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(B) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.Taking the natural logarithm on both sides,$\ln k = \ln A - \frac{E_a}{RT}$.Differentiat

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$A$ high value of activation energy

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(B) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.Taking the natural logarithm on both sides,$\ln k = \ln A - \frac{E_a}{RT}$.Differentiating with respect to temperature,$\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}$.This equation shows that the rate constant $k$ is highly sensitive to temperature changes when the activation energy $E_a$ is high.Therefore,a small increase in temperature leads to a significant increase in the rate of reaction if the activation energy is high.

For a gaseous reaction,a large increase in the rate of reaction with a small increase in temperature indicates:

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