Answer:
ΔH > 0; ΔS >0; ΔG = 0
Not spontaneous when T < 100 °C;
Equilibrium when T = 100 °C
Spontaneous when T > 100 °C
Step-by-step explanation:
The process is
H₂O(ℓ) ⇌ H₂O(g)
ΔH > 0 (positive), because we must <em>add heat</em> to boil water
ΔS > 0 (positive), because changing from a liquid to a gas i<em>ncreases the disorder
</em>
ΔG = 0, because the liquid-vapour equilibrium process is at <em>equilibrium</em> at 100 °C
ΔG = ΔH – TΔS
Both ΔH and ΔS are positive.
If T = 100 °C, ΔG =0. ΔH = TΔS, and the system is at equilibrium.
If T < 100 °C, the ΔH term will predominate, because T has decreased below the equilibrium value.
ΔG > 0. The process is not spontaneous below 100 °C.
If T > 100 °C, the TΔS term will predominate, because T has increased above the equilibrium value.
ΔG < 0. The process is spontaneous above 100 °C.
