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Derivation of Kp and Kc

13.8 g of N...

Question

$13.8 g$ of $N_{2} O_{4}$ was placed in a $1 L$ reaction vessel at $400 K$ and allowed to attain equilibrium $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
The total pressure at equilibrium was found to be $9.15 B a r$ . Calculate $K_{c}, K_{p}$ and partial pressure at equilibrium.

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Solution

$∵ P V = n R T$
$∴ P_{i n i t i a l} = \frac{n R T}{V} = \frac{(\frac{13.8}{92}) \times 0.083 \times 400}{1} = 4.98 b a r$
$N_{2} O_{4 (g)} ⇌ 2 N O_{3 (g)} I n i t i a l : 4.98 0 E q u i l i b r i u m : 4.98 - x 2 x$

$P_{t o t a l} = 9.15 = 4.98 - x + 2 x ⟹ x = 4.17 b a r$
$∴ P_{N O_{2}} = 2 x = 8.34 b a r, P_{N_{2} O_{4}} = 4.98 - 4.17 = 0.81$
$∴ K_{P} = \frac{(P_{N O_{2}})^{2}}{(P_{N_{2} O_{4}})} = \frac{(8.34)^{2}}{0.81} = 85.871$
Here, $△ n g =$ Difference in gaseous moles of reactants and products $= 2 - 1 = 1$
$∴$ $K_{P} = K_{C} [R T]^{1} \dots [∵ K_{P} = K_{C} (R T)^{△ n g}]$
$∴ K_{C} = \frac{K_{P}}{R T} = \frac{85.871}{0.083 \times 400} = 2.586$

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