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 $\ln \left(\dfrac{k _2}{k _1}\right)$ is equal to
$(R=8.314$ $ J $ $mol^{-1} K^{-1})$
(2017 Main)
(a) $8$
(b) $12$
(c) $6$
(d) $4$
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Answer:
Correct Answer: 19. (d)
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}$$\quad$ …….. (i)
For reaction $R _2, k _2=A e^{-E _{a _2} / R T}$$\quad$ …….. (ii)
On dividing Eq. (ii) by Eq. (i), we get
$ \dfrac{k _2}{k _1}=e^{-\dfrac{\left(E _{a _2}-E _{a _1}\right)}{R T}} $$\quad$ …….. (iii)
Taking $\ln$ on both the sides of Eq. (iii), we get
$ \ln \left(\dfrac{k _2}{k _1}\right)=\dfrac{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 \dfrac{k _2}{k _1}=\dfrac{10,000 J mol^{-1}}{8.314 J mol^{-1} K^{-1} \times 300 K}=4 $
BETA
