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Reaction Quotient

The K P value...

Question

The $K_{P}$ value for the reaction $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ at $460^{o} C$ is 49. If the initial pressure of $H_{2}$ and $I_{2}$ is $0.5 a t m$ respectively, determine the partial pressure of $H_{2}$ in atm at equilibrium.(give your answer upto 2 decimal)

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Solution

According to the given reaction,

$\begin{matrix} H_{2} & + & I_{2} & ⇌ & 2 H I I n i t i a l p r e s s u r e & 0.5 & 0.5 & 0 A t e q u i l i b r i u m & (0.5 - x) & (0.5 - x) & 2 x (from stoichiometry of equation) \end{matrix}$

using the formula,
$K_{P} = \frac{P_{H I}^{2}}{P_{H_{2}} \times P_{I_{2}}}$
$49 = \frac{(2 x)^{2}}{(0.5 - x)^{2}}$
$x = 0.39 a t m$
$∴$ Pressure of $H_{2}$ at equilibrium $= (0.5 - x) = 0.11 a t m$

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