Three gases A,B and C were taken at partial pressures of 1 bar each along with excess of liquid D so that following equation was established.
$A (g) + B (g) ⇌ C (g) + D (g)$
Calculate partial pressure of C (in Pascal) when equilibrium gets established in the container at 300 K.
Given:
$\begin{array}{l} Δ G_{f}^{\circ} A (g) = - 200 kcal / mole \\ Δ G_{f}^{\circ} D (l) = - 49.58 kcal / mole \\ Δ G_{f}^{\circ} B (g) = - 100 kcal / mole \\ Δ G_{f}^{\circ} D (g) = - 49.58 kcal / mole \\ Δ G_{f}^{\circ} C (g) = - 250 kcal / mole \end{array}$
$R = 2 cal / mole K$ , In $2 = 0.7, \sqrt{5} = 2.24$ All data are given at 300 K

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answer is 7600.

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Detailed Solution

$\begin{array}{l} \Rightarrow K_{p} = 1 \\ or V.P of D = 1 bar \\ A (g) + B (g) ⇌ C (g) + D (g) \\ Δ_{R x^{n}} G^{\circ} = - 250 - 49.58 + 100 + 200 = 0.42 \\ K_{P} = 10^{- \frac{0.42 \times 10^{3}}{2.303 \times 2 \times 300}} \\ = 0.5 \\ 0.5 = \frac{(1 - x) \times 1}{(1 + x)^{2}} \end{array}$

$\begin{array}{l} \Rightarrow x = - 2 + \sqrt{5} \\ \Rightarrow P_{C} = 3 - \sqrt{5} \\ = 0.76 bar \\ P_{C} = 0.76 \times \frac{10^{5}}{10} \\ = 7600 \end{array}$