JEE MAIN - Chemistry (2021 - 16th March Morning Shift - No. 22)
The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0 $$\times$$ 10$$-$$3 s$$-$$1 and the activation energy Ea = 11.488 kJ mol$$-$$1, the rate constant at 200 K is ____________ $$\times$$ 10$$-$$5 s$$-$$1. (Round off to the Nearest Integer). (Given : R = 8.314 J mol$$-$$1 K$$-$$1)
Answer
10
Explanation
$$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$$
k1 (at 200 K) = ?
k2 (at 300 K) = $$1 \times {10^{ - 3}}{s^{ - 1}}$$
$$\log {{1 \times {{10}^{ - 3}}} \over {{k_1}}} = {{11.488 \times {{10}^3}} \over {2.303 \times 8.314}}\left[ {{1 \over {600}}} \right] = 1$$
$$ \Rightarrow $$ $${{1 \times {{10}^{ - 3}}} \over {{k_1}}} = 10$$
$$ \Rightarrow $$ $${k_1} = 10 \times {10^{ - 5}}{s^{ - 1}}$$
k1 (at 200 K) = ?
k2 (at 300 K) = $$1 \times {10^{ - 3}}{s^{ - 1}}$$
$$\log {{1 \times {{10}^{ - 3}}} \over {{k_1}}} = {{11.488 \times {{10}^3}} \over {2.303 \times 8.314}}\left[ {{1 \over {600}}} \right] = 1$$
$$ \Rightarrow $$ $${{1 \times {{10}^{ - 3}}} \over {{k_1}}} = 10$$
$$ \Rightarrow $$ $${k_1} = 10 \times {10^{ - 5}}{s^{ - 1}}$$
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