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Chemistry · Chemical Kinetics

JEE Advanced 2021 — Paper 2 — Question 11

For the following reaction 2X+Y→kP2 \mathrm{X}+\mathrm{Y} \xrightarrow{\mathrm{k}} \mathrm{P} the rate of reaction is d[P]dt=k[X]\frac{\mathrm{d}[\mathrm{P}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{X}]. Two moles of X\mathbf{X} are mixed with one mole of Y\mathbf{Y} to make 1.0 L of solution. At 50 s,0.550 \mathrm{~s}, 0.5 mole of Y\mathbf{Y} is left in the reaction mixture. The correct statement(s) about the reaction is(are) (Use: ln⁡2=0.693\ln 2=0.693 )

  1. Option A:

    The rate constant, k , of the reaction is 13.86×10−4 s−113.86 \times 10^{-4} \mathrm{~s}^{-1}

  2. Option B:

    Half-life of X is 50 s

    Correct
  3. Option C:

    At 50 s,−d[X]dt=13.86×10−3 mol L−1 s−150 \mathrm{~s},-\frac{\mathrm{d}[\mathrm{X}]}{\mathrm{dt}}=13.86 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}.

    Correct
  4. Option D:

    At 100 s,−d[Y]dt=3.46×10−3 mol L−1 s−1100 \mathrm{~s},-\frac{\mathrm{d}[\mathrm{Y}]}{\mathrm{dt}}=3.46 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}.

    Correct

Answer: B, C, D

Step-by-step solution

2X+yt=01Pt=502−2×0.51−0.51 mole 0.5 rate =−12dxdt=−dydt=dPdt=K[X]−12dxdt=K[X]−dxdt=2 K[X]=K1[X]\begin{aligned} & \quad \begin{array}{lll} 2 \mathrm{X} & + & \mathrm{y} \mathrm{t} & =0 \quad 1 \\& \mathrm{P} \mathrm{t} \\& =50 \quad 2-2 \times 0.5 & 1-0.5 1 \text { mole } \\& 0.5 \end{array} \\& \text { rate }=-\frac{1}{2} \frac{\mathrm{dx}}{\mathrm{dt}}=-\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{K}[\mathrm{X}] \\& -\frac{1}{2} \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{K}[\mathrm{X}] \\& -\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{~K}[\mathrm{X}]=\mathrm{K}^{1}[\mathrm{X}] \end{aligned} Half life is t=50t=50 sec 2 K=0.653 L502 \mathrm{~K}=\frac{0.653 \mathrm{~L}}{50}

K=0.6932100=6.332×10−3K=\frac{0.6932}{100}=6.332 \times 10^{-3}

t=50 sec\mathbf{t}=\mathbf{5 0} \mathbf{~ s e c}

−dxdt=2 K[X]-\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{~K}[\mathrm{X}]

−dxdt=2×6.332×10−3×1=13.864×10−3 mole/L/Sec-\frac{\mathrm{dx}}{\mathrm{dt}}=2 \times 6.332 \times 10^{-3} \times 1=13.864 \times 10^{-3} \mathrm{~mole} / \mathrm{L} / \mathrm{Sec}

−dydt=K[X]=6.332×10−3(12)=3.46×10−3 mole/L/Sec−1-\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{K}[\mathrm{X}]=6.332 \times 10^{-3}\left(\frac{1}{2}\right)=3.46 \times 10^{-3} \mathrm{~mole} / \mathrm{L} / \mathrm{Sec}^{-1}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2021
Paper
Paper 2
Subject
Chemistry
Chapter
Chemical Kinetics
Topic
Integrated Rate Laws