Bond enthalpy is the energy required to break a chemical bond in a gaseous state at 298K. It is usually expressed in terms of KJ / mol. In other words, it can be stated as the energy needed to break one mole of the bond into its constituent atoms in a gaseous state. After the cleavage of the bond, the molecules dissociate into its free radicals. Bond enthalpy is also referred to as bond energy.
Enthalpy is the amount of heat energy contained in a system. The cleavage of the bond also requires some amount of energy. The process which absorbs the heat is called the endothermic process. Similarly, the breaking of bonds is an endothermic process as it requires energy to dissociate and form the constituent radicals. In short, we can say that breaking a bond is an endothermic process, while the formation of bonds is exothermic. To break a bond a positive change in enthalpy is required, while a negative change in enthalpy is accompanied by the formation of a bond.
The various factors that affect the bond enthalpy are:
Hess’ law of constant heat summation states is given by a Russian chemist Germain Hess. The Hess law is based on two other popular laws, the first law of thermodynamics and state function character of enthalpy. It states that “regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes”. It is the manifestation that enthalpy is a state function.
The enthalpy change for the overall process is the sum of the enthalpy change of the steps in the process:
Answer
Bond enthalpy is the energy required to break a chemical bond in a gaseous state at 298K. It is usually expressed in terms of KJ / mol. In other words, it can be stated as the energy needed to break one mole of the bond into its constituent atoms in a gaseous state.
Answer
The factors affecting bond enthalpy are:
Answer
It states that “regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes”. It is the manifestation that enthalpy is a state function.
2H2(g) + O2(g) ——> H2O(g) , DH = –115.4 kcal the bond energy of H–H and O = O bond respectively is 104 kcal and 119 kcal, then the calculate the O–H bond energy in water vapour?
Answer
We know that heat of reaction
ΔH = ΣB.E. (reactant) – ΣB.E (product)
For the given reaction,
Given data are:
ΔH = –115.4 kcal
B.E. of H–H = 104 kcal
B.E. of O=O = 119 kcal
Since water molecule contains two O–H bonds
–115.4 = (2 × 104) + 119 – 4 (O–H) bond energy
∴ 4 (O–H) bond energy = (2 × 104) + 119+115.4
Hence, O–H bond energy = (2×104) + 119 + 115.4 / 4 = 110.6 kcal mol–1
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