For the reaction $N H_{4} H S_{(g)} ⇌ N H_{3 (g)} + H_{2} S_{(g)}$ in a closed flask, the equilibrium pressure is P atm. The standard free energy of the reaction would be :

-RT In p

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-RT(In p - In 2)

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-2 RT In p

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-2 RT (In p - In 2)

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Solution

The correct option is D
-2 RT (In p - In 2)

$P_{N H_{3}} = P_{H_{2} S} = \frac{P}{2} a t m K_{p} = P_{N H_{3}} P_{H_{2} S} = (\frac{P}{2})^{2} = \frac{P^{2}}{4} Δ G = - R T I n k_{p} = - R T I n (\frac{p}{2})^{2} = - 2 R T [I n p - I n 2]$

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For the reaction $N H_{4} H S_{(g)} ⇌ N H_{3 (g)} + H_{2} S_{(g)}$ in a closed flask, the equilibrium pressure is P atm. The standard free energy of the reaction would be :

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For the reaction NH4HS(g)⇌NH3(g)+H2S(g) in a closed flask, the equilibrium pressure is P atm. The standard free energy of the reaction would be :

The correct option is D -2 RT (In p - In 2) PNH3=PH2S=P2atmKp=PNH3PH2S=(P2)2=P24ΔG=−RT In kp=−RT In(p2)2=−2RT[In p−In 2]

For the reaction $N H_{4} H S_{(g)} ⇌ N H_{3 (g)} + H_{2} S_{(g)}$ in a closed flask, the equilibrium pressure is P atm. The standard free energy of the reaction would be :

The correct option is D
-2 RT (In p - In 2)

$P_{N H_{3}} = P_{H_{2} S} = \frac{P}{2} a t m K_{p} = P_{N H_{3}} P_{H_{2} S} = (\frac{P}{2})^{2} = \frac{P^{2}}{4} Δ G = - R T I n k_{p} = - R T I n (\frac{p}{2})^{2} = - 2 R T [I n p - I n 2]$