The rate constant,k for a first order reactio | vedclass

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Question

(A) Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1T_2} \right)$ Given: $k_1 = x$,$k_2 = 4x$,$T_1

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(A) Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1T_2} ight)$ Given: $k_1 = x$,$k_2 = 4x$,$T_1 = 600 \ K$,$T_2 = 700 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$. Substituting the values: $\log \left( \frac{4x}{x} ight) = \frac{E_a}{2.303 	imes 8.314} \left( \frac{700 - 600}{700 	imes 600} ight)$ $\log(4) = \frac{E_a}{19.147} \left( \frac{100}{420000} ight)$ $0.602 = \frac{E_a}{19.147} 	imes 2.381 	imes 10^{-4}$ $E_a = \frac{0.602 	imes 19.147}{2.381 	imes 10^{-4}} \approx 48415 \ J \ mol^{-1} = 48.41 \ kJ \ mol^{-1}$. The closest option is $48.16 \ kJ \ mol^{-1}$.

The rate constant,$k$ for a first order reaction,$C_2H_5I_{(g)} \rightarrow C_2H_{4(g)} + HI_{(g)}$ is $x \ s^{-1}$ at $600 \ K$ and $4x \ s^{-1}$ at $700 \ K$. The energy of activation of the reaction (in $kJ \ mol^{-1}$) is (in $.16$)

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