Chemical Kinetics - Result Question 16

####19. Two reactions $R _1$ and $R _2$ have identical pre- exponential factors. Activation energy of $R _1$ exceeds that of $R _2$ by $10 kJ$ $mol^{-1}$. If $k _1$ and $k _2$ are rate constants for reactions $R _1$ and $R _2$, respectively at $300 K$, then $\ln \left(\frac{k _2}{k _1}\right)$ is equal to $\left(R=8.314 J mol^{-1} K^{-1}\right)$

(2017 Main)

(a) 8

(b) 12

(c) 6

(d) 4

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Answer:

Correct Answer: 19. (c)

Solution:

  1. According to Arrhenius equation

$$ k=A e^{-E _a / R T} $$

where, $A=$ collision number or pre-exponential factor.

$R=$ gas constant

$T=$ absolute temperature

$E _a=$ energy of activation

For reaction $R _1, k _1=A e^{-E _{a _1} / R T}$

For reaction $R _2, k _2=A e^{-E _{a _2} / R T}$

On dividing Eq. (ii) by Eq. (i), we get

$$ \frac{k _2}{k _1}=e^{-\frac{\left(E _{a _2}-E _{a _1}\right)}{R T}} $$

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Taking $\ln$ on both the sides of Eq. (iii), we get

$$ \ln \left(\frac{k _2}{k _1}\right)=\frac{E _{a _1}-E _{a _2}}{R T} $$

Given, $\quad E _{a _1}=E _{a _2}+10 kJ mol^{-1}=E _{a _2}+10,000 J mol^{-1}$

$$ \therefore \quad \ln \frac{k _2}{k _1}=\frac{10,000 J mol^{-1}}{8.314 J mol^{-1} K^{-1} \times 300 K}=4 $$



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