The rate of a reaction quadruples when the te | vedclass

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(A) Using the Arrhenius equation: $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$ Given: $\frac{k_2}{k_1} = 4$,$T_1

Vedclass · Accepted Answer

$107.2$

Vedclass · Answer

(A) Using the Arrhenius equation: $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} ight)$ Given: $\frac{k_2}{k_1} = 4$,$T_1 = 300 \, K$,$T_2 = 310 \, K$,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$,$\ln 2 = 0.693$. $\ln 4 = \frac{E_a}{8.314} \left( \frac{310 - 300}{310 	imes 300} ight)$ $2 \ln 2 = \frac{E_a}{8.314} \left( \frac{10}{93000} ight)$ $2 	imes 0.693 = \frac{E_a}{8.314} 	imes \frac{1}{9300}$ $E_a = 1.386 	imes 8.314 	imes 9300 \, J \, mol^{-1} \approx 107200 \, J \, mol^{-1} = 107.2 \, kJ \, mol^{-1}$.

Similar Questions

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