Question

A dilute solution contains 'm' mole of solute A in one kg solvent (molal depression constant is $K_f$). The solute A undergoes dimerisation as $2A\rightleftharpoons A_2$. If the equilibrium constant for dimerisation $K_c$ is $K_c=\frac{K_f[K_f\times m-\Delta T_f]}{[x\Delta T_f-K_f\times m{]}^y}$.

Find the value of $x$ and $y$.

• A. $x=1,y=1$
• B. $x=1,y=2$
• C. $x=2,y=2$
• D. $x=2,y=1$

Answer 1

The correct option is C $x=2,y=2$

Equilibrium constant for dimerisation is $K_c=\frac{\left[A_2\right]}{{\left[A\right]}^2}$

$2A\leftrightharpoons A_2$

Initial: 1 0

After time t:$1-\alpha \frac{\alpha }{2}$

So,$K_c=\frac{\left[\frac{\alpha }{2}\right]}{{[1-\alpha ]}^2}$

Now, $i=1-\alpha +\frac{\alpha }{2}=1-\frac{\alpha }{2}$

As we know that, $\Delta T_f=i{K}_{f}mi=\frac{\Delta T_f}{K_{f}m}i=1-\frac{\alpha }{2}=\frac{\Delta T_f}{K_{f}m}\frac{\alpha }{2}=1-\frac{\Delta T_f}{K_{f}m}1-\alpha =1-2(1-\frac{\Delta T_f}{K_{f}m})=2\times \frac{\Delta T_f}{K_{f}m}-1$

Put in above equation, we get $K_c=\frac{K_f[K_f\times m-\Delta T_f]}{{[2\Delta T_f-K_f\times m]}^2}$

So we get x=2 and y=2.