Question

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) 0.027 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)

Answer 1

The standard entropy change of a reaction can be calculated as follows: ASn = sum of ASⓇ (products) - sum of AS (reactants) ...... (1) The reaction given in the question is as follows: 3NO, (g)+H2O(1) + 2HNO3(1) + NO(g) The standard thermodynamic values are as follows: AS NO2(g) = 240.0 J/Kmol ASⓇH,0(1) = 69.9 J/K.mol AS HNO, (1) =155.60J/K.mol AS°NO(g) = 210.7 J/Kmol Substitute these values in equation (1) AS® an = 2xAS HNO3 (1)+AS NO(g)] -[3xAS NO2(g) +ASº H,0(1) = [2x155.60J/K.mol + 210.7 J/Kmol]-[3x240.0 J/K.mol +69.9 J/K.mol] = [311.2J/K.mol+210.7 J/K.mol – 720J/K.mol - 69.9 J/K.mol] =-268 J/K.mol Therefore, the entropy change for the given reaction is found to be =-268 J/K.mol. (b) The reaction given in the question is as follows: N, (g)+372(g) + 2NF, (g) The standard thermodynamic values are as follows: AS NF, (g) = 260.8 J/K.mol ASF, (g) = 202.78 J/K mol ASON, (g) =191.6 J/K.mol Substitute these values in equation (1) AS® an = [2xAS°NF, (g)] -[AS°N, (g) +3XAS° F (8)] = [2x 260.8 J/K.mol]-[191.6 J/K.mol +3Ⓡ 202.78 J/K.mol] = [521.6J/Kmol-191.6J/K.mol -608.34J/Kmol] AS an=-278.34J/K.mol Therefore, the entropy change for the given reaction is found to be = -278.34 J/K.mol. The reaction given in the question is as follows: C.H20. (s) +602(g) +6C02(g) + 6H2O(g) The standard thermodynamic values are as follows: AS°CH,0, (s)=212.1 J/K mol AS O, (g) = 205.1 J/Kmol AS CO2 (g) = 213.7 J/K.mol AS°H,0(g)=188.8 J/Kmol Substitute these values in equation (1) As an DT6XAS CO2(g) +6XAS® H, 0(g)-TAS°CH,O. (s)+6x48°02 = [6x213.7 J/K.mol +6x188.8 J/Kmol]-[212.1 J/K.mol +6×205.1 J/K.mol] = [1282.2J/K.m ol+1132.8 J/Kmol - 212.1J/Kmol-1230.6J/K.mol] AS n =972.3J/K.mol Therefore, the entropy change for the given reaction is found to be = 972.3 J/K.mol.