Question

Answer the following questions related to H2O.

Substance ΔG°f at 298K(kJ/mol)
H2O(l) −237.2
H2O(g) −228.4

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

Question 2
(b) Considering your answer to part (a), indicate whether the process is thermodynamically favorable at 298K. Justify your answer.

Answer 1

Here are the answers to the questions related to H2O:

(a) Using the ΔG°f values given for H2O(l) and H2O(g) at 298K:

ΔG°(H2O(l) ⇄ H2O(g)) = ΔG°f(H2O(g)) - ΔG°f(H2O(l))

= -228.4 - (-237.2) kJ/mol

= +8.8 kJ/mol

(b) The ΔG° value for the process H2O(l) ⇄ H2O(g) is +8.8 kJ/mol, which is positive.

Therefore, the process is not thermodynamically favorable at 298K.

A negative ΔG° indicates a thermodynamically favorable process while a positive ΔG° means the process proceeds in the opposite direction.

The positive ΔG° value shows that at 298K, the equilibrium lies on the left side favoring the liquid state.

In summary, the melting of H2O is not spontaneous at 298K due to the positive ΔG° value.

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