JEE MAIN - Chemistry (2022 - 24th June Morning Shift - No. 16)
The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700 - 1000 K. The data has been analysed by plotting ln k vs $${{{{10}^3}} \over T}$$ graph. The value of activation energy for the reaction is ___________ kJ mol$$-$$1. (Nearest integer)
(Given : R = 8.31 J K$$-$$1 mol$$-$$1)
_24th_June_Morning_Shift_en_16_1.png)
Answer
154
Explanation
_24th_June_Morning_Shift_en_16_2.png)
$\ln \mathrm{k}=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}$
$\therefore$ Slope of the graph $=-\frac{E_{a}}{R \times 10^{3}}=-18.5$
$\therefore E_{a}=18.5 \times 8.31 \times 1000 \simeq 154 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Comments (0)
