**Practice Problem 11**

Use the following data to determine the activation energy for the decomposition of HI:

**Temperature (K)
Rate Constant ( M/s)**

573
2.91 x 10^{-6}

673
8.38 x 10^{-4}

773
7.65 x 10^{-2}

**Solution**

We can determine the activation energy for a reaction from a plot of the natural log of the rate constants versus the reciprocal of the absolute temperature. We therefore start by calculating 1/T and the natural logarithm of the rate constants:

ln k |
1/T (K^{-1}) |

-12.75 | 0.00175 |

-7.08 | 0.00149 |

-2.57 | 0.00129 |

When we construct a graph of these data, we get a straight line with a slope of -22,200 K, as seen in Figure 1.

Figure 1. A plot of the natural log of the rate
constant for the reaction at different temperatures versus the inverse of the temperature
in kelvin is a straight line with a slope equal to -E._{a}/R |

According to the Arrhenius equation, the slope of this line is equal to -*E _{a}/R*:

When this equation is solved, we get the follwoing value for the activation energy for this reaction:

E_{a} = 183 kJ/mol