Question

For the following reaction $2 X+Y \xrightarrow{k} P$ the rate of reaction is $\frac{d[P]}{d t}=k[X]$. 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 : $\ln\,2 = 0.693)$

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

Answer 1

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

rate $=-\frac{1}{2} \frac{ d x }{ dt }=-\frac{ dy }{ dt }=\frac{ dP }{ dt }= K [ X ]$
$-\frac{1}{2} \frac{ d x }{ dt }= K [ X ]$
$-\frac{ dx }{ dt }=2 K [ X ]= K ^{1}[ X ]$
Half life is $t =50 \,sec$
$2 K =\frac{0.653 L }{50}$
$K =\frac{0.6932}{100}=6.332 \times 10^{-3}$
$t = 5 0 \,sec$
$-\frac{ dx }{ dt }=2 K [ X ]$
$-\frac{ dx }{ dt }=2 \times 6.332 \times 10^{-3} \times 1$
$=13.864 \times 10^{-3} mole / L / Sec$
$-\frac{ dy }{ dt }= K [ X ]=6.332 \times 10^{-3}\left(\frac{1}{2}\right)$
$=3.46 \times 10^{-3} mole / L / Sec ^{-1}$