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Chemistry · Ch 7 — Thermodynamics

Criteria for Spontaneity of a Process

7.11.1

Criteria for Spontaneity of a Process

Spontaneity of any process depends on three interacting factors, all of which are pulled together in the criterion at the end:

  • If a process's enthalpy change is NEGATIVE (exothermic), the process MAY be spontaneous.
  • If a process's entropy change is POSITIVE, the process MAY occur spontaneously.
  • Gibbs free energy combines both of these into the single necessary-and-sufficient condition: for a reaction to be spontaneous, ΔH−TΔS\Delta H-T\Delta S must be negative, i.e. ΔG<0\Delta G<0.

Table 7.5 works through all four sign combinations of ΔHr\Delta H_r and ΔSr\Delta S_r (assuming both stay roughly constant across the temperature range considered):

  • ΔHr\Delta H_r negative, ΔSr\Delta S_r positive ⇒\Rightarrow ΔGr\Delta G_r negative at ALL temperatures -- spontaneous at all temperature (example: 2O3(g)→3O2(g)2O_3(g)\rightarrow3O_2(g)).
  • ΔHr\Delta H_r negative, ΔSr\Delta S_r negative ⇒\Rightarrow ΔGr\Delta G_r negative at LOW TT, positive at HIGH TT -- spontaneous at low temperature, non-spontaneous at high temperature (example: adsorption of gases).
  • ΔHr\Delta H_r positive, ΔSr\Delta S_r positive ⇒\Rightarrow ΔGr\Delta G_r positive at LOW TT, negative at HIGH TT -- non-spontaneous at low temperature, spontaneous at high temperature (example: melting of a solid).
  • ΔHr\Delta H_r positive, ΔSr\Delta S_r negative ⇒\Rightarrow ΔGr\Delta G_r positive at ALL temperatures -- non-spontaneous at all temperatures (example: 2H2O(g)+O2(g)→2H2O2(l)2H_2O(g)+O_2(g)\rightarrow2H_2O_2(l)).

Two caveats the book is careful to add: this table assumes ΔH\Delta H and ΔS\Delta S genuinely stay the way indicated across the whole temperature range -- which is not always true in practice. And "spontaneity" of a reaction only describes the reaction's THERMODYNAMIC POTENTIAL to proceed as written; the actual RATE at which such a process happens is governed by kinetic factors that lie entirely outside thermodynamic prediction. …

Table 7.5Effect of Temperature on Spontaneity of Reactions
ΔHr\Delta H_rΔSr\Delta S_rΔGr=ΔHr−TΔSr\Delta G_r=\Delta H_r-T\Delta S_rDescriptionExample
−-++−- (at all T)Spontaneous at all temperature2O3(g)→3O2(g)2O_3(g)\rightarrow3O_2(g)
−-−-−- (at low T), ++ (at high T)Spontaneous at low temperature; non-spontaneous at high temperatureAdsorption of gases
++++++ (at low T), −- (at high T)Non-spontaneous at low temperature; spontaneous at high temperatureMelting of a solid
Misc 7.8Problem 7.8 -- spontaneity of CO oxidation from $\Delta G_f^0$

Worked out. CO+12O2→CO2CO+\tfrac{1}{2}O_2\rightarrow CO_2 at 300 K; ΔGf0(CO2)=−394.4\Delta G_f^0(CO_2)=-394.4, ΔGf0(CO)=−137.2\Delta G_f^0(CO)=-137.2 kJ mol−1^{-1} (O2O_2 zero). ΔGreaction0=ΔGf0(CO2)−[ΔGf0(CO)+0]=−394.4−(−137.2)=−257.2\Delta G_{reaction}^0=\Delta G_f^0(CO_2)-[\Delta G_f^0(CO)+0]=-394.4-(-137.2)=-257.2 kJ mol−1^{-1} -- negative, so the reaction is spontaneous at 300 K. …

Misc evaluate-yourself-8Evaluate Yourself 8 -- $\Delta G$ at 300 K and 600 K

Worked out. Book's practice box (no printed solution): ΔH=−10\Delta H=-10 kJ mol−1^{-1}, ΔS=−20\Delta S=-20 J deg−1^{-1}mol−1^{-1} at 300 K; find ΔG\Delta G at 300 K, and at 600 K (assuming ΔH,ΔS\Delta H,\Delta S constant); predict the reaction's nature. Working it through: at 300 K, ΔG=−10,000−300(−20)=−4000\Delta G=-10{,}000-300(-20)=-4000 J =−4=-4 kJ mol−1^{-1} (spontaneous); at 600 K, ΔG=−10,000−600(−20)=+2000\Delta G=-10{,}000-600(-20)=+2000 J =+2=+2 kJ mol−1^{-1} (non-spontaneous). Since ΔH<0\Delta H<0 and ΔS<0\Delta S<0, the reaction is spontaneous only at LOW temperature -- matching Table 7.5's second row (own solution, not …