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(B) Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} [\frac{T_2 - T_1}{T_1 T_2}]$.For reaction $A$: $\ln(2) = \frac{E_{a,A}}{R} [\f

Vedclass · Accepted Answer

$4.92$

Vedclass · Answer

(B) Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} [\frac{T_2 - T_1}{T_1 T_2}]$.For reaction $A$: $\ln(2) = \frac{E_{a,A}}{R} [\frac{310 - 300}{300 	imes 310}] = \frac{E_{a,A}}{R} [\frac{10}{93000}]$.For reaction $B$: $\ln(2) = \frac{E_{a,B}}{R} [\frac{\Delta T}{300(300 + \Delta T)}]$.Given $E_{a,B} = 2 E_{a,A}$,we substitute $\ln(2)$ from reaction $A$ into reaction $B$:$\frac{E_{a,A}}{R} [\frac{10}{93000}] = \frac{2 E_{a,A}}{R} [\frac{\Delta T}{300(300 + \Delta T)}]$.$\frac{10}{93000} = \frac{2 \Delta T}{300(300 + \Delta T)}$.$\frac{1}{9300} = \frac{\Delta T}{150(300 + \Delta T)}$.$150(300 + \Delta T) = 9300 \Delta T$.$45000 + 150 \Delta T = 9300 \Delta T$.$9150 \Delta T = 45000$.$\Delta T = \frac{45000}{9150} \approx 4.92 \, K$.

The rate of a reaction $A$ doubles on increasing the temperature from $300 \, K$ to $310 \, K$. By how much should the temperature of reaction $B$ be increased from $300 \, K$ so that its rate doubles,if the activation energy of reaction $B$ is twice that of reaction $A$ (in $, K$)?

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