δh > 0; δs > 0; δg = 0
not spontaneous when t < 100 °c;
equilibrium when t = 100 °c
spontaneous when t > 100 °c
the process is
h₂o(ℓ) ⇌ h₂o(g)
δh > 0 (positive), because we must add heat to boil water
δs > 0 (positive), because changing from a liquid to a gas increases the disorder
δg = 0, because the liquid-vapour equilibrium process is at equilibrium 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.
i have know clue i guess aluminum
i think lol