2 B (s) + 3 H2 (g) → B2H6 (g) ΔrHº = ??? Use the following information to solve for the enthalpy of reaction of the above equation: 4 B (s) + 3 O2 (g) → 2 B2O3 (g) ΔrHº = -2544 kJ/mol H2 (g) + 1/2 O2 (g) → H2O (g) ΔfHº = -242 kJ/mol B2H6 (g) + 3 O2 (g) → B2O3 (g) + 3 H2O (g) ΔrHº = -2033 kJ/mol a) 35 kJ/mol b) -4019 kJ/mol c) -269 kJ/mol d) 753 kJ/mol
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.
2 B (s) + 3 H2 (g) → B2H6 (g) ΔrHº = ???
Use the following information to solve for the enthalpy of reaction of the above equation:
4 B (s) + 3 O2 (g) → 2 B2O3 (g) ΔrHº = -2544 kJ/mol
H2 (g) + 1/2 O2 (g) → H2O (g) ΔfHº = -242 kJ/mol
B2H6 (g) + 3 O2 (g) → B2O3 (g) + 3 H2O (g) ΔrHº = -2033 kJ/mol
a) 35 kJ/mol
b) -4019 kJ/mol
c) -269 kJ/mol
d) 753 kJ/mol
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