Question

For the $1^{st}$ order reaction, $A\left(g\right)\to 2B\left(g\right)+C\left(s\right),t_{1/2}=24min.$ The reaction is carried out by taking a certain mass of 'A' enclosed in a vessel in which it exerts a pressure of $400mmHg.$ The pressure of the reaction mixture after the expiry of $48min$ will be :
Take
$\text{Antilog}\left(0.6\right)=4$

• A. $700mmHg$
• B. $600mmHg$
• C. $500mmHg$
• D. $1000mmHg$

Answer 1

The correct option is A $700mmHg$

Given:
$\text{Initial pressure,}{P}_0=400mmHg$

$t_{1/2}=24min;k=\frac{0.693}{t_{1/2}}=\frac{0.693}{24}$

$A\left(g\right)\to 2B\left(g\right)+C\left(s\right)$
$\text{At time,}t=0P_{0}0-$
$\text{At time,}t=t{P}_0-P2P-$

$\text{Total pressure}=P_0+P$
For $1^{st}$ order reaction

$k=\frac{2.303}{t}\log \frac{P_0}{P_0-P}$

At time, $t=48min$

$\frac{0.693}{24}=\frac{2.303}{48}\log \frac{400}{400-P}$
$0.6=\log \frac{400}{400-P}$
$\text{Antilog}\left(0.6\right)=\frac{400}{400-P}$
$4=\frac{400}{400-P}$

$P=300mmHg$
$\therefore$
The pressure of the reaction mixture after expiry time is,
$=P_0+P$
$=400+300=700mmHg$