Chemistry 2006 JAMB Past Questions - Question 49: If the heat of combustion of hydrogen is-285.8kJ, what is the heat of formation of water?

Correct Answer

C

Explanation

The heat of formation (\( \Delta H_f \)) of a compound is the enthalpy change that occurs when one mole of the compound is formed from its elements in their standard states. The standard state for hydrogen is \( H_2(g) \) and for oxygen is \( O_2(g) \), both at 1 atmosphere pressure and 25°C.

The balanced chemical equation for the combustion of hydrogen to form water is:

\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]

The given heat of combustion of hydrogen (\( \Delta H_c \)) is -285.8 kJ. This value represents the energy released when one mole of hydrogen is burned.

For the reaction given above, the heat of combustion is equal to the enthalpy change of formation of water minus the enthalpy change of formation of the reactants:

\[ \Delta H_c = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants}) \]

In this case, the enthalpy change of formation of water (\( \Delta H_f(\text{water}) \)) is what we're looking for.

The enthalpy change of formation of \( H_2O \) on the product side is -285.8 kJ/mol (negative because it's an exothermic reaction). The enthalpy change of formation of \( H_2 \) and \( O_2 \) on the reactant side is zero by definition because they are in their standard states.

So, \[ \Delta H_c = \Delta H_f(\text{water}) \]

Therefore, the heat of formation of water (\( \Delta H_f(\text{water}) \)) is -285.8 kJ.