Question

Enthalpy change for the reaction $ 4H(g) \to 2{H_2}(g) $ is $ - 869.6kJ $ . The dissociation energy of $ H - H $ bond is?

Answer 1

Hint: On reversing a chemical equation, the enthalpy associated with the reaction becomes opposite in sign but remains same in magnitude and on multiplying an equation with a particular number, the magnitude of enthalpy is also multiplied with the same number to get the enthalpy of the new reaction produced.

Complete answer:
The enthalpy of reaction is the total heat change involved in the conversion of reactants to the products carried out at a constant pressure.
The dissociation enthalpy is the total amount of heat needed to break apart a particular bond at constant pressure.
The $ H - H $ bond consists of two atoms belonging to the same element connected to each other through a strong covalent bond. The enthalpy of dissociation needed to segregate the two hydrogen atoms from each other will be called the dissociation energy of this bond. Hence, the chemical equation that represents dissociation can be written as follows:
 $ {H_2}(g)\xrightarrow{{dissociation}}2H(g) $
The above equation can be derived from the chemical equation given in the question (for which the enthalpy of reaction is known) by reversing the equation and multiplying a stoichiometric factor of $ \dfrac{1}{2} $ . On doing so we get the same equation as that of the dissociated reaction needed. Thus, the enthalpy becomes positive is sign (sign reversed) and half in magnitude.
Therefore, the enthalpy of bond dissociation of hydrogen-hydrogen bond is,
 $ {H_{H - H}} = 869.6kJ \times \dfrac{1}{2} = 434.8kJ $ .

Note:
Enthalpy is the heat changes involved during a reaction at constant pressure and internal energy is associated with the heat changes measured at constant volume. The enthalpy measurements are always preferred as the thermodynamic data rather than internal energy because reactions are easier to carry out under constant pressure conditions as compared to reactions carried out at constant volume.