \label{eq1}\tag{1}$$

For these second-order reactions, the integrated rate law is: $$\frac{1}{[A]_t}=kt+\frac{1}{[A]_0} \label{eq2}\tag{2}$$

where the terms in the equation have their usual meanings as defined when discussing rate laws earlier.

### The Integrated Rate Law for a Second-Order Reaction

The reaction of butadiene gas (C_{4}H_{6}) with itself produces C_{8}H_{12} gas as follows: $$2\;C_4 H_6 (g)\rightarrow C_8 H_{12} (g)$$

The reaction is second order with a rate constant equal to $ 5.76 \times 10^{-2}\; \frac{L}{mol\cdot min} $ (or $ 5.76 \times 10^{-2}\; \frac{1}{M\cdot min} $) under certain conditions. If the initial concentration of butadiene is 0.200 *M*, what is the concentration remaining after 10.0 min?

**Solution: **

We use the integrated form of the rate law to answer questions regarding time. For a second-order reaction, this is Equation $\ref{eq2}$ above.

We know three variables in this equation: [*A*]_{0} = 0.200 mol/L, $k= 5.76 \times 10^{-2}\; \frac{1}{M\cdot min} $ , and *t* = 10.0 min. Therefore, we can solve for the remaining variable, [*A*]_{t}:

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

\frac{1}{[A]_t}=( 5.76 \times 10^{-2}\; \frac{1}{M\cdot min} )(10\;min)+\frac{1}{0.200\;M}\\

\frac{1}{[A]_t}=(5.76 \times 10^{−1}\; \frac{1}{M})+5.00 \frac{1}{M} \\

\frac{1}{[A]_t}=5.58 \frac{1}{M} \\

[A]_t =1.79\times 10^{−1}\;M$$

Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L that was originally present.

### Check Your Learning

If the initial concentration of butadiene is 0.0200 M, what is the concentration remaining after 20.0 min? (Following the reaction described in the example above).

**Answer: **

Using Equation $\ref{eq2}$ and the data provided (*k* is given in the example above):
$$\frac{1}{[A]_t}=kt+\frac{1}{[A]_0} \\
\frac{1}{[A]_t}=\left( 5.76 \times 10^{-2}\; \frac{L}{mol\cdot min} \right)\, (20.0\: \text{min}) + \frac{1}{0.0200\: M} \\
\frac{1}{[A]_t} = 51.1_{52}\: M^{-1} \\
[A]_t = 0.0195_{495}\: M \\
[A]_t = 0.0195\: M $$

The integrated rate law for our second-order reactions has the form of the equation of a straight line: $$\frac{1}{[A]_t}=kt+\frac{1}{[A]_0}\\y=mx+b$$

A plot of $\frac{1}{[A]_t}$ versus *t* for a second-order reaction is a straight line with a slope of *k* and an intercept of $\frac{1}{[A]_0}$. If the plot is not a straight line, then the reaction is not second order.

### Determination of Reaction Order by Graphing

The data below are for the same reaction described in the previous example. Test these data to confirm that this dimerization reaction is second-order.

**Solution**

Trial | Time (s) | [C_{4}H_{6}] (M) |
---|---|---|

1 | 0 | 1.00 × 10^{−2} |

2 | 1600 | 5.04 × 10^{−3} |

3 | 3200 | 3.37 × 10^{−3} |

4 | 4800 | 2.53 × 10^{−3} |

5 | 6200 | 2.08 × 10^{−3} |

In order to distinguish a first-order reaction from a second-order reaction, we plot ln[C_{4}H_{6}] versus *t* and compare it with a plot of $\frac{1}{[C_4 H_ 6]}$ versus *t*. The values needed for these plots follow.

Time (s) | $\frac{1}{[C _4 H_6]} $ (M ^{−1}) | ln[C_{4}H_{6}] |
---|---|---|

0 | 100 | −4.605 |

1600 | 198 | −5.289 |

3200 | 296 | −5.692 |

4800 | 395 | −5.978 |

6200 | 481 | −6.175 |

The plots are shown below. As you can see, the plot of ln[C_{4}H_{6}] versus *t* is not linear, therefore the reaction is not first order. The plot of $\frac{1}{[C_4 H_6 ]}$ versus *t* is linear, indicating that the reaction is second order.

### Check Your Learning

Does the following data describe a reaction with second-order kinetics?

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

1 | 5 | 0.952 |

2 | 10 | 0.625 |

3 | 15 | 0.465 |

4 | 20 | 0.370 |

5 | 25 | 0.308 |

6 | 35 | 0.230 |

## Answer:

Yes, this reaction is second-order in [A]. The plot of $\frac{1}{[A]}$ vs $t$ is linear: