Calclulate h for the reaction described by the equation.
ba(oh)2.8h2o + 2nh4scn ?
ba(scn)2 + 2nh3 +10h2o

ΔH° = 118.486 Kcal/mol

Explanation:

Ba(OH)2.8H2O + 2NH4SCN ↔ Ba(SCN)2 + 2NH3 + 10H2O

∴ ΔH°f H2O = - 68.3174 Kcal/mol frome literature

∴ ΔH°f NH3 = -19.82 Kcal/mol

∴ ΔH°f SCN-(aq) = 17.2 Kcal/mol

∴ ΔH°f NH4SCN = - 12.3 Kcal/mol

∴ ΔH°f Ba(OH)2.8H2O = - 799.5 Kcal/mol

⇒ ΔH° = 10ΔH°f H2O + 2ΔH°f NH3 + ΔH°f SCN-(aq) - 2ΔH°f NH4SCN - ΔH°f Ba(OH)2.8H2O

⇒ ΔH° = - 683.174 - 39.64 + 17.2 + 24.6 + 799.5

⇒ ΔH° = 118.486 Kcal/mol