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

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