Question

If the bond energies of $H-H$,$Br-Br$ and $H-Br$ are 433,192 and 364 $k{{J}_{{}}}mo{{l}^{-1}}$ respectively. Then $\Delta {{H}^{o}}$ for the reaction ${{\text{H}}_{2}}\text{ + B}{{\text{r}}_{2(g)}}\text{ }\to \text{ 2HB}{{\text{r}}_{(g)}}$ is:

Answer 1

Hint: $\Delta {{H}^{o}}$ for a reaction can be calculated by finding the difference between the energy possessed by the products and the energy possessed by the reactants. In this case the sum of energies possessed by the individual bonds of the molecules is the total energy for both products and reactants.

Complete answer:
Enthalpy ($\Delta H$) is a state function widely used in thermodynamic systems. It is defined as the sum of the internal energy of a system and product of its volume and pressure. It is a state function and has a unit similar to energy. Hence enthalpy is calculated in joules.
Let us now try to calculate the change in enthalpy for the above chemical reaction:
Energy of $H-H$ bond = 433 kJ
Energy of $Br-Br$ bond = 192 kJ
Energy of $H-Br$ bond = 364 kJ
The total change in enthalpy is equal to the difference between the bond energy of reactants and bond energy of products.
$\Delta {{\text{H}}^{o}}\text{ = }\sum{\text{ bond energy(reactants) - }}\sum{\text{ bond energy(products)}}$
$\Delta {{H}^{o}}$ = 433 + 192 - (2x 364) kJ = -103 kJ
The enthalpy for the above reaction is -103kJ.

Therefore, the correct answer is option (D).

Note:
The enthalpy for a chemical reaction is the difference between the enthalpy of formation of products and enthalpy of formation of reactants. The enthalpy of formation for naturally existing molecules like ${{O}_{2}}$,${{H}_{2}}$ is taken as 0. The magnitude of change in enthalpy of a chemical reaction is equal to the amount of heat released in the reaction.