Question

In a system $A\left(s\right)\rightleftharpoons 2B\left(g\right)+3C\left(g\right)$. If the concentration of $C$ at equilibrium is increased by a factor $2$, it will cause the equilibrium concentration of $B$ to change to:

• A. two times of its original value
• B. one half of its original value
  
• C. $3\sqrt{2}$ times of its original value
  
• D. $\frac{1}{2\sqrt{2}}$ times of its original value
  

Answer 1

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

$A\left(s\right)\rightleftharpoons 2B\left(g\right)+3C\left(g\right)$

$\Rightarrow K_P=(P_B{)}^2\times (P_C{)}^3$

$\frac{K_P}{(P_C{)}^3}=(PB{)}^2⟶(1)$

If we increase concentration of $C$ by a factor of $2$, $\frac{K_P}{(2P_C{)}^3}=({P^1}_B{)}^2⟶(2)$

From (1) & (2)

$\Rightarrow \frac{K_P}{2^3(P_C{)}^3}=({P^1}_B{)}^2$

$\Rightarrow \frac{K_P}{2^3(P_C{)}^3}=({P_B}^1{)}^2$

$\Rightarrow \frac{1}{2^3}(P_B{)}^2=({P^1}_B{)}^2$

$\therefore \left({P_B}^1\right)=\frac{P_B}{2\sqrt{2}}$

$\therefore$ Pressure of B will reduce by $2\sqrt{2}$ i.e; $\frac{1}{2\sqrt{2}}$ times of its original value.