Be sure to answer all parts. For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2 (g) + H2O (l) → 2 HNO3 (l) + NO (g) J/K (b) N2 (g) + 3 F2 (g) → 2 NF3 (g) J/K (c) C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (g) J/K

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