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For the reaction
$A_{2} (g) + 2 B_{2} (g) ⇌ 2 C_{2} (g)$
the partial pressure of $A_{2}$ and $B_{2}$ at equilibrium are 0.80 atm and 0.40 atm respectively. The pressure of the system is 2.80 atm. The equilibrium constant $K_{p}$ (atm^-1) will be

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Detailed Solution

$A_{2} (g) + 2 B_{2} (g) ⇌ 2 C_{2} (g)$

$P_{A_{2}} = 0.80 atm$

$P_{B_{2}} = 0.4 atm$

Total pressure of the system = 2.8 atm

$∴ P_{C_{2}}$ = 2.8 - 0.8 - 0.4 = 1.6

$K_{p} = \frac{P_{C_{2}}^{2}}{P_{A_{2}}^{} \times P_{B_{2}}^{3}}$ = $\frac{{(1.6)}^{2}}{0.8 \times {(0.4)}^{2}} = 20$

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