Answer:
a. -268.13 J/K
b. -279.95 J/K
c. + 972.59 J/K
Explanation:
The value of the change in entropy (ΔS°) can be calculated by:
ΔS° = ∑n*S° products - ∑n*S° reactants, where n is the stoichiometric number of moles.
The values of S° for each substance can be found on a thermodynamic table.
a. 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
S°, NO2(g) = 240.06 J/mol.K
S°, H2O(l) = 69.91 J/mol.K
S°, HNO3(l) = 155.60 J/mol.K
S°, NO(g) = 210.76 J/mol.K
ΔS° = (210.76 + 2*155.60) - (3*240.06 + 69.91)
ΔS° = -268.13 J/K
b. N2(g) + 3F2(g) → 2NF3(g)
S°, N2(g) = 191.61 J/mol.K
S°, F2(g) = 202.78 J/mol.K
S°, NF3(g) = 260.0 J/mol.K
ΔS° = (2*260.0 ) - (191.61 + 3*202.78)
ΔS° = -279.95 J/K
c. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)
S°, C6H12O6(s) = 212 J/mol.K
S°, O2(g) = 205.138 J/mol.K
S°, CO2(g) = 213.74 J/mol.K
S°, H2O(g) = 188.83 J/mol.K
ΔS° = (6*213.74 + 6*188.83) - (212 + 6*205.138)
ΔS° = +972.59 J/K
The change in entropy is obtained from;ΔS° = ∑n*S° products - ∑n*S° reactants
Entropy is the degree of disorderliness in a system. To obtain the change in entropy (ΔS°) we have: ΔS° = ∑n*S° products - ∑n*S° reactants, where n is the stoichiometric coefficient of species in the reaction.
From the thermodynamic table, we have the values to use in the following calculations:
i. 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
NO2(g) = 240.06 J/mol.K
H2O(l) = 69.91 J/mol.K
HNO3(l) = 155.60 J/mol.K
NO(g) = 210.76 J/mol.K
ΔS° = (210.76 + 2*155.60) - (3*240.06 + 69.91)
ΔS° = -268.13 J/K
ii. N2(g) + 3F2(g) → 2NF3(g)
N2(g) = 191.61 J/mol.K
F2(g) = 202.78 J/mol.K
NF3(g) = 260.0 J/mol.K
ΔS° = (2*260.0 ) - (191.61 + 3*202.78)
ΔS° = -279.95 J/K
iii. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)
C6H12O6(s) = 212 J/mol.K
O2(g) = 205.138 J/mol.K
CO2(g) = 213.74 J/mol.K
H2O(g) = 188.83 J/mol.K
ΔS° = (6*213.74 + 6*188.83) - (212 + 6*205.138)
ΔS° = +972.59 J/K
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