Question

For the following reaction $2X+Y\stackrel{k}{\to }P$ the rate of reaction is $\frac{d\left[P\right]}{dt}=k\left[X\right]$. Two moles of $X$ are mixed with one mole of $Y$ to make $1.0L$ of solution. At $50s,0.5\text{mole}$ of $Y$ is left in the reaction mixture. The correct statement(s) about the reaction is(are)
(Use: $ln2=0.693$)

• A. The rate constant, $k$, of the reaction is $13.86\times {10}^{-4}{s}^{-1}$
• B. Half-life of $X$ is $50s$
• C. At $50s$, $-\frac{d\left[X\right]}{dt}=13.86\times {10}^{-4}molL{s}^{-1}$
• D. At $100s$, $-\frac{d\left[Y\right]}{dt}=3.46\times {10}^{-3}molL{s}^{-1}$

Answer 1

Answer: (B), (C), (D)

At $100s$, $-\frac{d\left[Y\right]}{dt}=3.46\times {10}^{-3}molL{s}^{-1}$

$\begin{array}{ccccc}& 2X+& Y& \stackrel{k}{\to }& P\\ \text{t=0}& 2& 1& & 0\\ \text{t=50}& 1& 0.5& & 1\end{array}$

$2X+Y\underset{}{\overset{}{\to }}Pt=0210t=502-2\times 0.51-0.5$
$rate=-\frac{1}{2}\frac{d\left[X\right]}{dt}=-\frac{d\left[Y\right]}{dt}=\frac{dP}{dt}=K\left[X\right]-\frac{1}{2}\frac{d\left[X\right]}{dt}=k\left[X\right]\Rightarrow -\frac{d\left[X\right]}{dt}=2k\left[X\right]=k^{\prime }\left[X\right]$
Half life is $t=50s$
$2k=\frac{0.693L}{50}\text{Rate Constant},\left(k\right)=\frac{0.6932}{100}=6.932\times {10}^{-3}{t}_{\frac{1}{2}}=50s$
$-\frac{d\left[X\right]}{dt}=2k\left[X\right]\text{At 50s}\to -\frac{d\left[X\right]}{dt}=2\times 6.932\times {10}^{-3}\times 1=13.864\times {10}^{-3}moleL{s}^{-1}\text{At 50s}\to -\frac{d\left[Y\right]}{dt}=k\left[X\right]=6.932\times {10}^{-3}molL{s}^{-1}$
So, $\text{At 100 s}\to -\frac{d\left[Y\right]}{dt}=\frac{6.932\times {10}^{-3}}{2}=3.46\times {10}^{-3}molL{s}^{-1}$
Thus the correct options are B, C, D.