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(D) The rate constant $K$ is given by the Arrhenius equation: $K = A e^{\frac{-E_{a}}{RT}}$.Let $K$ be the rate constant without the enzyme and $K^{\p

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$-6(2.303) RT$

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(D) The rate constant $K$ is given by the Arrhenius equation: $K = A e^{\frac{-E_{a}}{RT}}$.Let $K$ be the rate constant without the enzyme and $K^{\prime}$ be the rate constant with the enzyme.Given $K^{\prime} = 10^{6} K$.Substituting the Arrhenius equation:$A e^{\frac{-E^{\prime}_{a}}{RT}} = 10^{6} 	imes A e^{\frac{-E_{a}}{RT}}$.Taking the natural logarithm on both sides:$\frac{-E^{\prime}_{a}}{RT} = \frac{-E_{a}}{RT} + \ln(10^{6})$.Multiplying by $-RT$:$E^{\prime}_{a} = E_{a} - RT \ln(10^{6})$.Therefore,the change in activation energy $\Delta E_{a} = E^{\prime}_{a} - E_{a} = -RT \ln(10^{6})$.Since $\ln(10^{6}) = 6 \ln(10) = 6 	imes 2.303$,the change is $-6(2.303) RT$.

The rate of a certain biochemical reaction at physiological temperature $T$ occurs $10^{6}$ times faster with an enzyme than without. The change in the activation energy upon adding the enzyme is

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