Question

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

• A. Twice the original pressure
• B. Half of its original pressure
• C. $\frac{1}{2\sqrt{2}}$ times, the original pressure
• D. $2\sqrt{2}$ times its original pressure

Answer 1

Answer: (C)

$\frac{1}{2\sqrt{2}}$ times, the original pressure

Solution:- (C) $\frac{1}{2\sqrt{2}}$ times, the original pressure

$A_{\left(s\right)}\rightleftharpoons 2B_{\left(g\right)}+3C_{\left(g\right)}$

$K_P={\left(P_B\right)}^2{\left(P_C\right)}^3.....\left(1\right)$

If we double the partial pressure of $C$, i.e., $P_C^{\prime }=2P_C$

$\therefore K_P^{\prime }={\left(P_B^{\prime }\right)}^2{\left(P_C^{\prime }\right)}^3$

$\Rightarrow K_P^{\prime }={\left(P_B^{\prime }\right)}^2{\left(2P_C\right)}^3$

$\Rightarrow K_P^{\prime }=8{\left(P_B^{\prime }\right)}^2{\left(P_C\right)}^3$

Since $K_P$ is constant,

$\therefore K_P=K_P^{\prime }$

$\Rightarrow {\left(P_B\right)}^2{\left(P_C\right)}^3=8{\left(P_B^{\prime }\right)}^2{\left(P_C\right)}^3$

$\Rightarrow P_B^{\prime }=\sqrt{\frac{P_B}{8}}$

$\Rightarrow P_B^{\prime }=\frac{P_B}{2\sqrt{2}}$

Hence the partial pressure of $B$ will be $\frac{1}{2\sqrt{2}}$ times of its original pressure.