Question

Bond dissociation enthalpy of H2 , Cl2 and HCl are 434, 242 and 431 kJ mol–1 respectively. Enthalpy of formation of HCl is -

Answer 1

Answer:-

The correct option is: (a) -93 kJ mol-1

Explanation:

H2 + Cl2⇒2HCl

ΔHreaction = ∑(B.E)reactant - ∑(B.E)product

=[(B.E)H-H + (B.E)Cl-Cl]-[2B.E(H-Cl)]

=434 +242 - (431 × 2)

=676 - 862

=-186 KJ

Heat of formation is the amount of heat absorbed or evolved when one mole of substance is directly obtained from its constituent element.

Hence, enthalpy of formation of HCI = -186/2 kJ

                                                               = -93 kJ mol-1

Answer 2

Answer:

The enthalpy of the formation of HCl is -93kJmol⁻¹

Explanation:

Bond dissociation enthalpy of H-H=434 kJmol⁻¹

Bond dissociation enthalpy of Cl-Cl=242 kJmol⁻¹

Bond dissociation enthalpy of H-Cl=431 kJmol⁻¹

We know that,

ΔHreaction=∑ Bond energy of reactant - ∑ Bond energy of the product

∴ΔHreaction=$\frac{1}{2}$ΔH(H-H)+$\frac{1}{2}$ΔH(Cl-Cl) - ΔH(H-Cl)

                    = $\frac{1}{2}$*434+$\frac{1}{2}$*242 -431

                    =217+121-431

                    = -93kJmol⁻¹

Therefore, the enthalpy of the formation of HCl is -93kJmol⁻¹