For the reaction 2HI (g) ⇋ H2(g) + I2 (g)

Discussion Forum : Chemical Equilibrium

Question -

For the reaction $2 H I (g) ⇋ H_{2} (g) + I_{2} (g),$ the degree of dissociation α of $H I (g)$ is related to the equilibrium constant $(K_{p})$ by the expression

Options:

A .

\frac{1 + 2 \sqrt{K p}}{2}

B .

\frac{\sqrt{1 + 2 K p}}{\sqrt{2}}

C .

\frac{\sqrt{2 K p}}{\sqrt{2 + 2 K p}}

D .

\frac{2 \sqrt{K p}}{1 + 2 \sqrt{K p}}

Answer: Option D
:
D

\begin{matrix} 2 H I (g) ⇋ & H_{2} (g) + & l_{2} (g) A t e q u i l i b r i u m & 1 - α & α / 2 & α / 2 \end{matrix}

Total moles at equilibrium =1

P H I = (1 - α) \times P, P H_{2} = α / 2 \times P P l_{2} = α / 2 \times P

K_{p} = P H_{2} \times P l_{2} / (P H I)^{2}

Substituting above values we get,

K_{p} = a^{2} / [4 \times (1 - α^{2})]

solving,

α / 1 - α = 2 (K_{p})^{\frac{1}{2}}

α = \frac{2 \sqrt{K p}}{1 + 2 \sqrt{K p}}

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