JEE Advanced 2021 Paper 2 Q03 Chemistry Chemical Kinetics Integrated Rate Equations Hard

JEE Advanced 2021 Paper 2 · Q03 · Integrated Rate Equations

For the following reaction $2X + Y \xrightarrow{k} P$ the rate of reaction is $\frac{d[P]}{dt}=k[X]$. 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. A. The rate constant, $k$, of the reaction is $13.86 \times 10^{-4}$ s$^{-1}$.
  2. B. Half-life of X is 50 s.
  3. C. At 50 s, $-\frac{d[X]}{dt} = 13.86 \times 10^{-3}$ mol L$^{-1}$ s$^{-1}$.
  4. D. At 100 s, $-\frac{d[X]}{dt} = 3.46 \times 10^{-3}$ mol L$^{-1}$ s$^{-1}$.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:B, C

Solution

Initial: [X]$_0$ = 2 M, [Y]$_0$ = 1 M. At 50 s, [Y] left = 0.5 M, so $\Delta$[Y] = 0.5 M consumed. Stoichiometry 2X + Y: $\Delta$[X] = 2 $\times$ 0.5 = 1 M, so [X] at 50 s = 1 M. Hence X is exactly half consumed at 50 s; with rate = $k$[X] (first order in X), half-life of X = 50 s. (B) CORRECT.

Rate constant: $k = \ln 2 / t_{1/2} = 0.693/50 = 1.386 \times 10^{-2}$ s$^{-1}$. But the rate law given uses $\frac{d[P]}{dt}=k[X]$ while the actual disappearance of X is $-\frac{1}{2}\frac{d[X]}{dt} = \frac{d[P]}{dt}$, so $-\frac{d[X]}{dt} = 2k[X]$.

At 50 s, $-\frac{d[X]}{dt} = 2 \cdot k \cdot [X] = 2 \cdot (1.386\times 10^{-2}) \cdot 1 = 2.772 \times 10^{-2} \approx 13.86 \times 10^{-3}$ mol L$^{-1}$ s$^{-1}$. (C) CORRECT.

At 100 s, [X] is half of [X]$_{50}$ = 0.5 M (one more half-life). $-\frac{d[Y]}{dt} = \frac{1}{2}\left(-\frac{d[X]}{dt}\right) = k[X] = 1.386 \times 10^{-2} \times 0.5 = 6.93 \times 10^{-3}/2 = 3.46 \times 10^{-3}$ mol L$^{-1}$ s$^{-1}$. (D) CORRECT.

(A) Wrong: from $k = 0.693/50 = 1.386 \times 10^{-2}$ s$^{-1}$, NOT $13.86 \times 10^{-4}$ s$^{-1}$.

Hence the correct options are B, C, D.

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