Question

In a chemical system, $A\left(s\right)\rightleftharpoons 2B\left(g\right)+3C\left(g\right)$, if partial pressure of $C$ is doubled, then partial pressure of $B$ will be:

• A. $2\sqrt{2}$ times the original value
• B. $\frac{1}{2}$ times the original value
• C. $2$ times the original value
• D. $\frac{1}{2\sqrt{2}}$ times the original value

Answer 1

The correct option is D $\frac{1}{2\sqrt{2}}$ times the original value

In a chemical system $A\left(s\right)\rightleftharpoons 2B\left(g\right)+3C\left(g\right)$.
As per the initial conditions,
$K_p=p_{B_1}^2\times p_{C_1}^3$.....(1)
But $p_{C_2}=2\times p_{C_1}$ as partial pressure of $C$ is doubled.

As per the final conditions,

$K_p=p_{B_2}^2\times {\left(2p_{C_1}\right)}^3$.....(2)

Substitute this in equation (3).
$\therefore p_{B_1}^2\times p_{C_1}^3=p_{B_2}^2\times (2\times p_{C_1}{)}^3$
$\therefore p_{B_1}^2\times p_{C_1}^3=p_{B_2}^2\times 8p_{C_1}^3$
$\therefore p_{B_1}^2=p_{B_2}^2\times 8$
$\frac{p_{B_1}^2}{8}=p_{B_2}^2$
$\frac{p_{B_1}}{\sqrt{8}}=p_{B_2}$
$\frac{p_{B_1}}{2\sqrt{2}}=p_{B_2}$
if partial pressure of $C$ is doubled, then partial pressure of $B$ will be $\frac{1}{2\sqrt{2}}$ times the original value.