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(A) The Arrhenius equation is given by $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.Given: $T_1 = 27 + 273 = 300 \ K$,$T_2 =

Vedclass · Accepted Answer

$0.36$

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(A) The Arrhenius equation is given by $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.Given: $T_1 = 27 + 273 = 300 \ K$,$T_2 = 327 + 273 = 600 \ K$,$E_a = 24.942 \ kJ \ mol^{-1} = 24942 \ J \ mol^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Substituting the values: $\ln(\frac{k_2}{k_1}) = \frac{24942}{8.314} (\frac{1}{300} - \frac{1}{600}) = 3000 	imes (\frac{2-1}{600}) = 3000 	imes \frac{1}{600} = 5$.Therefore,$\frac{k_2}{k_1} = e^5 = 150$.Since the rate of a first-order reaction is proportional to the rate constant $(Rate = k[A])$,the ratio of rates is equal to the ratio of rate constants.$Rate_2 = 150 	imes Rate_1 = 150 	imes 2.4 	imes 10^{-3} = 360 	imes 10^{-3} = 0.36 \ mol \ L^{-1} \ s^{-1}$.

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