Question 12.13: Using the Arrhenius Equation Rate constants for the gas-phas......
Using the Arrhenius Equation
Rate constants for the gas-phase decomposition of hydrogen iodide, 2 HI(g) → H_2(g) + I_2(g), are listed in the following table:
(a) Find the activation energy (in kJ/mol) using all five data points.
(b) Calculate E_a from the rate constants at 283 °C and 508 °C.
(c) Given the rate constant at 283 °C and the value of E_a obtained in part (b), what is the rate constant at 293 °C?
STRATEGY
(a) The activation energy E_a can be determined from the slope of a linear plot of ln k versus 1/T.
(b) To calculate E_a from values of the rate constant at two temperatures, use the equation
\ln \left(\frac{k_2}{k_1}\right)=\left(\frac{-E_{\mathrm{a}}}{R}\right)\left(\frac{1}{T_2}-\frac{1}{T_1}\right)(c) Use the same equation and the known values of E_a and k_1 at T_1 to calculate k_2 at T_2.
| Temperature (°C) | k (M^{-1} s^{-1} ) |
| 283 | 3.52 \times 10^{-7} |
| 356 | 3.02 \times 10^{-5} |
| 393 | 2.19 \times 10^{-4} |
| 427 | 1.16 \times 10^{-3} |
| 508 | 3.95 \times 10^{-2} |
(a) Because the temperature in the Arrhenius equation is expressed in kelvin, we must first convert the Celsius temperatures to absolute temperatures. Ten calculate values of 1/T and ln k, and plot ln k versus 1/T. The results are shown in the following table and graph:
The slope of the straight-line plot can be determined from the coordinates of any two widely separated points on the line:
\text { Slope }=\frac{\Delta y}{\Delta x}=\frac{(-14.0)-(-3.9)}{\left(0.00175 \mathrm{~K}^{-1}\right)-\left(0.00130 \mathrm{~K}^{-1}\right)}=\frac{-10.1}{0.00045 \mathrm{~K}^{-1}}=-2.24 \times 10^4 \mathrm{~K}Finally, calculate the activation energy from the slope:
\begin{aligned}E_{\mathrm{a}} & =-R(\text { Slope })=-\left(8.314 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}\right)\left(-2.24 \times 10^4 \mathrm{~K}\right) \\& =1.9 \times 10^5 \mathrm{~J} / \mathrm{mol}=190 \mathrm{~kJ} / \mathrm{mol}\end{aligned}Note that the slope of the Arrhenius plot is negative and the activation energy is positive. The greater the activation energy for a particular reaction, the steeper the slope of the ln k versus 1/T plot and the greater the increase in the rate constant for a given increase in temperature.
(b) Substituting the values of k_1=3.52 \times 10^{-7} \mathrm{M}^{-1} \mathrm{~s}^{-1} at T_1=556 \mathrm{~K}\left(283^{\circ} \mathrm{C}\right) and k_2=3.95 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1} at T_2 = 781 K (508 °C) into the equation
\ln \left(\frac{k_2}{k_1}\right)=\left(\frac{-E_{\mathrm{a}}}{R}\right)\left(\frac{1}{T_2}-\frac{1}{T_1}\right)gives
\ln \left(\frac{3.95 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}}{3.52 \times 10^{-7} \mathrm{M}^{-1} \mathrm{~s}^{-1}}\right)=\left(\frac{-E_{\mathrm{a}}}{8.314 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}}\right)\left(\frac{1}{781 \mathrm{~K}}-\frac{1}{556 \mathrm{~K}}\right)Simplifying this equation gives
\begin{aligned}11.628 & =\left(\frac{-E_{\mathrm{a}}}{8.314 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}}\right)\left(\frac{-5.18 \times 10^{-4}}{\mathrm{~K}}\right) \\ E_{\mathrm{a}} & =1.87 \times 10^5 \mathrm{~J} / \mathrm{mol}=187 \ \mathrm{~kJ} / \mathrm{mol}\end{aligned}(c) Use the same equation as in part (b), but now the known values are
\begin{aligned}& k_1=3.52 \times 10^{-7} \mathrm{M}^{-1} \mathrm{~s}^{-1} \text { at } T_1=556 \mathrm{~K}\left(283^{\circ} \mathrm{C}\right) \\& E_{\mathrm{a}}=1.87 \times 10^5 \mathrm{~J} / \mathrm{mol}\end{aligned}and k_2 at T_2 = 566 K (293 °C) is the unknown.
\ln \left(\frac{k_2}{3.52 \times 10^{-7} \mathrm{M}^{-1} \mathrm{~s}^{-1}}\right)=\left(\frac{-1.87 \times 10^5 \frac{\mathrm{J}}{\mathrm{mol}}}{8.314 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}}\right)\left(\frac{1}{566 \mathrm{~K}}-\frac{1}{556 \mathrm{~K}}\right)=0.715Taking the antiln of both sides gives
\begin{aligned}& \frac{k_2}{3.52 \times 10^{-7} \mathrm{M}^{-1} \mathrm{~s}^{-1}}=e^{0.715}=2.04 \\& k_2=7.18 \times 10^{-7} \mathrm{M}^{-1} \mathrm{~s}^{-1}\end{aligned}In this temperature range, a temperature increase of 10 K doubles the rate constant.
| Temperature (°C) | T (K) | k (M^{-1} \ s^{-1} ) | 1/T (1/K) | ln k |
| 283 | 556 | 3.52 \times 10^{-7} | 0.001 80 | -14.86 |
| 356 | 629 | 3.02 \times 10^{-5} | 0.001 59 | -10.408 |
| 393 | 666 | 2.19 \times 10^{-4} | 0.001 50 | -8.426 |
| 427 | 700 | 1.16 \times 10^{-3} | 0.001 43 | -6.759 |
| 508 | 781 | 3.95 \times 10^{-2} | 0.001 28 | -3.231 |
