Chapter 12

Q. 12.13

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?


(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}


Verified Solution

(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)


\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)=0.715

Taking 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
worked example 12.13