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Step-by-Step Solution
Step 1: Identify the Relevant Relationship (Arrhenius Equation)
For a reaction, the rate constant $k$ is given by the Arrhenius equation:
$k = A \, e^{-\frac{E_a}{RT}}$
where:
• $E_a$ is the activation energy,
• $R$ is the gas constant,
• $T$ is the temperature in Kelvin,
• $A$ is the frequency factor.
Step 2: Write the Rate Constants With and Without Enzyme
Let:
• $k$ be the rate constant without the enzyme,
• $k'$ be the rate constant with the enzyme.
Then:
$k = A \, e^{-\frac{E_a}{RT}}$
$k' = A \, e^{-\frac{E'_a}{RT}}$
where $E_a$ is the activation energy without enzyme and $E'_a$ is the activation energy with enzyme.
Step 3: Form the Ratio of Rate Constants
We are told that the reaction is $10^6$ times faster with the enzyme. Thus:
$\frac{k'}{k} = 10^6$
Substitute the expressions for $k'$ and $k$:
$\frac{k'}{k} = \frac{A \, e^{-\frac{E'_a}{RT}}}{A \, e^{-\frac{E_a}{RT}}} = e^{-\frac{E'_a - E_a}{RT}}$
Therefore:
$e^{-\frac{E'_a - E_a}{RT}} = 10^6$
Step 4: Take Natural Logarithms
Take the natural logarithm on both sides:
$-\frac{E'_a - E_a}{RT} = \ln(10^6)$
Recall that $\ln(10^x) = x \ln(10)$, hence:
$\ln(10^6) = 6 \ln(10)$
So:
$-\frac{E'_a - E_a}{RT} = 6 \ln(10)$
Step 5: Rearrange to Find the Change in Activation Energy
$\quad E'_a - E_a = -6 \cdot R \, T \cdot \ln(10)$
And since $\ln(10) \approx 2.303$:
$E'_a - E_a = -6 \, R \, T \, (2.303)$
Thus, the change in activation energy upon adding the enzyme is:
$\boxed{-6 \,(2.303)\, R \, T}$
