Using the values for∆H and ∆S in the worksheet, Calculate ∆G for the melting of water at 24°C.?

Question

The value ∆H = 6.01 kJ mol^-1

The value ∆S = 22.0 J K^-1

Answer 1

Assume you are melting 1 mole of water

∆G = ∆H - T∆S

T = 24ºC + 273.15 = 297.15K

∆G = 6010 J/mole * 1 mole - (297.15K)(22 J/K) = -527 J

ref: http://en.wikipedia.org/wiki/Gibbs_free_energy

Answer 2

Using the formula for Gibbs free energy at constant pressure:

ΔG = ΔH -TΔS

T is absolute temperature (kelvin)

First, convert ΔH into J mol^-1:
6.01 kJ mol^-1 = 6010 J mol^-1

And 24°C = 297K

∆G = 6010 - 297*22
∆G = -524 J mol^-1 = 0.524 kJ mol^-1

Answer 3

Maussy did the calculation at 0 C, and found DeltaG =0, which, as he correctly points out, is as it should be for a reversible process.

The other answerers are correct (the difference between them is a matter of how they round off numbers). The point is that DeltaG for melting is negative, showing that this is a thermodynamically favourable process, which of course it should be.

Answer 4

delta g = delta H-T delta S
T=273.15K
=6010-22.0*273.15=0
which is good since melting is a reversible process