The rate of a certain biochemical reaction at physiological temperature (T) occurs 10⁶ times faster with enzyme than without. The change in the activation energy upon adding enzyme is :

Solution

Chemistry ma’am · 4 steps

Same A, same T: the 10^6 speed-up comes entirely from the exponential, so take ln of the ratio.

1. Write both rates. Without enzyme k₁ = A e^(-E₁/RT), and with enzyme k₂ = A e^(-E₂/RT). The temperature is the same, and A is taken as unchanged.
2. Divide them. A cancels in the ratio, and k₂/k₁ = 10⁶.

   k₂/k₁ = e^((E₁ - E₂)/RT) = 10⁶

3. Take ln. (E₁ - E₂)/RT = ln 10⁶ = 6 ln 10 = 6 × 2.303. So E₁ - E₂ = 6(2.303)RT.
4. So. The change on adding the enzyme is Δ Eₐ = E₂ - E₁ = -6(2.303)RT. It is negative because a catalyst lowers the activation energy, so the answer is (b).

The idea to keep. A catalyst that speeds a reaction n-fold at fixed T lowers Eₐ by RT ln n. Turn ln into 2.303 log whenever the factor is a power of 10.

Answer from the NTA final answer key. Solution by PrepWiser.