Question

In a reaction, $A+2B\rightleftharpoons 2C$, $2.0$ mole of $A$ $3.0$ mole of $B$ and $2.0$ mole of $C$ are placed in a $2.0L$ flask and the equilibrium concentration of $C$ is $0.5mole/L$. The equilibrium constant ($K$) for the reaction is_________.

• A. $0.073$
• B. $0.147$
• C. $0.05$
• D. $0.026$

Answer 1

The correct option is C $0.05$

$A+2B\rightleftharpoons 2C$ Volume = 2 lit

At $2$ $3$ $2$

$t=0$

At $2-\alpha$ $3-2\alpha$ $2+2\alpha$

$t=t_{eq}$

$\left[C\right]=\frac{2+2\alpha }{2}=1+\alpha =0.5$

$\alpha =-0.5$

$\alpha$ is negative because reaction favoured in reverse direction

$\left[B\right]=\frac{3-2\alpha }{2}=\frac{3-2(-0.5)}{2}=2$

$\left[A\right]=\frac{2-\alpha }{2}=\frac{2-(-0.5)}{2}=1.25$

$K=\frac{{\left[C\right]}^2}{{\left[B\right]}^2\left[A\right]}=\frac{{\left(0.5\right)}^2}{2^2\times \left(1.25\right)}=0.05$