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Boiling point relation
T_b = ∆H_vap / ∆S_vap
∆G° and K
∆G° = −RT ln K
Non-standard ∆G
∆G = ∆H − T∆S; ∆G = ∆G° + RT ln Q
Spontaneity by ∆G
∆G < 0 spontaneous; ∆G > 0 non-spontaneous
Equilibrium value of ∆G
∆G = 0
Second Law condition
∆S_univ = ∆S_sys + ∆S_surr > 0 for spontaneous
Surroundings entropy formula
∆S_surr = −∆H_sys / T
Exothermic effect on surroundings
∆H < 0 → ∆S_surr > 0 (helps spontaneity)
Endothermic effect on surroundings
∆H > 0 → ∆S_surr < 0 (hurts spontaneity)
Temperature switch (T*)
T* = ∆H / ∆S (use J and K)
Always spontaneous combo
∆H < 0 and ∆S > 0
Never spontaneous combo
∆H > 0 and ∆S < 0
Low-T spontaneous combo
∆H < 0 and ∆S < 0
High-T spontaneous combo
∆H > 0 and ∆S > 0
Entropy: phase ↑ disorder
solid → liquid → gas gives +∆S
Entropy: phase ↓ disorder
gas → liquid/solid gives −∆S
More gas moles in products
∆S > 0 (entropy increases)
Compressing gases
∆S < 0 (entropy decreases)
Dissolving crystalline solid
Usually ∆S > 0
Unit rule for calculations
Convert kJ→J; use T in Kelvin
At T* what is ∆G?
∆G = 0 (equilibrium threshold)
Why engines aren't 100% efficient
Some energy disperses; ∆S_univ must increase
"Hot loves chaos" meaning
High temperature favors processes with +∆S
Quick ∆S_surr tip
Bigger |∆H| or lower T → larger |∆S_surr| magnitude