If the decomposition reaction $4 H {N O}_{2} (g) ⇌ 4 {N O}_{2} (g) + 2 H_{2} O (g) + O_{2} (g)$ is started with where $(P)$ is the total pressure at equilibrium and ( $P^{'}$ ) is the partial pressure of $O_{2}$ . Derive relationship for $K_{p}$ is terms of ( $P$ ) and ( $P^{'}$ ).

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Solution

Total pressure at equilibrium $= P_{0} - 4 P^{^{'}} + 4 P^{^{'}} + 2 P^{^{'}} + P^{^{'}}$
$= P_{0} + 3 P^{^{'}}$
$\Rightarrow P_{0} + 3 P^{^{'}} = P$ (given in question) $⟶ 1$
$K_{P} = \frac{(P^{^{'}}) {(2 P^{^{'}})}^{2} {(4 P^{^{'}})}^{4}}{(P_{0} - 4 P^{^{'}})}$
$= \frac{P^{^{'}} \times 4 P^{{^{'}}^{2}} \times 256 P^{{^{'}}^{4}}}{P - 3 P^{^{'}} - 4 P^{^{'}}}$
$K_{P} = \frac{(1024) P^{{^{'}}^{7}}}{P - 7 P^{^{'}}}$

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If the decomposition reaction 4HNO2(g)⇌4NO2(g)+2H2O(g)+O2(g) is started with where (P) is the total pressure at equilibrium and (P′) is the partial pressure of O2. Derive relationship for Kp is terms of (P) and (P′).

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