For the following reaction 2 X + Y {k}{} P 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 Lof 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: 2 = 0.693 ) [JEE Adv 2021 P2]

The rate constant, $k$ , of the reaction is $13.86 \times 10^{-4} \text{ s}^{-1}$ .
Half-life of $X$ is $50 \text{ s}$ .
At $50 \text{ s}, -\;\frac{d[X]}{dt} = 13.86 \times 10^{-3} \text{ mol} \text{ L}^{-1} \text{ s}^{-1}$ .
At $100 \text{ s}, -\;\frac{d[Y]}{dt} = 3.46 \times 10^{-3} \text{ mol} \text{ L}^{-1} \text{ s}^{-1}$ .

Solution:

Rate

= \;\frac{dP}{dt} = k[X]^{l}

2 x + y \rightarrow P

t = 0 \qquad 2 \text{ mol} \qquad 1 \text{ mol}

t = 50 \text{ s} \qquad 1 \text{ mol} \qquad 0.5 \text{ mol} 0.5 \text{ mol}

-\;\frac{1}{2} \;\frac{d[X]}{dt} = \;\frac{d[P]}{dt} = k[X]^{1}

-\;\frac{d[X]}{dt} = 2 k[X]^{1}

2 k = \;\frac{\ln 2}{50} \qquad \Rightarrow \qquad k = \;\frac{\ln 2}{100} = 6.93 \times 10^{-3} \text{ s}^{-1}

t_{1/2} = \;\frac{\ln 2}{2 k} = \;\frac{\ln 2 \times 50}{\ln 2} = 50 \text{ sec}

50 \text{ sec} \;\frac{-d[X]}{dt} = 2 k \times (1)^{1} = \;\frac{\ln 2}{50}

= 13.86 \times 10^{-3} \text{ mol} \text{ L}^{-1} \text{ s}^{-1}

100 \text{ sec} - \;\frac{1}{2} \;\frac{d[X]}{dt} = \;\frac{-d[Y]}{dt}

\Rightarrow -\;\frac{d[Y]}{dt} = \;\frac{\ln 2}{100} \times \;\frac{1}{2} \left\{ \;\frac{-d[Y]}{dt} = k[X]^{1} \right\}

\;\frac{d[Y]}{dt} = 3.46 \times 10^{-3} \text{ mol} \text{ L}^{-1} \text{ s}^{-1}

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