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

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 $50 \mathrm{~s}, 0.5$ mole of $Y$ is left in the reaction mixture. The correct statement(s) about the reaction is(are)\ (Use: $\ln 2=0.693$ )\

Professor PC Thomas Classes · Answer

$(\mathrm{b}, \mathrm{c}, \mathrm{d}): 2 X+Y \xrightarrow{k} P$\ $\frac{d[P]}{d t}=k[X]$\ \ $-\frac{d[X]}{d t}=k_{1}[X]=2 k[X] \Rightarrow 2 k=k_{1}$\ $-\frac{d[Y]}{d t}=k_{2}[X]=k[X] \Rightarrow k_{2}=k$\ $2 k=\frac{1}{50} \ln 2$\ $k=\frac{1}{100} \ln 2=\frac{0.693}{100}=6.93 	imes 10^{-3} \mathrm{~s}^{-1}$\ $t_{1 / 2}=\frac{\ln 2}{k_{1}}=\frac{\ln 2 	imes 100}{2 	imes 0.693}=50 \mathrm{sec}$\ At 50 sec\ $-\frac{d[X]}{d t}=2 k[X]=2 	imes \frac{0.693}{100} 	imes 1$\ $=13.86 	imes 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$\ At 100 sec\ $-\frac{d[Y]}{d t}=k_{2}[X]=k[X]=\frac{0.693}{100} 	imes \frac{1}{2}$\ $=3.46 	imes 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$\ $[\because$ concentration of $X$ after $$ \left.2 	ext { half lives }=\frac{1}{2} Might] $$