JEE MAIN - Chemistry (2023 - 24th January Morning Shift - No. 19)
The number of correct statement/s from the following is __________
A. Larger the activation energy, smaller is the value of the rate constant.
B. The higher is the activation energy, higher is the value of the temperature coefficient.
C. At lower temperatures, increase in temperature causes more change in the value of k than at higher temperature
D. A plot of $$\mathrm{\ln k}$$ vs $$\frac{1}{T}$$ is a straight line with slope equal to $$-\frac{E_a}{R}$$
Answer
3
Explanation
(A) $k=A e^{-\frac{E_{a}}{R T}}\left(E_{a} \uparrow k \downarrow\right)$
(B) $\ln k=\ln A-\frac{E_{a}}{R T}$
$$ \frac{1}{\mathrm{k}} \cdot \frac{\mathrm{dk}}{\mathrm{dT}}=\frac{+\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}^{2}} $$
$\mathrm{E}_{\mathrm{a}} \uparrow$ temp. coefficient $\uparrow$
(C)
_24th_January_Morning_Shift_en_19_1.png)
Option (C ) is wrong. $\Delta$k may be greater or lesser depending on temperature.
(D) $\ln k=\ln A-\frac{E_{a}}{R T}$
Slope of $\ln k$ vs $\frac{1}{T}$ is $\left(\frac{-E_{a}}{R}\right)$
(B) $\ln k=\ln A-\frac{E_{a}}{R T}$
$$ \frac{1}{\mathrm{k}} \cdot \frac{\mathrm{dk}}{\mathrm{dT}}=\frac{+\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}^{2}} $$
$\mathrm{E}_{\mathrm{a}} \uparrow$ temp. coefficient $\uparrow$
(C)
Option (C ) is wrong. $\Delta$k may be greater or lesser depending on temperature.
(D) $\ln k=\ln A-\frac{E_{a}}{R T}$
Slope of $\ln k$ vs $\frac{1}{T}$ is $\left(\frac{-E_{a}}{R}\right)$
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