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BOND DISSOCIATION ENTHALPY (BOND ENERGY) MCQ
IIT JEE

The bond dissociation enthalpy of gaseous H₂, Cl₂ and HCl are 435, 243 and 431 kJ mol^-1, respectively. Calculate the enthalpy of formation of HCl gas.

(IIT JEE 1985)

Logic & Solution:

Bond dissociation enthalpy is defined as the amount of heat absorbed when one mole of specific bonds are cleaved homolytically in the gaseous state.

Note: There are other terms like bond enthalpy or bond energy or mean bond enthalpy, used through out the literature. These terms represent the average of bond dissociation enthalpies of same type of bonds present in molecules.

The thermochemical equations for bond dissociation of H₂, Cl₂ and HCl can be written as follows.

(1) H₂(g) --------> 2H(g) ; ΔH₁ = +435 kJ

(2) Cl₂(g) --------> 2Cl(g) ; ΔH₂ = +243 kJ

(3) HCl(g) --------> H(g) + Cl(g) ; ΔH₃ = +431 kJ

Thermochemical equation for the formation of HCl is:

0.5H₂(g) + 0.5Cl₂(g) --------> HCl(g) ; Δ_fH = ?

Above equation can be obtained by subtracting equation (3) from the sum of (1)/2+(2)/2

i.e., (1)/2 + (2)/2 - (1)

Therefore the Δ_fH for HCl gas = (ΔH₁/2)+(ΔH₂/2)-ΔH₃ = (435/2) + (243/2) - (431) = -92 kJ

Alternatively you can logically solve the problem as follows:

The formation of HCl from H₂ and Cl₂ occurs in the following steps:

i) the bonds between H-H and Cl-Cl are broken that require energy 

ii) formation of new bond between H and Cl, which involve liberation of heat.

Hence, according to Hess's law, the sum of enthalpy changes in above steps will give the enthalpy change for formation of HCl from H₂ and Cl₂.

i.e.,

Δ_fH = (435/2) + (243/2) + (-431) = -92 kJ

Note: -431 kJ is the negative of bond dissociation enthalpy as shown in above diagram, since heat is liberated during the formation of H-Cl bond. It is reverse of bond dissociation of HCl and hence is exothermic.

Related questions

1) Calculate the O-H bond energy from the following data.

0.5H₂(g) + 0.5O₂(g) ------> OH(g) ; ΔH = +42 kJ

H₂(g) --------> 2H(g) ; ΔH = +435 kJ

O₂(g) --------> 2O(g) ; ΔH = +495 kJ

(IIT JEE-1981) 

Solution:

The thermochemical equation for bond energy or bond formation enthalpy for O-H radical can be written as follows:

O(g) + H(g) -------> OH(g)

We can get this equation as follows:

[0.5H₂(g) + 0.5O₂(g) ------> OH(g)] - 0.5[H₂(g) --------> 2H(g)] - 0.5[O₂(g) --------> 2O(g)] 

Therefore:

bond energy =  +42 - 0.5(435) - 0.5(495) = -423 kJ

Alternatively, the bond energy can be calculated as follows:

O(g) + H(g) -------> OH(g) is the last step involved in the formation of OH from H₂ and O₂ as shown below.

According to Hess's law: The enthalpy change is same whether the reaction occurs in one step or in several steps. i.e. the enthalpy change in direct step is equal to the sum of enthalpy changes in indirect method.

435/2 kJ + 495/2 kJ + ΔH_OH = 42 kJ

or

ΔH_OH = 42 kJ - 435/2 kJ - 495/2 kJ = -423 kJ

Homework

1) What is enthalpy?

2) Why the enthalpy change in one reaction path is equal to that in another path?