Calculate the enthalpy of the reaction 4 B (s) + 3 O2 (g) → 2 B203 (s) given the following pertinent information: 1. B203 (s) + 3 H20 (g) → 3 02 (g) + B2H6 (g), AH°A = +2035 kJ 2. 2 B (s) + 3 H2 (g) → B2H6 (g), AH°B = +36 kJ 3. H2 (g) + 02 (g) → H20 (1), AH°C = -285 kJ 4. H20 (I) → H20 (g), AH°D = +44 kJ Express your answer numerically, to four significant figures and in terms of 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.
Target equation is
4B(s) + 3O2(g) ------> 2B2O3(s) ; ∆H° = ?
Given thermo chemical reactions are
B2O3(s) + 3H2O(g) ------> 3O2(g) + B2H6(g); ∆H°A = +2035 kJ ------(i)
2 B(s) + 3 H2(g) ------> B2H6(g) ;
∆H°B = +36 kJ ------(ii)
H2(g) + 1/2 O2(g) -------> H2O(l) ; ∆H°C = -285 kJ----(iii)
H2O(l) ------> H2O(g) ; ∆H°D = +44 kJ -----(iv)
We should get Target equation by using these four thermo chemical reactions .
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