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, $[C]=0.5mol{L}^{-1}$
$\because$ Vessel is of $1L$
$\Rightarrow$ 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.
$K_c=\frac{[C]}{[A][B]}$
$=\frac{(2+x)}{(1-x)(1.5-x)}$
$=\frac{(2-1.5)}{[1-(-1.5)][1.5-(-1.5)]}$
$=\frac{0.5}{2.5\times 3}=0.067$