Without the enzyme, we have:

$\overline{){{\mathbf{k}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{A}}{\mathbf{\xb7}}{{\mathbf{e}}}^{\frac{\mathbf{-}{{\mathbf{E}}_{\mathbf{a}}}_{\mathbf{1}}}{\mathbf{RT}}}}$

In the presence of the enzyme, we have:

$\overline{){{\mathbf{k}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{A}}{\mathbf{\xb7}}{{\mathbf{e}}}^{\frac{\mathbf{-}{{\mathbf{E}}_{\mathbf{a}}}_{\mathbf{2}}}{\mathbf{RT}}}}$

Getting the ratio of k_{2}/k_{1}:

$\frac{{\mathbf{k}}_{\mathbf{2}}}{{\mathbf{k}}_{\mathbf{1}}}\mathbf{=}\frac{{{\mathbf{e}}}^{\frac{\mathbf{-}{{\mathbf{E}}_{\mathbf{a}}}_{\mathbf{2}}}{\mathbf{RT}}}}{{{\mathbf{e}}}^{\frac{\mathbf{-}{{\mathbf{E}}_{\mathbf{a}}}_{\mathbf{1}}}{\mathbf{RT}}}}\mathbf{=}{\mathbf{e}}^{\frac{{\mathbf{E}}_{\mathbf{a}\mathbf{1}}\mathbf{-}{\mathbf{E}}_{\mathbf{a}\mathbf{2}}}{\mathbf{RT}}}$

Getting ln of both sides:

$\mathbf{ln}\left(\frac{{k}_{2}}{{k}_{1}}\right)\mathbf{=}\mathbf{ln}\left({e}^{\frac{{E}_{a1}-{E}_{a2}}{\mathrm{RT}}}\right)$

$[\mathrm{ln}\left(\frac{{k}_{2}}{{k}_{1}}\right)=\frac{{E}_{a1}-{E}_{a2}}{\overline{)\mathrm{RT}}}]\overline{)\mathbf{RT}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\overline{){{\mathbf{E}}}_{\mathbf{a}\mathbf{1}}{\mathbf{-}}{{\mathbf{E}}}_{\mathbf{a}\mathbf{2}}{\mathbf{=}}{\mathbf{RTln}}\mathbf{\left(}\frac{{\mathbf{k}}_{\mathbf{2}}}{{\mathbf{k}}_{\mathbf{1}}}\mathbf{\right)}}$

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. If an enzyme increases the rate of the hydrolysis reaction by a factor of 1 million, how much lower does the activation barrier have to be when sucrose is in the active site of the enzyme? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical and that the temperature is 298 K.)

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