Chemical Kinetics - Result Question 27
27. For a first order reaction, $A \rightarrow P$, the temperature ( $T$ ) dependent rate constant $(k)$ was found to follow the equation :
$ \log k=\dfrac{2000}{T}+6.0 $
the pre-exponential factor $A$ and the activation energy $E _a$, respectively, are
(2009)
(a) $1.0 \times 10^{6} s^{-1}$ and $9.2 kJ mol^{-1}$
(b) $6.0 s^{-1}$ and $16.6 kJ mol^{-1}$
(c) $1.0 \times 10^{6} s^{-1}$ and $16.6 kJ mol^{-1}$
(d) $1.0 \times 10^{6} s^{-1}$ and $38.3 kJ mol^{-1}$
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Answer:
Correct Answer: 27. (d)
Solution:
- The logarithmic form of Arrhenius equation is $\log k=\log A-\dfrac{E _a}{2.303 R T}$
Given : $\quad \log k=6-\dfrac{2000}{T}$
Comparing the above two equations :
$ \begin{aligned} & \log A=6 \Rightarrow A=10^{6} \\ & \text { and } \quad \dfrac{E _a}{2.303 R}=2000 \\ & \Rightarrow \quad E _a=2000 \times 2.303 \times 8.314 J \\ & =38.3 kJ \quad mol^{-1} \end{aligned} $
BETA
