Question

Consider a reaction $A\left(g\right)\to 3B\left(g\right)+2C\left(g\right)$with rate constant is $1.386\times 10^{-2}mi{n}^{-1}$ in a non-rigid closed container starting with $2$ moles of $A$ in $12.5L$ vessel initially, if reaction is allowed to take place at constant pressure and at $298K$ the conc, of $B$ after $100min$ is

• A. $0.18M$
• B. $0.03M$
• C. $0.09M$
• D. $0.01M$

Answer 1

Answer: (C)

$A\to 3B+2C$
$K=\frac{2.303}{t}log\frac{{\left[A\right]}_0}{\left[A\right]}\Rightarrow \left[A\right]=0.0016$
$.0016=\frac{n_A}{Vol}$
$.0016\times 12.5=n_A$
$n_A=.02$
$x_A$ decomposed $=2-0.02=1.98$
Moles of $A$ decomposed $=1.98$
So moles of $B$ formed $=5.94$
Assuming $n\propto v$ final $vol.=62L$
So find cone, of $B=\frac{5.94}{62}=0.09mol/L$