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IIT JEE

(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 H_{2}, Cl_{2} and HCl can be written as follows.

(1) H_{2}(g) --------> 2H(g) ; ΔH_{1} = +435 kJ

(2) Cl_{2}(g) --------> 2Cl(g) ; ΔH_{2} = +243 kJ

(3) HCl(g) --------> H(g) + Cl(g) ; ΔH_{3} = +431 kJ

Thermochemical equation for the formation of HCl is:

0.5H_{2}(g) + 0.5Cl_{2}(g) --------> HCl(g) ; Δ_{f}H = ?

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_{f}H for HCl gas = (ΔH_{1}/2)+(ΔH_{2}/2)-ΔH_{3} = (435/2) + (243/2) - (431) = -92 kJ

**Alternatively you can logically solve the problem as follows:**

The formation of HCl from H_{2} and Cl_{2} 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 H_{2} and Cl_{2}.

i.e.,

Δ_{f}H = (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.

** H _{2}(g) --------> 2H(g) ; ΔH = +435 kJ**

**O _{2}(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:

[0.5H_{2}(g) + 0.5O_{2}(g) ------> OH(g)] - 0.5[H_{2}(g) --------> 2H(g)] - 0.5[O_{2}(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 H_{2} and O_{2} 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

1) What is enthalpy?

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

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Author: Aditya vardhan Vutturi