Question

For, $A+B\rightleftharpoons C$, the equilibrium concentration of $A$ and $B$ at a temperature are $15mol{litre}^{-1}$. When volume is doubted the reaction has equilibrium concentration of $A$ as $10mol{litre}^{-1}$. The value $K_c$ is X ${mol}^{-1}litre$. Then, 10X is :

Answer 1

$A+B\rightleftharpoons C$
Concentration at I equilibrium $1515a$ (Volume = $V$ litre)
Concentration when volume is doubled $\frac{15}{2}\frac{15}{2}\frac{a}{2}$ (Volume = $2V$ litre)
Concentration at II equilibrium $\left(\frac{15}{2}+x\right)\left(\frac{15}{2}+x\right)\left(\frac{a}{2}-x\right)$
On increasing volume, the reaction proceeds in the direction where an increase in mole is noticed, i.e., backward reaction
$\therefore \frac{15}{2}+x=10;\therefore x=\frac{5}{2}$
Since, for I equilibrium
$K_c=\frac{\left[C\right]}{\left[A\right]\left[B\right]}=\frac{a}{15\times 15}$ ...(i)
For II equilibrium
$K_c=\frac{\left[C\right]}{\left[A\right]\left[B\right]}=\frac{[\left(a/2\right)-\left(5/2\right)]}{[\left(15/2\right)+\left(5/2\right)][\left(15/2\right)+\left(5/2\right)]}$ ...(ii)
Since, $K_c$ are same and therefore by equations (i) and (ii)
$a=45M$
Also, by equation (i) $K_c=\frac{45}{15\times 15}=0.2{mol}^{-1}litre$