JEE MAIN - Chemistry (2016 (Offline) - No. 1)
Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases
from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be :
6.93 $$\times$$ 10-4 mol min-1
6.96 $$\times$$ 10-2 mol min-1
1.34 $$\times$$ 10-2 mol min-1
2.66 L minā1 at STP
Explanation
$${H_2}{O_2}\left( {aq} \right) \to {H_2}O\left( {aq} \right) + {1 \over 2}{O_2}\left( g \right)$$
For a first order reaction
$$k = {{2.303} \over t}\log {a \over {\left( {a - x} \right)}}$$
Given $$a = 0.5,\left( {a - x} \right) = 0.125,\,t = 50\,\,$$ min
$$\therefore$$ $$\,\,\,\,\,k = {{2.303} \over {50}}\log {{0.5} \over {0.125}}$$
$$ = 2.78 \times {10^{ - 2}}\,{\min ^{ - 1}}$$
$$r = k\left[ {{H_2}{O_2}} \right] = 2.78 \times {10^{ - 2}} \times 0.05$$
$$ = 1.386 \times {10^{ - 3}}\,\,mol\,{\min ^{ - 1}}$$
Now
$$ - {{d\left[ {{H_2}{O_2}} \right]} \over {dt}} = {{d\left[ {{H_2}O} \right]} \over {dt}} = {{2d\left[ {{O_2}} \right]} \over {dt}}$$
$$\therefore$$ $$\,\,\,\,\,{{2d\left[ {{O_2}} \right]} \over {dt}} - {{d\left[ {{H_2}{O_2}} \right]} \over {dt}}$$
$$\therefore$$ $$\,\,\,\,\,$$ $${{d\left[ {{O_2}} \right]} \over {dt}} = {1 \over 2} \times {{d\left[ {{H_2}{O_2}} \right]} \over {dt}}$$
$$ = {{1.386 \times {{10}^{ - 3}}} \over 2} = 6.93 \times {10^{ - 4}}\,mol\,{\min ^{ - 1}}$$
For a first order reaction
$$k = {{2.303} \over t}\log {a \over {\left( {a - x} \right)}}$$
Given $$a = 0.5,\left( {a - x} \right) = 0.125,\,t = 50\,\,$$ min
$$\therefore$$ $$\,\,\,\,\,k = {{2.303} \over {50}}\log {{0.5} \over {0.125}}$$
$$ = 2.78 \times {10^{ - 2}}\,{\min ^{ - 1}}$$
$$r = k\left[ {{H_2}{O_2}} \right] = 2.78 \times {10^{ - 2}} \times 0.05$$
$$ = 1.386 \times {10^{ - 3}}\,\,mol\,{\min ^{ - 1}}$$
Now
$$ - {{d\left[ {{H_2}{O_2}} \right]} \over {dt}} = {{d\left[ {{H_2}O} \right]} \over {dt}} = {{2d\left[ {{O_2}} \right]} \over {dt}}$$
$$\therefore$$ $$\,\,\,\,\,{{2d\left[ {{O_2}} \right]} \over {dt}} - {{d\left[ {{H_2}{O_2}} \right]} \over {dt}}$$
$$\therefore$$ $$\,\,\,\,\,$$ $${{d\left[ {{O_2}} \right]} \over {dt}} = {1 \over 2} \times {{d\left[ {{H_2}{O_2}} \right]} \over {dt}}$$
$$ = {{1.386 \times {{10}^{ - 3}}} \over 2} = 6.93 \times {10^{ - 4}}\,mol\,{\min ^{ - 1}}$$
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