Question

For the reaction $A_{\left(g\right)}+B_{\left(g\right)}\rightleftharpoons C_{\left(g\right)}+D_{\left(g\right)}$, the degree of dissociation $\alpha$ would be:

• A. $\frac{\sqrt{K}}{\sqrt{K}+1}$
• B. $\sqrt{K}+1$
• C. $\sqrt{K}\pm 1$
• D. $\sqrt{K}-1$

Answer 1

The correct option is A $\frac{\sqrt{K}}{\sqrt{K}+1}$

Given : $A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right)$

initial no. of moles :$1100$ (let A and B are in equal quality)

At equilibrium $1-\alpha 1-\alpha \alpha \alpha$

Total no. of moles at equilibrium = $1-\alpha +1-\alpha +\alpha +\alpha =2$

Let total pressure be p.

Now,

$R=\frac{P_C.P_D}{P_A.P_B}$

equilibrium constant

$P_A=\frac{1-\alpha }{2}.P$ $P_B=\frac{1-\alpha }{2}P$

$P_C=\frac{\alpha }{2}.P$ $P_D=\frac{\alpha }{2}.P$

$\therefore K=\frac{\frac{\alpha }{2}.P\times \frac{\alpha }{2}.P}{\frac{(1-\alpha )}{2}P.\frac{(1-\alpha )}{2}}P=\frac{{\alpha }^2}{(1-\alpha {)}^2}\Rightarrow \sqrt{K}=\frac{\alpha }{1-\alpha }$

$\Rightarrow \frac{1-\alpha }{\alpha }=\frac{1}{\sqrt{K}}\Rightarrow \frac{1}{\alpha }-2=\frac{1}{\sqrt{K}}\Rightarrow \frac{1}{\alpha }=\frac{1+\sqrt{K}}{\sqrt{K}}\Rightarrow \alpha =\frac{\sqrt{K}}{1+\sqrt{K}}$

Correct Ans (A)