Question

Substance ΔG°f at 298K(kJ/mol)

H₂O(l) −237.2
H₂O(g) −228.4

Using the information in the table above, determine the value of ΔG° at 298K for the process represented by the equation H₂O(l)⇄H₂Og).

Answer 1

The standard Gibbs free energy change (ΔG°) at 298K for the conversion of water from liquid to gas is calculated to be 8.8 kJ/mol, indicating that the process is non-spontaneous under standard conditions.

Step-by-step explanation:

The question asks us to determine the standard Gibbs free energy change (ΔG°) at 298K for the phase transition of water from liquid (H₂O(l)) to gas (H₂O(g)). Using the provided values for the standard free energy of formation (ΔG°f) of liquid water and gaseous water, we can calculate ΔG° for the process using the formula:

ΔG° = ΔG°f (H₂O(g)) - ΔG°f (H₂O(l))

Substituting the values into the formula, we get:

ΔG° = (-228.4 kJ/mol) - (-237.2 kJ/mol)

ΔG° = 8.8 kJ/mol

This positive value of ΔG° indicates that the process of converting water from liquid to gas at 298K under standard conditions is non-spontaneous.