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(A) Given: $T_{1} = 27^{\circ} C = 300 \ K$,$T_{2} = 42^{\circ} C = 315 \ K$.Since the number of molecules with energy greater than the threshold ener

Vedclass · Accepted Answer

$84297.5$

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(A) Given: $T_{1} = 27^{\circ} C = 300 \ K$,$T_{2} = 42^{\circ} C = 315 \ K$.Since the number of molecules with energy greater than the threshold energy increases five-fold,the rate constant $k$ increases by a factor of $5$,i.e.,$k_{2} = 5k_{1}$.Using the Arrhenius equation: $\ln \left(\frac{k_{2}}{k_{1}}ight) = \frac{E_{a}}{R} \left(\frac{1}{T_{1}} - \frac{1}{T_{2}}ight)$.Substituting the values: $\ln(5) = \frac{E_{a}}{8.314} \left(\frac{1}{300} - \frac{1}{315}ight)$.$\ln(5) = \frac{E_{a}}{8.314} \left(\frac{315 - 300}{300 	imes 315}ight) = \frac{E_{a}}{8.314} \left(\frac{15}{94500}ight)$.$1.6094 = \frac{E_{a}}{8.314} 	imes \frac{1}{6300}$.$E_{a} = 1.6094 	imes 8.314 	imes 6300 = 84297.47 \ J \ mol^{-1}$.

The number of molecules with energy greater than the threshold energy for a reaction increases five-fold by a rise of temperature from $27^{\circ} C$ to $42^{\circ} C$. Its energy of activation in $J \ mol^{-1}$ is.............. (Take $\ln 5 = 1.6094$; $R = 8.314 \ J \ mol^{-1} K^{-1}$)

Similar Questions

According to the collision theory of chemical reaction,which of the following is correct?

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The rate constant for a first order reaction is given by the following equation:
$\ln k = 33.24 - \frac{2.0 \times 10^{4} \, K}{T}$
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