Chemical Kinetics Ques 41
41. For a reaction, activation energy $E_a=0$ and the rate constant at $200 \mathrm{~K}$ is $1.6 \times 10^6 \mathrm{~s}^{-1}$. The rate constant at $400 \mathrm{~K}$ will be
[NEET Odisha 2019]
[Given that gas constant, $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]
(a) $3.2 \times 10^6 \mathrm{~s}^{-1}$
(b) $3.2 \times 10^4 \mathrm{~s}^{-1}$
(c) $1.6 \times 10^6 \mathrm{~s}^{-1}$
(d) $1.6 \times 10^3 \mathrm{~s}^{-1}$
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Answer:
Correct Answer: 41.(c)
Solution: (c) From Arrhenius equation
$ \log \frac{k_{400}}{k_{200}}=\frac{E_a}{2.303 \mathrm{R}}\left[\frac{T_2-T_1}{T_1 T_2}\right] $
Since, given that $E_a=0$
$ \begin{aligned} & \therefore \log \frac{k_{400}}{k_{200}}=0 \\ & \Rightarrow \frac{k_{400}}{k_{200}}=1 \end{aligned} $
So, $k_{400}=k_{200}$
So rate constant at $400, k=1.6 \times 10^6 \mathrm{~s}^{-1}$
BETA
