Chapter 13
Q. 13.5
Exploring the Link between Equilibrium and Kinetics
The equilibrium constant Kc for the reaction of hydrogen with iodine is 57.0 at 700 K, and the reaction is endothermic (ΔE = 9 kJ).
\mathrm{H}_2(g)+\mathrm{I}_2(g) \underset{k_{\mathrm{r}}}{\stackrel{k_{\mathrm{f}}}{\rightleftharpoons}} 2 \mathrm{HI}(g) \quad K_{\mathrm{c}}=57.0 \text { at } 700 \mathrm{~K}(a) Is the rate constant k_f for the formation of HI larger or smaller than the rate constant k_r for the decomposition of HI?
(b) The value of k_r at 700 K is 1.16 × 10^{-3} M^{-1} \ s^{-1} . What is the value of k_f at the same temperature?
(c) How are the values of k_f, \ k_r, and K_c affected by an increase in temperature?
STRATEGY
To answer these questions, make use of the relationship K_c = k_f/k_r. Also, recall that the temperature dependence of a rate constant increases with increasing value of the activation energy (Section 12.10).
Step-by-Step
Verified Solution
(a) Because K_c = k_f/k_r = 57.0, the rate constant for the formation of HI (forward reaction) is larger than the rate constant for the decomposition of HI (reverse reaction) by a factor of 57.0.
(b) Because K_c = k_f/k_r,
k_{\mathrm{f}}=\left(K_c\right)\left(k_{\mathrm{r}}\right)=(57.0)\left(1.16 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\right)=6.61 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}(c) Because the reaction is endothermic, E_a for the forward reaction is greater than E_a for the reverse reaction. Consequently, as the temperature increases, k_f increases by more than k_r increases, and therefore K_c = k_f/k_r increases