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Enthalpy of Combustion

Value of f Kp...

Question

Value of f $K_{p}$ for the equilibrium reaction
$N_{2} O_{4 (g)} ⇌ 2 N O_{2 (g)}$ at 288 K is 47.9. The $K_{C}$ for this reaction at same temperature is ........(Nearest integer)

$(R = 0.083 L b a r K^{- 1} m o l^{- 1})$

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Solution

$N_{2} O_{4 (g)} ⇌ 2 N O_{2 (g)}$
$Δ n_{g} = 2 - 1 = 1$

$K_{p} = K_{c} {(R T)}^{Δ n_{g}}$

$K_{p} = K_{c} (R T)^{1}$

$K_{c} = \frac{K_{p}}{R T} = \frac{47.9}{0.083 \times 288} = 2$

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