JEE Main 2016 · 3 Apr · Shift 1 · Chemistry Q39

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ChemistrySingle correct+4 / -1Question 39 of 90

Decomposition of $\ce{H2O2}$ follows a first order reaction. In fifty minutes the concentration of $\ce{H2O2}$ decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of $\ce{H2O2}$ reaches 0.05 M, the rate of formation of $\ce{O2}$ will be :

6.93\times10^{-2}

mol min

^{-1}

6.93\times10^{-4}

mol min

^{-1}

2.66 L min

^{-1}

at STP

1.34\times10^{-2}

mol min

^{-1}

Answer and worked solution

Official answer: B

0.5 → 0.125 M is two half-lives in 50 min, so $t_{1/2}=25$ min and $k=\frac{0.693}{25}=0.0277$ min⁻¹. At 0.05 M, $\ce{H2O2}$ is used at $0.0277\times0.05=1.39\times10^{-3}$ mol L⁻¹ min⁻¹. Since $\ce{2H2O2 -> 2H2O + O2}$ , oxygen forms at half that: $6.93\times10^{-4}$ mol min⁻¹ (per litre).

Topics:Integrated rate equations Half-life

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JEE Main 2016 · 3 Apr · Shift 1, Chemistry question 39