JEE Advance - Chemistry (1995 - No. 1)
At 380oC, the half-life period for the first order decomposition of H2O2 is 360 min. Calculate the time required for 75% decomposition at 450oC.
10.17 min
20.34 min
30.51 min
40.68 min
5.08 min
Explanation
For the first order reaction half-life = $${{0.693} \over k}$$
or, $$360 = {{0.693} \over k}$$ $$\therefore$$ $${k_{380^\circ C}} = {{0.693} \over {360}}$$
We know, $$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$$
$$\therefore$$ $$\log {{{k_{450^\circ C}}} \over {{{0.693} \over {360}}}} = {{200 \times 1000} \over {2.303 \times 8.314}}\left[ {{1 \over {653}} - {1 \over {723}}} \right]$$
or, $${k_{450^\circ C}} = 6.18 \times {10^{ - 2}}$$ min$$-$$1
For 75% decomposition at 723K (450$$^\circ$$C),
$${k_{450^\circ C}} = {{2.303} \over t}\log {a \over {a - x}}$$
or, $$6.81 \times {10^{ - 2}} = {{2.303} \over t}\log {{100} \over {(100 - 75)}}$$ or, t = 20.358 min
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