Question

$1$ mole of '$A$', $1.5$ mole of ' $B$ ' and $2$ mole of ' $C$ ' are taken in a vessel of volume one litre. At equilibrium concentration of $C$ is $0.5$ mole/L. Equilibrium constant for the reaction
$A _{(g)}+ B _{(g)} \rightleftharpoons C _{(g)}$ is

• A. 0.66
• B. 0.066
• C. 66
• D. 6.6

Answer 1

Answer: (B)

At equilibrium, $[ C ]=0.5 mol L ^{-1}$
$\because$ Vessel is of $1 L$
$\Rightarrow $ moles of $C=0.5$
$\therefore 2+x=0.5$
$\Rightarrow x =-1.5$
Which means $C$ decreases and $A$ and $B$ are formed i.e. reaction shifts backwards.
$K _{ c } =\frac{[ C ]}{[ A ][ B ]}$
$=\frac{(2+ x )}{(1- x )(1.5- x )} $
$=\frac{(2-1.5)}{[1-(-1.5)][1.5-(-1.5)]} $
$=\frac{0.5}{2.5 \times 3}=0.067$