Q68 / 180 · NEET 2019 (Odisha) · Chemistry
For a reaction, activation energy $\mathrm{E}_{\mathrm{a}}=0$ 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
[Given that gas constant $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ ]
- A$3.2 \times 10^4 \mathrm{~s}^{-1}$
- Correct answerB$1.6 \times 10^6 \mathrm{~s}^{-1}$
- C$1.6 \times 10^3 \mathrm{~s}^{-1}$
- D$3.2 \times 10^6 \mathrm{~s}^{-1}$
Solution
Key Idea Rate constants at two different temperatures is given by Arrhenius equation as follows:
$$
\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right], T_2>T_1
$$
Given, $\mathrm{E}_{\mathrm{a}}=0, \mathrm{~T}_1=200 \mathrm{~K}, \mathrm{k}_1=1.6 \times 10^6 \mathrm{~s}^{-1}$
$$
\mathrm{T}_2=400 \mathrm{~K}, \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
$$
According to Arrhenius equation,
$$
\log \frac{\mathrm{k}_2}{\mathrm{k}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]
$$
On substituting the given values in above equation, we get
$$
\begin{aligned}
& \log \frac{\mathrm{k}_2}{1.6 \times 10^6}=\frac{0}{2.303 \times 8.314}\left[\frac{\mathrm{l}}{200}-\frac{\mathrm{l}}{400}\right] \\
& \log \left(\frac{\mathrm{k}_2}{1.6 \times 10^6}\right)=0 \\
& \frac{\mathrm{k}_2}{1.6 \times 10^6}=10^0=1 \\
& \mathrm{k}_2=1.6 \times 10^6 \mathrm{~s}^{-1} \text { at } 400 \mathrm{~K}
\end{aligned}
$$
Thus, option (2) is correct.
Hence, the correct answer is option: 'B'.