For the adsorption of hydrogen on platinum,th | vedclass

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(D) The rate of chemisorption follows the Arrhenius equation: $k = Ae^{-\frac{E_a}{RT}}$.Let $k_1$ be the rate on platinum and $k_2$ be the rate on ni

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) The rate of chemisorption follows the Arrhenius equation: $k = Ae^{-\frac{E_a}{RT}}$.Let $k_1$ be the rate on platinum and $k_2$ be the rate on nickel.$\frac{k_2}{k_1} = \frac{Ae^{-\frac{(E_a)_2}{RT}}}{Ae^{-\frac{(E_a)_1}{RT}}} = e^{\frac{(E_a)_1 - (E_a)_2}{RT}}$.Taking the logarithm to the base $10$:$\log_{10} \left( \frac{k_2}{k_1} ight) = \frac{(E_a)_1 - (E_a)_2}{2.303 RT}$.Given $(E_a)_1 = 30 \ kJ \ mol^{-1} = 30000 \ J \ mol^{-1}$,$(E_a)_2 = 41.4 \ kJ \ mol^{-1} = 41400 \ J \ mol^{-1}$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,and $\ln 10 \approx 2.3$.$\log_{10} \left( \frac{k_2}{k_1} ight) = \frac{30000 - 41400}{2.3 	imes 8.3 	imes 300} = \frac{-11400}{5727} \approx -1.99$.Since the question asks for the logarithm of the ratio of rates (usually implying the magnitude or the ratio of $k_1/k_2$ to get a positive value),the nearest integer is $2$.

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