Question

A closed vessel initially contains a mixture of two gases $A_{2}B$ and an inert gas. During the thermal decomposition of gas $A_{2}B$ as per given $A_{2}B\left(g\right)\to 2A\left(g\right)+B\left(g\right)$ the pressure changed from an initial value of $2atm$ to $5.2atm$ at the end of reaction. The rate constant (in min) if total pressure was measured as $4.8atm$ after $10min$ is:

• A. $0.3log2$
• B. $3ln\frac{1.6}{1.6-1.4}$
• C. $\frac{ln2}{200}$
• D. $\frac{3ln2}{10}$

Answer 1

The correct option is D $\frac{3ln2}{10}$

Let pressure due to to $A_{2}B\left(g\right)$ be ${}^{\prime }{x}^{\prime }$ atm
$\therefore$ pressure due to inert gas$=(2-x)$atm
$t=0$
$t=t_{end}$
$\begin{array}{ccc}{A}_{2}B\left(g\right)\to & 2A\left(g\right)+& B\left(g\right)\\ x& 0& 0\\ 0& 2x& x\end{array}$
$\therefore$ Total pressure $=2x+x(2-x)=5.2$
$\therefore x=1.6atm$
After 10 mins:
$t=10min$
$\begin{array}{ccc}{A}_{2}B\left(g\right)\to & 2A\left(g\right)+& B\left(g\right)\\ 1.6-y& 2y& y\end{array}$
Total pressure $=(1.6-y)+2y+y+(2+1.6)=4.8$
$\therefore 2y+2=4.8$
$\therefore y=1.4atm$
$K=\frac{1}{10}ln\frac{x}{x-y}=\frac{1}{10}ln\left(\frac{1.6}{0.2}\right)=\frac{3ln2}{10}$