Question

At a given temperature the following reaction is allowed to reach equilibrium in a vessel of volume $V_1$ litre. The degree of dissociation is ${\alpha }_1$. If by keeping the temperature fixed the volume of the reaction vessel is doubled (assuming the degrees of dissociation to be small) the new degree of dissociation shall be :
$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

• A. $2{\alpha }_1$
• B. $\sqrt{\frac{{\alpha }_1}{2}}$
• C. $\sqrt{2{\alpha }_1}$
• D. $2\sqrt{{\alpha }_1}$

Answer 1

Answer: (A)

$2{\alpha }_1$

$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

$a_o(1-{\alpha }_1)$ $a_o{\alpha }_1$ $a_o{\alpha }_1$

$△n_g=1$ (change in gaseous moles)

$K_{C_1}=\frac{a_o{\alpha }_1^2}{1-{\alpha }_1}\left(\frac{1}{V_1}\right)$

$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

$a_o(1-{\alpha }_2)$ $a_o{\alpha }_2$ $a_o{\alpha }_2$

$K_{C_2}=\frac{a_o{\alpha }_2^2}{1-{\alpha }_2}\left(\frac{1}{2V_1}\right)$

$K_{C_1}=K_{C_2}$ (as equilibrium constant only depends upon temperature)

$\frac{2{\alpha }_1^2}{1-{\alpha }_1}=\frac{{\alpha }_2^2}{1-{\alpha }_2}\Rightarrow (1-{\alpha }_1){\alpha }_2^2+2{\alpha }_1^2{\alpha }_2-2{\alpha }_1^2$

$\Rightarrow {\alpha }_1<<1$ and ${\alpha }_2<<1$ (Let)

${\alpha }_2=2{\alpha }_1$