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Derivation of Kp and Kc

Assume that t...

Question

Assume that the decomposition of ${H N O}_{3}$ can be represented by the following equation
$4 {H N O}_{3} (g) ⇌ 4 {N O}_{2} (g) + 2 H_{2} O (g) + O_{2} (g)$

The reaction approaches equilibrium at $400 K$ temperature and $30$ atm pressure. At equilibrium partial pressure of ${H N O}_{3}$ is $2$ atm.

Calculate $K_{c}$ in ${(m o l / L)}^{3}$ at $400 K$ :
(Use : $R = 0.08 a t m - L / m o l - K$ )

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Solution

The correct option is D 32
We know, $P_{t o t a l} = P_{H N O_{3}} + P_{N O_{2}} + P_{H_{2} O} + P_{O_{2}}$
$P_{N O_{2}} = 4 P_{O_{2}}$ and $P_{H_{2} O} = 2 P_{O_{2}}$
So, $P_{t o t a l} = P_{H N O_{3}} + 7 P_{O_{2}}$
$30 = 2 + 7 P_{O_{2}}$

$7 P_{O_{2}}$ = 28 , $P_{O_{2}} = 4$

$K_{p} = \frac{P_{N O_{2}}^{4} P_{H_{2} O}^{2} P_{O_{2}}}{P_{H N O_{3}}^{4}}$
$K_{p} = \frac{(4 \times 4)^{4} \times (2 \times 4)^{2} \times 4}{2^{4}} = 2^{20}$
$K_{p} = K_{c} (R T)^{Δ n} = K_{c} (0.08 \times 400)^{3}$
$K_{c} = \frac{2^{20}}{(32)^{3}} = 32$

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