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QJEEM16Ch7e7c: JEE Main 2016 Chemistry - Class 12 JEEM16Ch7e7c Solution: Decomposition of H 2 O 2 follows a first order reaction. In…

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 :

Multiple Choice Options

  • A 6.93 ×\times 10-4 mol min-1
  • B 6.96 ×\times 10-2 mol min-1
  • C 1.34 ×\times 10-2 mol min-1
  • D 2.66 L min–1 at STP

Step-by-Step Verified Solution

H2O2(aq)→H2O(aq)+12O2(g){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=2.303tlog⁡a(a−x)k = {{2.303} \over t}\log {a \over {\left( {a - x} \right)}}

Given a=0.5,(a−x)=0.125, t=50  a = 0.5,\left( {a - x} \right) = 0.125,\,t = 50\,\, min

∴\therefore      k=2.30350log⁡0.50.125\,\,\,\,\,k = {{2.303} \over {50}}\log {{0.5} \over {0.125}}

=2.78×10−2 min⁡−1= 2.78 \times {10^{ - 2}}\,{\min ^{ - 1}}

r=k[H2O2]=2.78×10−2×0.05r = k\left[ {{H_2}{O_2}} \right] = 2.78 \times {10^{ - 2}} \times 0.05

=1.386×10−3  mol min⁡−1= 1.386 \times {10^{ - 3}}\,\,mol\,{\min ^{ - 1}}

Now

−d[H2O2]dt=d[H2O]dt=2d[O2]dt- {{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[O2]dt−d[H2O2]dt\,\,\,\,\,{{2d\left[ {{O_2}} \right]} \over {dt}} - {{d\left[ {{H_2}{O_2}} \right]} \over {dt}}

∴\therefore      \,\,\,\,\, d[O2]dt=12×d[H2O2]dt{{d\left[ {{O_2}} \right]} \over {dt}} = {1 \over 2} \times {{d\left[ {{H_2}{O_2}} \right]} \over {dt}}

=1.386×10−32=6.93×10−4 mol min⁡−1= {{1.386 \times {{10}^{ - 3}}} \over 2} = 6.93 \times {10^{ - 4}}\,mol\,{\min ^{ - 1}}

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