Calculate enthalpy of neutralization when 0.60 mole of HNO3 neutralize 0.30 mole of NaNO3.

To calculate the enthalpy of neutralization for the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH), we first need to clarify the reaction involved. The neutralization reaction can be represented as follows:

Understanding the Reaction

The balanced chemical equation for the neutralization of nitric acid with sodium hydroxide is:

HNO3 (aq) + NaOH (aq) → NaNO3 (aq) + H2O (l)

In this reaction, one mole of HNO3 reacts with one mole of NaOH to produce one mole of sodium nitrate (NaNO3) and one mole of water (H2O). However, in your question, you mentioned 0.60 moles of HNO3 and 0.30 moles of NaNO3, which seems to indicate a misunderstanding since NaNO3 is a product, not a reactant. Instead, we should focus on the moles of HNO3 and NaOH involved in the reaction.

Identifying the Limiting Reactant

Given that you have 0.60 moles of HNO3, we need an equal amount of NaOH for complete neutralization. If we assume we have enough NaOH, the reaction will proceed as follows:

• 0.60 moles of HNO3 will require 0.60 moles of NaOH.
• Since you mentioned 0.30 moles of NaNO3, it seems we might be looking at a scenario where only half of the available HNO3 is neutralized.

Calculating Enthalpy of Neutralization

The enthalpy of neutralization (\( \Delta H_{neut} \)) is typically defined as the heat change when one mole of acid reacts with one mole of base to form water and a salt. This value is generally negative, indicating that the reaction is exothermic.

To find the enthalpy change, we can use the formula:

\( \Delta H_{neut} = \frac{q}{n} \)

Where:

• q = heat absorbed or released (in joules or kilojoules)
• n = number of moles of water produced (which is equal to the moles of acid or base that reacted)

Example Calculation

Assuming we have a standard enthalpy of neutralization for strong acids and bases, which is approximately -57.1 kJ/mol, we can calculate the enthalpy change for the reaction:

Since 0.30 moles of HNO3 would react with 0.30 moles of NaOH to produce 0.30 moles of water, we can calculate the enthalpy change for this amount:

\( \Delta H_{neut} = -57.1 \, \text{kJ/mol} \times 0.30 \, \text{mol} = -17.13 \, \text{kJ} \)

Final Thoughts

The enthalpy of neutralization for the reaction involving 0.30 moles of HNO3 and NaOH would be approximately -17.13 kJ. This indicates that the reaction releases energy, consistent with the exothermic nature of neutralization reactions. If you have any further questions or need clarification on any part of this process, feel free to ask!