Given that ?Hf for OH- ions is -229.6 kJ/mol, calculate the enthalpy of neutralization when 1 mol of a strong monoprotic acid (such as HCl) is titrated by 1 mol of a strong base (such as KOH) at 25 degrees C
To calculate the enthalpy of neutralization, we first need to understand the concept. The enthalpy of neutralization is the heat evolved or absorbed when one mole of a substance reacts with one mole of a base to form a neutral product, usually water. In this case, we are considering the reaction between a strong monoprotic acid (HCl) and a strong base (KOH).
The balanced chemical equation for the neutralization reaction is:
HCl (aq) + KOH (aq) → KCl (aq) + H2O (l)
Now, the enthalpy change of the reaction (∆Hrxn) can be determined using Hess's law, which states that the enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states.
Here's how we can calculate the enthalpy of neutralization:
1. Calculate the enthalpy change for the formation of water (∆Hf) using the enthalpy of formation of hydroxide ions (∆Hf[OH-]). The balanced equation shows that one mole of water is formed in the reaction, so the enthalpy change of formation for water is equal to ∆Hf[OH-].
2. Since we know that ∆Hf[OH-] = -229.6 kJ/mol, we can take ∆Hf[H2O] = -229.6 kJ/mol.
3. As per the balanced equation, the stoichiometric coefficient of water is 1, so the enthalpy change of neutralization is ∆Hrxn = ∆Hf[H2O] = -229.6 kJ/mol.
Therefore, the enthalpy of neutralization when 1 mol of a strong monoprotic acid is titrated by 1 mol of a strong base at 25 degrees C is -229.6 kJ/mol.