For thedecomposition reaction 4 $H N O_{3}$ (g) $⇋$ 4 $N O_{2}$ (g) + 2 $H_{2}$ O (g) + $O_{2}$ (g) if we start with only the reactant $H N O_{3}$ and p = equilibrium pressure, the equilibrium constant $K_{p}$ is:

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= 1024

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= 7

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Solution

The correct option is B
= 1024

Let the initial pressure be P. For constant T and volume, equal moles of all ideal gases contribute equal partial pressures.

4 $H N O_{3}$ (g) $⇋$ 4 $N O_{2}$ (g) + 2 $H_{2}$ O (g) + $O_{2}$ (g)

$\begin{matrix} 4 H N O_{3} (g) & 4 N_{O_{2}} (g) & 2 H_{2} O (g) & O_{2} (g) t = 0 P a r t i a l p r e s s u r e s & P & O & O & O E q u i l i b r i u m P a r t i a l p r e s s u r e s & P - 4 p_{O_{2}} & 4 p_{O_{2}} & 2 p_{O_{2}} & p_{O_{2}} \end{matrix}$
Total pressure at equilibrium = sum of all partial pressures = p = P + $3 p_{O_{2}}$

$K_{p} = \frac{[4 p (O_{2})]^{4} [2 p (O_{2})]^{2} p (O_{2})}{[p - 3 p (O_{2}) - 4 p (O_{2})]^{4}} = \frac{1024 [p (O_{2})]^{7}}{[p - 7 p (O_{2})]^{4}}$

This kind of playing around with partial pressures is convenient when only either reactants or products are present initially. Otherwise we have to take existing individual partial pressures into account as well.

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For thedecomposition reaction 4HNO3 (g) ⇋ 4NO2 (g) + 2H2 O (g) + O2 (g) if we start with only the reactant HNO3 and p = equilibrium pressure, the equilibrium constant Kp is:

For thedecomposition reaction 4 $H N O_{3}$ (g) $⇋$ 4 $N O_{2}$ (g) + 2 $H_{2}$ O (g) + $O_{2}$ (g) if we start with only the reactant $H N O_{3}$ and p = equilibrium pressure, the equilibrium constant $K_{p}$ is: