Question

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

• 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 $50s,-\frac{d[X]}{dt}=13.86\times 10^{-3}mol{L}^{-1}{s}^{-1}$.
• D. At $100s,-\frac{d[Y]}{dt}=3.46\times 10^{-3}mol{L}^{-1}{s}^{-1}$.

Answer 1

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

rate $=-\frac{1}{2}\frac{dx}{dt}=-\frac{dy}{dt}=\frac{dP}{dt}=K[X]$
$-\frac{1}{2}\frac{dx}{dt}=K[X]$
$-\frac{dx}{dt}=2K[X]=K^1[X]$
Half life is $t=50sec$
$2K=\frac{0.653L}{50}$
$K=\frac{0.6932}{100}=6.332\times 10^{-3}$
$t=50sec$
$-\frac{dx}{dt}=2K[X]$
$-\frac{dx}{dt}=2\times 6.332\times 10^{-3}\times 1$
$=13.864\times 10^{-3}mole/L/Sec$
$-\frac{dy}{dt}=K[X]=6.332\times 10^{-3}\left(\frac{1}{2}\right)$
$=3.46\times 10^{-3}mole/L/Se{c}^{-1}$