## 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

The 'Blue Check Mark' means that this solution was answered by an expert.

(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

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