The decomposition of formic acid on a gold su | vedclass

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(A) Given:$T_1 = 300 \ K, K_1 = 1.0 	imes 10^{-3} \ s^{-1}$$T_2 = 200 \ K, K_2 = ?$$E_a = 11.488 \ kJ \ mol^{-1} = 11488 \ J \ mol^{-1}$$R = 8.314 \

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) Given:$T_1 = 300 \ K, K_1 = 1.0 	imes 10^{-3} \ s^{-1}$$T_2 = 200 \ K, K_2 = ?$$E_a = 11.488 \ kJ \ mol^{-1} = 11488 \ J \ mol^{-1}$$R = 8.314 \ J \ mol^{-1} K^{-1}$Using the Arrhenius equation:$\ln \left(\frac{K_1}{K_2}ight) = \frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}ight)$$\ln \left(\frac{1.0 	imes 10^{-3}}{K_2}ight) = \frac{11488}{8.314} \left(\frac{1}{200} - \frac{1}{300}ight)$$\ln \left(\frac{1.0 	imes 10^{-3}}{K_2}ight) = \frac{11488}{8.314} \left(\frac{300 - 200}{60000}ight) = \frac{11488}{8.314} 	imes \frac{100}{60000} = \frac{11488}{8.314 	imes 600} \approx \frac{11488}{4988.4} \approx 2.303$$\frac{1.0 	imes 10^{-3}}{K_2} = e^{2.303} \approx 10$$K_2 = \frac{1.0 	imes 10^{-3}}{10} = 1.0 	imes 10^{-4} = 10 	imes 10^{-5} \ s^{-1}$Thus,the value is $10$.

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