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:
- 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
$$ \frac{k _2}{k _1}=e^{-\frac{\left(E _{a _2}-E _{a _1}\right)}{R T}} $$
reactions]
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, $\quad E _{a _1}=E _{a _2}+10 kJ mol^{-1}=E _{a _2}+10,000 J mol^{-1}$
$$ \therefore \quad \ln \frac{k _2}{k _1}=\frac{10,000 J mol^{-1}}{8.314 J mol^{-1} K^{-1} \times 300 K}=4 $$
