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Magazine Chapter 12: Q63 E (page 709)
Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, for the reaction was measured at several different temperatures, and the data are shown here:
Temperature(K) | k(M-1s-1) |
555 | 6.23*10-7 |
575 | 2.42*10-6 |
645 | 1.44*10-4 |
700 | 2.01*10-3 |
What is the value of the activation energy (in kJ/mol) for this reaction?
Short Answer
The energy of activation for the reaction is 178.79 kJ/mol.
Step by step solution
Plot of ln K vs 1/T
T (K) | 1/T (K-1) | k (M-1s-1) | ln k |
555 | 0.0018 | 6.23*10-7 | -14.2887 |
575 | 0.00174 | 2.42*10-6 | -12.9317 |
645 | 0.00155 | 1.44*10-4 | -8.84570 |
700 | 0.00143 | 2.01*10-3 | -6.20962 |
The plot of ln k vs 1/T gives a straight line.
Calculation of Activation energy
The activated energy of a reaction is the minimum energy that a reactant must possess to convert into products.
The Arrhenius equation is given as
\(\begin{align}k &= A{e^{\frac{{ - {E_a}}}{{RT}}}}\\\ln k &= \ln A + (\frac{{ - {E_a}}}{{RT}})\end{align}\)
Here k is the rate constant; A is the pre-exponential factor, Eais the activation energy, R is the universal gas constant, and T is the temperature.
The value of\({{\bf{E}}_{\bf{a}}}\)can be calculated from the slope of the plot.
\({\bf{Slope = }}\frac{{{\bf{ - }}{{\bf{E}}_{\bf{a}}}}}{{\bf{R}}}\)
Considering the second and third data points on the plot, the slope is
\(\begin{align}Slope &= \frac{{\Delta y}}{{\Delta x}}\\\frac{{ - {E_a}}}{R} &= \frac{{ - 12.9317 - \left( { - 8.8457} \right)}}{{0.00174 - 0.00155}}\\ - {E_a} &= - 21505.26 \times 8.314\\{E_a} &= 178794.7J/mol\\{E_a} &= 178.79kJ/mol\end{align}\)
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