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(C) Given that the rate of a chemical reaction doubles with every $10^{\circ}C$ rise in temperature,we have $\frac{k_2}{k_1} = 2$.Initial temperature

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$49.8 \ kJ \ mol^{-1}$

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(C) Given that the rate of a chemical reaction doubles with every $10^{\circ}C$ rise in temperature,we have $\frac{k_2}{k_1} = 2$.Initial temperature $T_1 = 22 + 273 = 295 \ K$.Final temperature $T_2 = 32 + 273 = 305 \ K$.Using the Arrhenius equation: $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left[ \frac{T_2 - T_1}{T_1 T_2} ight]$.Substituting the values: $0.69 = \frac{E_a}{8.3} \left[ \frac{305 - 295}{295 	imes 305} ight]$.$E_a = \frac{0.69 	imes 8.3 	imes 295 	imes 305}{10} \approx 49800 \ J \ mol^{-1} = 49.8 \ kJ \ mol^{-1}$.

The rate of a chemical reaction doubles with every $10^{\circ}C$ rise in temperature. If the reaction is carried out in the vicinity of $22^{\circ}C$,the activation energy of the reaction is (Given $R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$\ln 2 = 0.69$ and $\ln 3 = 1.1$)

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