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(A) Using the Arrhenius equation: $\ln(\frac{K_1}{K_2}) = \frac{E_a}{R} (\frac{1}{T_2} - \frac{1}{T_1})$Given: $T_1 = 300 \, K$,$T_2 = 280 \, K$,$E_a

Vedclass · Accepted Answer

$K_2 \approx 0.25 K_1$

Vedclass · Answer

(A) Using the Arrhenius equation: $\ln(\frac{K_1}{K_2}) = \frac{E_a}{R} (\frac{1}{T_2} - \frac{1}{T_1})$Given: $T_1 = 300 \, K$,$T_2 = 280 \, K$,$E_a = 1.157 	imes 10^4 \, cal \, mol^{-1}$,$R = 1.987 \, cal \, K^{-1} \, mol^{-1}$.$\ln(\frac{K_1}{K_2}) = \frac{1.157 	imes 10^4}{1.987} (\frac{1}{280} - \frac{1}{300})$$\ln(\frac{K_1}{K_2}) = 5822.85 	imes (\frac{20}{84000}) = 5822.85 	imes 0.000238 \approx 1.386$Since $\ln(4) \approx 1.386$,we have $\frac{K_1}{K_2} = 4$,which implies $K_1 = 4 K_2$ or $K_2 = 0.25 K_1$.

$A$ chemical reaction was carried out at $300 \, K$ and $280 \, K$. The rate constants were found to be $K_1$ and $K_2$ respectively. The energy of activation is $1.157 \times 10^4 \, cal \, mol^{-1}$ and $R = 1.987 \, cal \, K^{-1} \, mol^{-1}$. Then:

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