1. Use Arrhenius equation

For a reaction, $k = A e^{-E_a/(RT)}$

With and without enzyme, assuming the pre-exponential factor does not change significantly, $\frac{k_{\text{enzyme}}}{k_{\text{without}}} = e^{-(E_{a,\text{enzyme}}-E_{a,\text{without}})/(RT)}$

2. Given rate enhancement

The reaction occurs $10^6$ times faster with enzyme, so $\frac{k_{\text{enzyme}}}{k_{\text{without}}} = 10^6$

Thus, $10^6 = e^{-\Delta E_a/(RT)}$ where $\Delta E_a = E_{a,\text{enzyme}} - E_{a,\text{without}}$

3. Take natural logarithm

$\ln(10^6) = -\frac{\Delta E_a}{RT}$

Since $\ln(10^6) = 6\ln 10 = 6(2.303)$

we get $-\frac{\Delta E_a}{RT} = 6(2.303)$

So, $\Delta E_a = -6(2.303)RT$

4. Interpretation

The negative sign means the activation energy decreases on adding enzyme.

5. Check options

• A: $+6RT$ ❌
• B: $-6(2.303)RT$ ✅
• C: $-6RT$ ❌
• D: $+6(2.303)RT$ ❌

Therefore, the correct option is B.