2nd and 3rd Law of Thermodynamics Summary

Key Concepts and Summary

The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, Suniv > 0. If ΔSuniv < 0, the process is nonspontaneous, and if ΔSuniv = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process.

Practice Questions

What is the difference between ΔS and ΔS° for a chemical change?

Calculate ΔS° for the following changes.

(a) SnCl4(l)⟶SnCl4(g)

(b) CS2(g)⟶CS2(l)

(c) Cu(s)⟶Cu(g)

(d) H2O(l)⟶H2O(g)  

(e) 2H2(g)+O2(g)⟶2H2O(l)

(f) 2HCl(g)+Pb(s)⟶PbCl2(s)+H2(g)  

(g) Zn(s)+CuSO4(s)⟶Cu(s)+ZnSO4(s)  


(a) 107 J/K; (b) −86.4 J/K; (c) 133.2 J/K; (d) 118.8 J/K; (e) −326.6 J/K; (f) −171.9 J/K; (g) −7.2 J/K

Determine the entropy change for the combustion of liquid ethanol, C2H5OH, under the standard conditions to give gaseous carbon dioxide and liquid water.

Determine the entropy change for the combustion of gaseous propane, C3H8, under the standard conditions to give gaseous carbon dioxide and water.


100.6 J/K

“Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is: $$Fe_2O_3 (s)+2Al(s)\rightarrow Al_2O_3 (s)+2 Fe(s)$$

Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat.

Using the relevant S° values listed in this Appendix, calculate ΔS°298 for the following changes:

(a) N2(g)+3H2(g)⟶2NH3(g)

(b) N2 (g) + 5/2 O2 (g)⟶N2 O5 (g)


(a) −198.1 J/K; (b) −348.9 J/K

From the following information, determine ΔS° for the following: $$N(g)+O(g)\rightarrow NO(g)\qquad ΔS°=? $$ $$N_2(g)+O_2(g)⟶2NO(g)\qquad ΔS°=24.8 J/K$$ $$N_2(g)⟶2N(g)\qquad ΔS°=115.0 J/K$$ $$O_2(g)⟶2O(g)\qquad ΔS°=117.0 J/K$$

By calculating ΔSuniv at each temperature, determine if the melting of 1 mole of NaCl(s) is spontaneous at 500 °C and at 700 °C.

$$ \Delta S_{NaCl(s)}°=72.11\; Jmol·K \\ \Delta S_{NaCl(l)}°=95.06\; Jmol·K \\ \Delta H_{fusion}°=27.9\; kJ/mol$$


What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?


As ΔSuniv < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.

Use the standard entropy data in the Appendix to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and 25 °C.

(a) MnO2(s)⟶Mn(s)+O2(g)

(b) H2(g)+Br2(l)⟶2HBr(g)

(c) Cu(s)+S(g)⟶CuS(s)

(d) 2LiOH(s)+CO2(g)⟶Li2CO3(s)+H2O(g)

(e) CH4(g)+O2(g)⟶C(s,graphite)+2H2O(g)

(f) CS2(g)+3Cl2(g)⟶CCl4(g)+S2Cl2(g)