Using data in table below and S °(NH3(g))= 192.5 J·mol-1·K-1, calculate ÄfS° for NH3(g) in J·mol-1·K-1.
Compound S °, J·mol-1·K-1 Compound S °, J·mol-1·K-1
C(s) 5.69 H2(g) 130.6
N2(g) 191.5 O2(g) 205.0
Na(s) 51.0 Cl2(g) 223.0
Ca(s) 154.8 S(s) 31.8
To calculate ÄfS° for NH3(g), we need to use the standard molar entropies (S °) of the reactants and products.
The reactants are N2(g) and H2(g), and the product is NH3(g).
From the table:
S °(N2(g)) = 191.5 J·mol-1·K-1
S °(H2(g)) = 130.6 J·mol-1·K-1
S °(NH3(g)) = 192.5 J·mol-1·K-1
The change in entropy (ÄS°) for the reaction is given by the formula:
ÄS° = S °(products) - S °(reactants)
For NH3(g), the change in entropy is:
ÄS°(NH3(g)) = S °(NH3(g)) - [S °(N2(g)) + 3*S °(H2(g))]
Substituting the values:
ÄS°(NH3(g)) = 192.5 - [191.5 + 3*130.6]
ÄS°(NH3(g)) = 192.5 - [191.5 + 391.8]
ÄS°(NH3(g)) = 192.5 - 583.3
ÄS°(NH3(g)) = -390.8 J·mol-1·K-1
Therefore, the ÄfS° for NH3(g) is -390.8 J·mol-1·K-1.
To calculate the standard molar entropy change (ΔfS°) for NH3(g), we need to use the following equation:
ΔfS° = ΣnΔS°(products) - ΣmΔS°(reactants)
Where:
ΔfS° is the standard molar entropy change for NH3(g)
ΣnΔS°(products) is the sum of the standard molar entropies of the products, each multiplied by their respective stoichiometric coefficients
ΣmΔS°(reactants) is the sum of the standard molar entropies of the reactants, each multiplied by their respective stoichiometric coefficients
Given the reactant and product equation for the formation of NH3(g) is:
N2(g) + 3H2(g) → 2NH3(g)
We can calculate ΣnΔS°(products) by using the standard molar entropy values for NH3(g) and applying the stoichiometric coefficient:
ΣnΔS°(products) = 2 × S°(NH3(g))
Similarly, we can calculate ΣmΔS°(reactants) by using the standard molar entropy values for N2(g) and H2(g) and applying their respective stoichiometric coefficients:
ΣmΔS°(reactants) = S°(N2(g)) + 3 × S°(H2(g))
Now, let's plug in the given values:
S°(NH3(g)) = 192.5 J·mol-1·K-1
S°(N2(g)) = 191.5 J·mol-1·K-1
S°(H2(g)) = 130.6 J·mol-1·K-1
ΣnΔS°(products) = 2 × 192.5 J·mol-1·K-1 = 385 J·mol-1·K-1
ΣmΔS°(reactants) = 191.5 J·mol-1·K-1 + 3 × 130.6 J·mol-1·K-1 = 583.3 J·mol-1·K-1
Finally, we can calculate ΔfS° for NH3(g):
ΔfS° = ΣnΔS°(products) - ΣmΔS°(reactants)
= 385 J·mol-1·K-1 - 583.3 J·mol-1·K-1
= -198.3 J·mol-1·K-1
Therefore, the ΔfS° for NH3(g) is -198.3 J·mol-1·K-1.