For the equilibriumPCl5 (g) == PCl3 (g) + Cl2 (g)at 298 K, K = 1.8 x 1

1.	For the equilibriumPCl5 (g) == PCl3 (g) + Cl2 (g)at 298 K, K = 1.8 x 10-7. What is ΔG° for the reaction?
Answer» ΔG° = -2.303 RT log K = -2.303 x 8.31 JK^-1 mol^-1 x 298 K x log (1.8 x 10^-7) = -2.303 x 8.31 JK^-1 mol^-1 x 298 K x (0.0334 - 7) = -2.303 x 8.31 x 298 K x (-6.9664) J mol^-1 = + 39730 J mol^-1 = + 39.73 kJ mol^-1

For the equilibriumPCl5 (g) == PCl3 (g) + Cl2 (g)at 298 K, K = 1.8 x 10-7. What is ΔG° for the reaction?

Answer»

ΔG° = -2.303 RT log K

= -2.303 x 8.31 JK^-1 mol^-1 x 298 K x log (1.8 x 10^-7)

= -2.303 x 8.31 JK^-1 mol^-1 x 298 K x (0.0334 - 7)

= -2.303 x 8.31 x 298 K x (-6.9664) J mol^-1

= + 39730 J mol^-1

= + 39.73 kJ mol^-1

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For the equilibriumPCl5 (g) == PCl3 (g) + Cl2 (g)at 298 K, K = 1.8 x 10-7. What is ΔG° for the reaction?

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