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AIIMS 2016
Exam
AIIMS
Year
2016
Subject
Chemistry
Chapter
Chemical Kinetics

AIIMS 2016 · Chemistry · Chemical Kinetics

During the decomposition of H_2 O_2 to give oxygen, 48 ~g O_2 is...

During the decomposition of H2O2\mathrm{H}_2 \mathrm{O}_2 to give oxygen, 48 gO248 \mathrm{~g} \mathrm{O}_2 is formed per minute at a certain point of time. The rate of formation of water at this point is

Options

  1. A0.75 mol min−10.75 \mathrm{~mol} \mathrm{~min}^{-1}
  2. B1.5 mol min−11.5 \mathrm{~mol} \mathrm{~min}^{-1}
  3. C2.25 mol min−12.25 \mathrm{~mol} \mathrm{~min}^{-1}
  4. D3.0 mol min−13.0 \mathrm{~mol} \mathrm{~min}^{-1}

Correct answer

D. 3.0 mol min−13.0 \mathrm{~mol} \mathrm{~min}^{-1}

Step-by-step solution

2H2O2⟶2H2O+O22 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 Rate =−12d[H2O2]dt=12d[H2O]dt=d[O2]dt=-\frac{1}{2} \frac{d\left[\mathrm{H}_2 \mathrm{O}_2\right]}{d t}=\frac{1}{2} \frac{d\left[\mathrm{H}_2 \mathrm{O}\right]}{d t}=\frac{d\left[\mathrm{O}_2\right]}{d t} Rate of formation of oxygen =48 g min−1=48 \mathrm{~g} \mathrm{~min}^{-1}=4832 mol min−1=1.5 mol min−1∴ Rate of formation of H2O⇒12d[H2O]dt=d[O2]dtd[H2O]dt=2×d[O2]dt=2×1.5=3 mol min\begin{aligned} & =\frac{48}{32} \mathrm{~mol} \mathrm{~min}^{-1}=1.5 \mathrm{~mol} \mathrm{~min}^{-1} \\ & \therefore \quad \text{ Rate of formation of } \mathrm{H}_2 \mathrm{O} \\ & \Rightarrow \quad \frac{1}{2} \frac{d\left[\mathrm{H}_2 \mathrm{O}\right]}{d t}=\frac{d\left[\mathrm{O}_2\right]}{d t} \\ & \frac{d\left[\mathrm{H}_2 \mathrm{O}\right]}{d t}=2 \times \frac{d\left[\mathrm{O}_2\right]}{d t}=2 \times 1.5=3 \mathrm{~mol} \mathrm{~min}\end{aligned}$$
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