Question

An equilibrium $A\left(g\right)\rightleftharpoons 2B\left(g\right)$ has been established in a one-litre container. At equilibrium, moles of A and moles of B are found to be equal (2 moles each). Now, if the volume of the reaction vessel is halved keeping the temperature constant and the equilibrium is allowed to restablish, then which of the following statements is/are true?

• A. The concentration of B at new equilibrium is 3 M
• B. The equilibrium can be said to be shifted in forward direction upon reducing the volume to half
• C. The equilibrium can be said to be shifted in backward direction upon reducing the volume to half
• D. The equilibrium will not shift in any direction upon reducing the volume to half as moles of A and B at initial equilibrium are equal (2 each)

Answer 1

Answer: (A), (C)

The correct options are
A The concentration of B at new equilibrium is 3 M
C The equilibrium can be said to be shifted in backward direction upon reducing the volume to half

$A\left(g\right)\rightleftharpoons 2B\left(g\right)$
I eq. conc. 2 2 $\therefore K_c=\frac{2^2}{2}=2$
Conc. after 4 4 $\therefore Q_c=\frac{4^2}{4}=\frac{16}{4}=4>K_c\therefore$ Eq. will shift in backward direction
Volume made half
II. eq. conc. 4+x 4-2x $\therefore K_c=\frac{(4-2x{)}^2}{(4+x)}=2$ (as T remains constant)
On solving we get x = 0.5
$\therefore \left[A\right]=4+0.5=4.5M,\left[B\right]=4-2\left(0.5\right)=3M$