Question

When chlorine gas and hydrogen react together to form hydrogen chloride, what will be the change of enthalpy of the reaction?
A.$+ 245\;{\text{kJ/mol}}$
B.$+ 185{\text{ kJ/mol}}$
C.$- 185{\text{ kJ/mol}}$
D.$- 1105{\text{ kJ/mol}}$
E.$+ 1105{\text{ kJ/mol}}$

Answer 1

The bond enthalpy of a reaction is defined as the enthalpy change that takes place due to the formation of one mole bonds from isolated atoms present in their gaseous states. We shall substitute the values in the equation given.

Formula: ${\Delta E = }{{\text{E}}_{\text{p}}} - {{\text{E}}_{\text{r}}}$

Complete step by step answer:
Whenever the bonds in any diatomic molecule are broken, the bond energy or the bond enthalpy is equal to the bond dissociation energy of the molecules. Both chlorine gas and hydrogen gas have single bonds which are non-polar in nature. The bond dissociation energy of chlorine is $+ 243{\text{ kJ/mol}}$, that for hydrogen is $+ 436{\text{ kJ/mol}}$, and that for hydrogen chloride is $+ 432{\text{ kJ/mol}}$. The reaction proceeds as follows:
${H_2} + C{l_2} \to 2HCl$
As the bond enthalpy is ${\Delta E = }{{\text{E}}_{\text{p}}}{\text{ - }}{{\text{E}}_{\text{r}}}$, so the enthalpy change is,
${\Delta E = }\left( {432 \times 2} \right) - \left( {436 + 243} \right)$= $+ 185{\text{ kJ/mol}}$.

Hence, the correct answer is option B.

Notes: The bond dissociation energy is the energy required to break one mole of bonds to give separated atoms in the gaseous state. The point about everything being in the gaseous state is very important., because energy is also required to change the state and this energy of this enthalpy is called the heat of atomization or the enthalpy of atomization. Hence it is important that the material be already present in the gaseous state. For complex molecules like methane where there are different types of bonds, the mean of the bond dissociation energy is calculated.