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

Answer: (B)

$0.066$

$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)molesatt=011.52at{eqb}^{m}1+x1.5+x2-x$

$Q_C=\frac{127}{\left(1.5\right)\left(1.7\right)}=1.33$

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

$K_{eq}=\frac{\left[C\right]}{\left[A\right]\left[B\right]}=\frac{\left[0.5\right]}{[1+1.5][1.5+1.5]}=\frac{\left[0.5\right]}{\left[2.5\right]\left[3\right]}=0.0666[\because 2-x=0.5⟹x=0.5]$

As $Q_C>K_{eq}$ thus reaction will go in the backward direction.