Question

In the reaction, $A_{\left(s\right)}+B_{\left(g\right)}+heat\rightleftharpoons 2C_{\left(s\right)}+2D_{\left(g\right)}$, equilibrium is established. If the partial pressure of B is doubled then the partial pressure of D will be?

• A. $2$
• B. $3$
• C. $\sqrt{2}$
• D. $\sqrt{3}$

Answer 1

The correct option is C $\sqrt{2}$

$\begin{array}{cc}\text{Given:}& -A\left(s\right)+B\left(g\right)+\text{heat}\rightleftharpoons 2C\left(s\right)+2D\left(g\right)\text{(equilibrium)}\\ \therefore & Kp=\frac{{\left(P_C\right)}^2{\left(P_D\right)}^2}{\left(PA\right)\left(P_B\right)}-\left(1\right)\end{array}$

if partial pressure of $B$ is doubled

Let partial pressure of $D=x$

$\text{Now,}kp=\frac{{\left(P_c\right)}^2(x{)}^2}{P_A\left(2P_B\right)}-\left(2\right)$

equating equ: (1) 4 (2)

$\frac{{\text{(PC)}}^2(PD{)}^2}{\left(P_A\right)\left(PB\right)}=\frac{\left(PC{)}^2\right(x{)}^2}{\left(PA\right)\left(2P_B\right)}$

$2{\left(P_D\right)}^2=(x{)}^2$

$\Rightarrow u=\sqrt{2}{P}_D$

Corvect answer iss C) $\sqrt{2}$