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JEE Advanced 2021Paper 2
Exam
JEE Advanced
Year
2021
Paper
Paper 2
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Advanced 2021 · Paper 2 · Chemistry · Chemical Kinetics

For the following reaction 2X + Y k → P the rate of reaction is (d[...

For the following reaction 2X+Y→kP2X + Y\overset{k}{\to}P the rate of reaction is d[P]dt=k[X]\dfrac{d[ P ]}{dt} = k[ X ]. Two moles of XX are mixed with one mole of YY to make 1.0L1 . 0L of solution. At 50s,0.550s, 0 . 5 mole of YY is left in the reaction mixture. The correct statement(s) about the reaction is(are) ((Use: ln⁡2=0.693)\ln 2 = 0 . 693 )

Options

  1. AThe rate constant, kk, of the reaction is 13.86×10−4s−113 . 86 \times {10}^{- 4}{s}^{- 1}.
  2. BHalf-life of XX is 50s50s.
  3. CAt 50s,−d[x]dt=13.86×10−3molL−1s−150s, - \dfrac{d[ x ]}{dt} = 13 . 86 \times {10}^{- 3}mol{L}^{- 1}{s}^{- 1}.
  4. DAt 100s,−d[Y]dt=3.46×10−3molL−1s−1100s, - \dfrac{d\left[Y\right]}{dt} = 3 . 46 \times {10}^{- 3}mol{L}^{- 1}{s}^{- 1}

Correct answer

B. Half-life of XX is 50s50s.

Step-by-step solution

(2X+y⟶Pt=0210t=502−2×0.51−0.5 1mole 0.5\begin{array}{lllll} & 2 \mathrm{X}+ & \mathrm{y} & \longrightarrow & \mathrm{P} \\ \mathrm{t}=0 & 2 & 1 & & 0 \\ \mathrm{t}=50 & 2-2 \times 0.5 & 1-0.5 & & \\ & \text{ 1mole } & 0.5 & & \end{array}) rate (=-\frac{1}{2} \frac{d x}{d t}=-\frac{d y}{d t}=\frac{d P}{d t}=K[X]) (-\frac{1}{2} \frac{d x}{d t}=K[X]) (-\frac{d x}{d t}=2 K[X]=K^1[X]) Half life is (t=50 \mathrm{sec}) (2 K=\frac{0.653 L}{50}) (K=\frac{0.6932}{100}=6.332 \times 10^{-3}) (t=50 \mathrm{sec}) (-\frac{d x}{d t}=2 K[X]) (-\frac{d x}{d t}=2 \times 6.332 \times 10^{-3} \times 1) (=13.864 \times 10^{-3} \mathrm{ mole} / \mathrm{L} / \mathrm{Sec}) (-\frac{d y}{d t}=K[X]=6.332 \times 10^{-3}\left(\frac{1}{2}\right)) (=3.46 \times 10^{-3} \mathrm{ mole} / \mathrm{L} / \mathrm{Sec}^{-1})
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