For the following reaction 2 X + Y → k P the rate of reaction is d [ P ] dt = k [ X ] . Two moles of X…

For the following reaction 2X+Y→kP the rate of reaction is d[P]dt=k[X]. Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are) (Use: ln 2=0.693)
  1. The rate constant, k, of the reaction is 13.86×10-4 s-1.
  2. Half-life of X is 50 s.
  3. At 50 s, -d[x]dt=13.86×10-3 mol L-1 s-1.
  4. At 100 s,-dYdt=3.46×10-3 mol L-1 s-1

Solution

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

Asked in: JEE Advanced 2021 (Paper 2)

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