Example 5.6. Predicting the sign of entropy change

Problem: The reaction given is:

2 SO_2(g) + O_2(g) = 2 SO_3(g)

What should be the sign of the S for these reactions?

Solution: On the left side of the equation there are 3 moles of gas. On the right side of the equation there are only 2 moles of gas, so, the total amount of matter of gases in the system decreased and S is negative.

Answer: S  0.

Gibbs free energy. When a chemical reaction occurs, both enthalpy and entropy are changed. Therefore, both H and S should be considered to predict, whether a given reaction will occur spontaneously or not under a given set of conditions. A thermodynamic function, called Gibbs free energy combines enthalpy and entropy:

G = H  TS

A change in Gibbs free energy is calculated as

G = H  TS

ΔG is negative for spontaneous processes, positive for nonspontaneous processes and zero for equilibrium processes.

If a reaction is favorable for both enthalpy (H < 0) and entropy (S > 0) changes, then the reaction will be spontaneous (G < 0) at any temperature. If a reaction is unfavorable for both enthalpy (H > 0 ) and entropy (S < 0) changes, then the reaction will be non- spontaneous (G > 0) at any temperature.

Temperature and free energy. If a reaction is favorable for only one of either entropy or enthalpy, the free energy calculation must be used to determine whether the reaction is spontaneous or not. If a reaction is favorable for enthalpy (H < 0), but unfavorable for entropy (S < 0), then in the equation

G = H  TS

the factor TS will be negative. As the temperature increases, the TS factor increases as well. Therefore, G for this reaction will be negative at lower temperature.

Hr	Sr	Gr	Reaction
< 0	> 0	< 0	spontaneous at any temperature
> 0	< 0	> 0	non-spontaneous at any temperature
< 0	< 0		spontaneous at low temperature
> 0	> 0		spontaneous at high temperature

Example 5.7. Calculating H S and G for a reaction from H_fand S

Problem: Given the values of H_fand S

Substance	Hf, kJmol1.	Sf, Jmol1K1.
NaHCO3(s)	 947.7	155
Na2CO3(s)	 1131	136
CO2(g)	 394	213.6
H2O (g)	 242	188.7

calculate H S and G for the reaction

2 NaHCO_3(s) = Na₂CO_3(s) + CO_2(g) + H₂O_(g).

Is the reaction spontaneous at standard conditions?

Solution: To calculate H_r we use the equation

H°_r = ΣΔH_f °(products) – ΣΔH_f °(reactants).

H°_r = ΔH_f °Na₂CO_3(s) + ΔH_f °CO_2(g) + ΔH_f °H₂O_(g) – 2ΔH_f °NaHCO_3(s) =

=  1131 +( 394) + ( 242) 2( 947.7) = 128 kJ

To calculate S_r we use the equation

S°_r = ΣS°(products) – ΣS°(reactants).

S°_r = S°Na₂CO_3(s) + S°CO_2(g) + S° H₂O_(g) – 2S °NaHCO_3(s) = 136 + 213.6 + 188.7  2155 = = 228 J = 0.228 kJK^1

We converted obtained S°_r value to kJK^1 because we have H_r measured in kilojoules. To calculate G_r we use the equation

G_r° = H_r°  TS_r° = 128  2980.228 = 60 kJ

Answer: H°_r = 128 kJ, S°_r = 228 JK^1, G_r° = 60 kJ. The reaction is nonspontaneous at standard conditions.