The molar entropy values of substances in their standard states are known as standard molar entropies and are denoted So. The standard state for any substance is defined as the pure substance at 1 atm pressure.
Sample Molar Entropies of Some Selected Substances at 298 K
Substance So, J/mol-K
Gases
H2(g) 130.7
N2(g) 191.6
O2(g) 205.2
H2O(g) 188.8
NH3(g) 192.5
CH3OH(g) 237.6
C6H6(g) 269.2
Liquids
H2O (l) 69.9
CH3OH(l) 126.8
C6H6(l) 172.8
Solids
Li(s) 29.1
Na(s) 51.3
K(s) 64.7
Fe(s) 27.3
FeCl3(s) 142.3
NaCl(s) 72.3
Al(s) 28.3
Al2O3(s) 51.0
Based from the sample of molar entropies above, observations about So are the following:
1. Unlike enthalpies of formation, the standard molar entropies of elements at the reference temperature at 298 K are not zero.
2. The standard molar entropies of gases are greater than those of liquids and solids, consistent with the interpretation of experimental observation.
3. Standard molar entropies generally increase with increasing molar mass.
4. Standard molar entropies generally increase with an increasing number of atoms in the formula of the substance.
In summary, the number of significance of the vibrational degrees of freedom of molecules increase with the increasing mass and increasing number of atoms.
The entropy change in a chemical reaction can be calculated by the sum of the entropies of the products less the sum of the entropies of the reactants:
∆S = ∑nSo(products) - ∑mSo(reactants)
The coefficient n and m are the coefficient in the balanced chemical equation.
Example 1
Calculate the ∆So for the synthesis of ammonia from N2(g) and H2(g) at 298 K.
N2(g) + 3H2(g) → 2NH3(g)
Solution:
Using the formula above, substitute the value of the molar entropies of the substances given in the table above:
∆S = ∑nSo(products) - ∑mSo(reactants)
= 2So(NH3) - [So(N2) + 3So(H2)
= (2 mol) (192.5 J/mol-K) - [(1 mol) (191.6 J/mol-K) + (3 mol)(130.7 J/mol-K)]
= - 198.4 J/K
= (2 mol) (192.5 J/mol-K) - [(1 mol) (191.6 J/mol-K) + (3 mol)(130.7 J/mol-K)]
= - 198.4 J/K
Example 2
Using the standard entropies in the table presented above, calculate the standard entropy change, ∆So, for the following reaction at 298 K.
Al2O3(s) + 3H2(g) → 2Al(s) + 3H2O(g)
Solution:
∆S = ∑nSo(products) - ∑mSo(reactants)
= [2So(Al) + 3So (H2O)] - [So(Al2O3) + 3So (H2)]
= [(2 mol)(28.3 J/mol-K) + (3 mol)(188.8 J/mol-K)] - [(1 mol)(51.0 J/mol-K) +(3 mol) (130.7 J/mol-K)]
= +180.39 J/K
= [(2 mol)(28.3 J/mol-K) + (3 mol)(188.8 J/mol-K)] - [(1 mol)(51.0 J/mol-K) +(3 mol) (130.7 J/mol-K)]
= +180.39 J/K
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