The rate constant for a first order reaction | vedclass

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(B) The Arrhenius equation is given by: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} [\frac{T_2 - T_1}{T_1 T_2}]$.Given: $k_1 = 0.026 \ s^{-1}$ at $T_1 = 290

Vedclass · Accepted Answer

$224.55 \ kJ \ mol^{-1}$

Vedclass · Answer

(B) The Arrhenius equation is given by: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} [\frac{T_2 - T_1}{T_1 T_2}]$.Given: $k_1 = 0.026 \ s^{-1}$ at $T_1 = 290 \ K$,$k_2 = 0.58 \ s^{-1}$ at $T_2 = 300 \ K$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Substituting the values: $\ln(\frac{0.58}{0.026}) = \frac{E_a}{8.314} [\frac{300 - 290}{300 	imes 290}]$.$\ln(22.307) = \frac{E_a}{8.314} [\frac{10}{87000}]$.$3.1048 = \frac{E_a}{8.314} 	imes 1.1494 	imes 10^{-4}$.$E_a = \frac{3.1048 	imes 8.314}{1.1494 	imes 10^{-4}} \approx 224600 \ J \ mol^{-1} = 224.6 \ kJ \ mol^{-1}$.The closest option is $224.55 \ kJ \ mol^{-1}$.

The rate constant for a first order reaction is $0.58 \ s^{-1}$ at $300 \ K$ and $0.026 \ s^{-1}$ at $290 \ K$. What is the energy of activation? $(R=8.314 \ J \ K^{-1} \ mol^{-1})$

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