The equilibrium constant at 25°C for the process. CO3+(aq) + 6NH3(aq

1.	The equilibrium constant at 25°C for the process. CO3+(aq) + 6NH3(aq) ⇌ [CO(NH3)6]3(aq) is 2.5 × 106. Calculate the value of ΔG° at 25°C. (R = 8.314 JK1 mol-1). In which direction is the reaction spontaneous under standard conditions?
Answer» ΔG° = -2.303 RT log K = -2.303 × 3.134 × 298 log (2.5 x 10⁶) = -5705.8 [0.3980 + 6.0000] = -5705.8 × 6.3980 = -36.505 k/mol. The reaction is spontaneous in forward direction under standard conditions.

The equilibrium constant at 25°C for the process. CO3+(aq) + 6NH3(aq) ⇌ [CO(NH3)6]3(aq) is 2.5 × 106. Calculate the value of ΔG° at 25°C. (R = 8.314 JK1 mol-1). In which direction is the reaction spontaneous under standard conditions?

Answer»

ΔG° = -2.303 RT log K

= -2.303 × 3.134 × 298 log (2.5 x 10⁶)

= -5705.8 [0.3980 + 6.0000]

= -5705.8 × 6.3980 = -36.505 k/mol.

The reaction is spontaneous in forward direction under standard conditions.

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The equilibrium constant at 25°C for the process. CO3+(aq) + 6NH3(aq) ⇌ [CO(NH3)6]3(aq) is 2.5 × 106. Calculate the value of ΔG° at 25°C. (R = 8.314 JK1 mol-1). In which direction is the reaction spontaneous under standard conditions?

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