In presence of a catalyst, the activation energy is lowered by 2 kcal at 127° C. The rate as compared to old rate

Question

In presence of a catalyst, the activation energy is lowered by 2 kcal at 127° C. The rate as compared to old rate (A) e 2 times (B) e -2 times (C) e 2.5 times (D) e -4.5 times

Tardigrade Admin · Accepted Answer

K 1= Ae - E a1 / RT , K 2= Ae - E a2 / RT .(K1/K2)=e(-Ea1-Ea2) / R T=e-[(2 × 103) /(2 × 400).] (K1/K2)=e-2.5=(r1/r2) ∴ ( r 1/ r 2)= e 2.5

Q. In presence of a catalyst, the activation energy is lowered by 2 kcal at $12 7^{\circ} C$ . The rate as compared to old rate

$e^{2}$ times

$e^{- 2}$ times

$e^{2.5}$ times

$e^{- 4.5}$ times

$K_{1} = A e^{- E_{a_{1}} / RT}, K_{2} = A e^{- E_{a_{2}} / RT}$

$\frac{K _{1}}{K _{2}} = e^{(- E_{a_{1}} - E_{a_{2}}) / RT} = e^{- [(2 \times 1 0^{3}) / (2 \times 400)}]$

$\frac{K _{1}}{K _{2}} = e^{- 2.5} = \frac{r _{1}}{r _{2}}$

$∴ \frac{r _{1}}{r _{2}} = e^{2.5}$

Q. In presence of a catalyst, the activation energy is lowered by 2 kcal at 127∘C. The rate as compared to old rate

e2 times

e−2 times

e2.5 times

e−4.5 times