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, $A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)$ is?

• A. $0.66$
• B. $0.066$
• C. $66$
• D. $6.6$

Answer 1

The correct option is B $0.066$

$A+B\rightleftharpoons C$

$A$ $B$ $C$

Initial $1$ $1.5$ $2$

change $1-x$ $1.5-x$ $2+x$

At equilibrium $0.5$

$2+x=0.5=>x=-1.5$.

$\therefore$ $A$= $2.5$; $B$ = $3$; $C$ = $0.5$.

Reaction will be backward as concentration of C is less at equilibrium than at initial.

$K_c=\frac{\left[C\right]}{\left[A\right]\left[B\right]}=\frac{0.5}{2.5\times 3}=0.066$

Hence,option B is correct.