For a general reaction A → 2 B + C ; Δ H = − 15 Kcal mol − 1 . Energy of activation of backward reaction is 20 kcal mol − 1 . In the presence of catalyst the energy of activation of forward reaction i

Question

For a general reaction A → 2 B + C ; Δ H = − 15 Kcal mol − 1 . Energy of activation of backward reaction is 20 kcal mol − 1 . In the presence of catalyst the energy of activation of forward reaction is 3 kcal mol − 1 . At 12 7 0 C the catalyst causes the rate of the reaction to increase by the number of times equal to e 2 x ​ , the value of ' x ' is ?

TestHub · Accepted Answer

Question Explanation: Calculate the value of $x$ in $e^{x /2}$ .

Concept:

This question is based on activation energy in the presence of a catalyst.

Solution:

Given: $T = 127℃ = 400 K$

,

R = 2 c a l / m o l = 0.002 Kc a l / m o l K

.

$Δ H = E_{a f} - E_{ab}$

$- 15 = E_{a f} - 20$

$E_{a f} = 5 Kc a l / m o l$

The activation energy for the catalyzed reaction is $E_{a f}^{'} = 3 Kc a l / m o l$ .

The Arrhenius equation is:

$K = A e^{- E_{a f} / RT}$

For the catalyzed reaction:

$K_{c a t} = A e^{- E_{a f}^{'} / RT}$

The ratio of the rate constants is:

$\frac{K _{c a t}}{K} = \frac{A e ^{- E_{a f}^{'} / RT}}{A e ^{- E_{a f} / RT}}$

$\frac{K _{c a t}}{K} = e^{(E_{a f} - E_{a f}^{'}) / RT}$

Substitute the given values:

$\frac{K _{c a t}}{K} = e^{(5 - 3) / (0.002 \times 400)}$

$\frac{K _{c a t}}{K} = e^{2/0.8}$

$\frac{K _{c a t}}{K} = e^{20/8}$

$\frac{K _{c a t}}{K} = e^{5/2}$

Since the rate of reaction is proportional to $K$ :

$\frac{r _{c a t}}{r} = e^{5/2}$

We are given that $\frac{r _{c a t}}{r} = e^{x /2}$ .

Therefore, by comparing the exponents, we get:

$x = 5$

Final Answer: The correct answer is 5.

Chemistry - Chemical Kinetics Question with Solution | TestHub

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