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(D) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$Given: $K_1 = 0.03 \, min^{-1}$

Vedclass · Accepted Answer

$2520$

Vedclass · Answer

(D) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} ight)$Given: $K_1 = 0.03 \, min^{-1}$ at $T_1 = 200 \, K$,$K_2 = 0.05 \, min^{-1}$ at $T_2 = 300 \, K$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$,$\ln 10 = 2.3$$\log \frac{0.05}{0.03} = \frac{E_a}{2.3 	imes 8.3} \left( \frac{300 - 200}{200 	imes 300} ight)$$\log \left( \frac{5}{3} ight) = \frac{E_a}{19.09} 	imes \frac{100}{60000}$$\log 5 - \log 3 = \frac{E_a}{19.09} 	imes \frac{1}{600}$$0.70 - 0.48 = \frac{E_a}{11454}$$0.22 = \frac{E_a}{11454}$$E_a = 0.22 	imes 11454 = 2519.88 \, J \approx 2520 \, J$

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