Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g)+O2(g)→NO2(g), ΔH∘A=33.2 kJ 12N2(g)+12O2(g)→NO(g), ΔH∘B=90.2 kJ Calculate the enthalpy of the reaction 4B(s)+3O2(g)→2B2O3(s) given the following pertinent information: B2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔH∘A=+2035 kJ 2B(s)+3H2(g)→B2H6(g), ΔH∘B=+36 kJ H2(g)+12O2(g)→H2O(l), ΔH∘C=−285 kJ H2O(l)→H2O(g), ΔH∘D=+44 kJ
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
Calculate the enthalpy of the reaction
2NO(g)+O2(g)→2NO2(g)
given the following reactions and enthalpies of formation:
- 12N2(g)+O2(g)→NO2(g), ΔH∘A=33.2 kJ
- 12N2(g)+12O2(g)→NO(g), ΔH∘B=90.2 kJ
Calculate the enthalpy of the reaction
4B(s)+3O2(g)→2B2O3(s)
given the following pertinent information:
- B2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔH∘A=+2035 kJ
- 2B(s)+3H2(g)→B2H6(g), ΔH∘B=+36 kJ
- H2(g)+12O2(g)→H2O(l), ΔH∘C=−285 kJ
- H2O(l)→H2O(g), ΔH∘D=+44 kJ
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Calculate the enthalpy of the reaction
4B(s)+3O2(g)→2B2O3(s)4B(s)+3O2(g)→2B2O3(s)
given the following pertinent information:
- B2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔH∘A=+2035 kJB2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔHA∘=+2035 kJ
- 2B(s)+3H2(g)→B2H6(g), ΔH∘B=+36 kJ2B(s)+3H2(g)→B2H6(g), ΔHB∘=+36 kJ
- H2(g)+12O2(g)→H2O(l), ΔH∘C=−285 kJH2(g)+12O2(g)→H2O(l), ΔHC∘=−285 kJ
- H2O(l)→H2O(g), ΔH∘D=+44 kJ
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