Question

At $300$ K, a gaseous reaction $A\to B+C$ was found to follow first-order kinetics. Starting with pure $A$, the total pressure at the end of $20$ minutes was $100$ mm of $Hg$. The total pressure after the completion of the reaction is $180$ mm of $Hg$. The partial pressure of $A$ (in mm of $Hg$) is :

• A. $100$
• B. $90$
• C. $180$
• D. $80$

Answer 1

Answer: (D)

$80$

Let $P_o$ be the initial pressure of $A$. The total pressure after the completion of the reaction is 180 mm of Hg. It is equal to $2P_o$ (because for every mole of $A$, one mole of $B$ and one mole of $C$ will be produced). Hence, total pressure at the end of the reaction will be twice the initial pressure of $A$.
$2P_A=180$ mm Hg.
$P_A=90$ mm Hg.
The total pressure at the end of 20 minutes was 100 mm of Hg. The total pressure is $P_o-x+x+x=P_o+x$ where $x$ is change in the partial of A to form the products at the end of 20 minutes .

$P_o+x=100$ mm Hg
But
$P_o=90$ mm Hg

Hence,
$90+x=100$
$x=10$ mm Hg
The partial pressure of A (in mm of Hg) is $P_o-x=90-10=80$ mmHg.