Question

For the reaction $2HI(g)\rightleftharpoons H_{2}(g)+I_{2}(g)$, the correct expression relating the degree of dissociation ($\alpha$) of HI and equilibrium constant $K_{p}$ is

• A. $\alpha=\frac{1+2\sqrt{K_{p}}}{2}$
• B. $\alpha=\sqrt{1+\frac{2K_{p}}{2}}$
• C. $\alpha=\frac{2\sqrt{K_{p}}}{1+2\sqrt{K_{p}}}$
• D. $\alpha=\sqrt{\frac{2K_{p}}{1+2K_{p}}}$

Hint:

When $\Delta n_{g} = 0$, total moles remain constant, making $K_{p}$ independent of pressure $P$, which simplifies solving for $\alpha$.

Answer 1

The Correct Option is C

Step 1: Concept
The equilibrium constant $K_{p}$ for a gaseous reaction is expressed in terms of the equilibrium partial pressures of the reactants and products, which can be linked to the degree of dissociation ($\alpha$).

Step 2: Meaning
Let the initial moles of $HI$ be $1$. At equilibrium, if $\alpha$ is the degree of dissociation, the remaining moles of $HI$ will be $1-\alpha$, while the moles of $H_{2}$ and $I_{2}$ formed will each be $\frac{\alpha}{2}$. The total number of moles at equilibrium is $(1-\alpha) + \frac{\alpha}{2} + \frac{\alpha}{2} = 1$.

Step 3: Analysis
Since the total number of moles is $1$, the partial pressures are equal to their respective mole fractions multiplied by the total pressure $P$: $p_{HI} = (1-\alpha)P$, $p_{H_{2}} = \frac{\alpha}{2}P$, and $p_{I_{2}} = \frac{\alpha}{2}P$. Substituting these values into the equilibrium expression: $K_{p} = \frac{p_{H_{2}} \cdot p_{I_{2}}}{(p_{HI})^{2}} = \frac{(\frac{\alpha}{2}P)(\frac{\alpha}{2}P)}{(1-\alpha)^{2}P^{2}} = \frac{\alpha^{2}}{4(1-\alpha)^{2}}$. Taking the square root on both sides gives $\sqrt{K_{p}} = \frac{\alpha}{2(1-\alpha)}$. Rearranging to solve for $\alpha$: $2\sqrt{K_{p}}(1-\alpha) = \alpha \implies 2\sqrt{K_{p}} - 2\sqrt{K_{p}}\alpha = \alpha \implies 2\sqrt{K_{p}} = \alpha(1 + 2\sqrt{K_{p}}) \implies \alpha = \frac{2\sqrt{K_{p}}}{1+2\sqrt{K_{p}}}$.

Step 4: Conclusion
This derived relationship match option (C) perfectly.

Final Answer: (C)