Chemistry - Chemical Kinetics Question with Solution | TestHub

A(g) $⟶$ 2B(g) + C(s)

At $t=0$: $400$ $0$ $0$

At $t=t$: $(400-P^{\prime })$ $2P^{\prime }$ $0$

Since the reaction is of the first order:

$k=\frac{2.303}{t}\log \left(\frac{400}{400-P^{\prime }}\right)$

$\frac{0.6932}{24}=\frac{2.303}{48}\log \left(\frac{400}{400-P^{\prime }}\right)$

On solving, $P^{\prime }=300$.

After 48 minutes, the pressure of the reaction vessel is:

$P_{total}=(400-P^{\prime })+2P^{\prime }=400+P^{\prime }$

$P_{total}=400+300=700mmHg$