Question

For a reaction at equilibrium
$A(g)\rightleftharpoons B(g)+\frac{1}{2}C(g)$
the relation between dissociation constant (K), degree of dissociation $(\alpha )$ and equilibrium pressure $(p)$ is given by :

• A. $K=\frac{{\alpha }^{\frac{1}{2}}{p}^{\frac{3}{2}}}{{\left(1+\frac{3}{2}\alpha \right)}^{\frac{1}{2}}(1-\alpha )}$
• B. $K=\frac{{\alpha }^{\frac{3}{2}}{p}^{\frac{1}{2}}}{(2+\alpha {)}^{\frac{1}{2}}(1-\alpha )}$
• C. $K=\frac{(\alpha p{)}^{\frac{3}{2}}}{{\left(1+\frac{3}{2}\alpha \right)}^{\frac{1}{2}}(1-\alpha )}$
• D. $K=\frac{(\alpha p{)}^{\frac{3}{2}}}{(1+\alpha )(1-\alpha {)}^{\frac{1}{2}}}$

Answer 1

Answer: (B)

Now, equilibrium pressure (p),
$P=P_i\times \left(1+\frac{\alpha }{2}\right)$
$\therefore P_A=\left(\frac{1-\alpha }{1+\frac{\alpha }{2}}\right)P$
$P_B=\left(\frac{\alpha }{1+\frac{\alpha }{2}}\right)P$
$P_C=\left(\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}\right)P$
$\therefore K=\frac{P_c^{\frac{1}{2}}\times P_B}{P_A}$
$K=\frac{{\alpha }^{\frac{3}{2}}{p}^{\frac{1}{2}}}{(2+\alpha {)}^{\frac{1}{2}}(1-\alpha )}$