Enthalpy change of neutralisation?

Question

Consider this problem.

The enthalpy of neutralisation is defined as the enthalpy change when one mole of water is formed through the react of an acid and base at standard conditions. As the coefficient of water in the products is 2. I feel like the answer is $\frac{x}{2}$. However, I feel confused. How do I know how many moles is formed. The reactions only give me the ratio. So even if this reaction produced one mole of water, it would still have the same ratio. Hence $x$ should be the answer. What's really going on?

Answer 1

Denoting $z$ the enthalpy value of the products, then, the enthalpy of neutralization can be calculated by the heights of the diagram \begin{align} \Delta_\pu{neut} H &= \text{Enthalpy of the products} - \text{Enthalpy of the reactants} \\ \Delta_\pu{neut} H &= z - (-x + z) \\ \Delta_\pu{neut} H &= x \end{align} By the information of the energy diagram, this corresponds to the following reaction $$2\ce{NaOH} + \ce{H2SO4} \rightarrow \ce{Na2SO4} + 2\ce{H2O} \hspace{1 cm} \Delta_\pu{neut} H $$ However, this reaction is not in agreement with your definition, so we have to "divide" the chemical equation by two $$ \ce{NaOH} + \frac{1}{2} \ce{H2SO4} \rightarrow \frac{1}{2}\ce{Na2SO4} + \ce{H2O} \hspace{1 cm} \frac{\Delta_\pu{neut} H}{2} $$ Therefore, the answer is $$ \boxed{\frac{\Delta_\pu{neut} H}{2} = \frac{x}{2}} $$