Question

Does the standard enthalpy of formation of \({{\bf{H}}_{\bf{2}}}{\bf{O}}\left( {\bf{g}} \right)\)differ from ΔH° for the reaction \({\bf{2}}{{\bf{H}}_{\bf{2}}}\left( {\bf{g}} \right){\bf{ + }}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right) \to {\bf{2}}{{\bf{H}}_{\bf{2}}}{\bf{O}}\left( {\bf{g}} \right)\)?

Short answer

Yes, standard enthalpy of formation of \({{\rm{H}}_{\rm{2}}}{\rm{O(g) }}\)differs from the change in enthalpy of the reaction, \({\rm{2H}}{}_{\rm{2}}{\rm{(g) + }}{{\rm{O}}_{\rm{2}}}{\rm{(g) }} \to {\rm{ 2}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\)_.

Step-by-step solution

Enthalpy change of the reaction

For answering the question, we first have to calculate the enthalpy of the given reaction.

2H₂O(g)+O₂(g) → 2H₂O(g)

We are going to use the formula, ΔH⁰_reaction = ∑ ΔH⁰_product-∑ ΔH⁰_reactant

⇒ΔH⁰_reaction = (2xΔH₂O(g)) - (2xΔH₂(g)+ΔO(g))

⇒ΔH⁰_reaction = (2 x -241.82kJmol^-1) - (2 x 0+0)

⇒ΔH⁰_reaction = -483.64kJmol^-1

Hence, the enthalpy change in the reaction is \( - 483.64k{\rm{J mo}}{{\rm{l}}^{ - 1}}\)_.

Enthalpy of formation

Now, we are going to evaluate the enthalpy of formation of \({{\rm{H}}_{\rm{2}}}{\rm{O(g) }}\)_.

\(\begin{array}{l}{\rm{The formation reaction of water, }}\\{{\rm{H}}_{\rm{2}}}{\rm{(g) + }}\frac{1}{2}{{\rm{O}}_{\rm{2}}}{\rm{(g) }} \to {\rm{ }}{{\rm{H}}_{\rm{2}}}{\rm{O(g) }}\\\therefore \Delta {{\rm H}_{reaction}} = \Delta {{\rm{H}}_{{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}}} - (\Delta {{\rm H}_{{{\rm{H}}_{\rm{2}}}{\rm{(g)}}}} + \frac{1}{2}\Delta {{\rm H}_{{{\rm{O}}_{\rm{2}}}{\rm{(g)}}}})\\ \Rightarrow \Delta {{\rm H}_{reaction}} = - 241.82{\rm{ KJ mo}}{{\rm{l}}^{ - 1}} - (0 + 0)\\ \Rightarrow \Delta {{\rm H}_{reaction}} = - 241.82{\rm{ KJ mo}}{{\rm{l}}^{ - 1}}\end{array}\)

Hence, the enthalpy of formation of \({{\rm{H}}_{\rm{2}}}{\rm{O(g) }}\)is \( - 241.82{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\)_.

Finally, we can say that the standard enthalpy of formation for\({{\bf{H}}_{\bf{2}}}{\bf{O(g) }}\)differs from the change in enthalpy of the reaction.