100 ml of 0.1 N H2so4react with 100ml of 0.15N NaoH then enthalpy of neutralisation is?

To determine the enthalpy of neutralization when 100 ml of 0.1 N H₂SO₄ reacts with 100 ml of 0.15 N NaOH, we first need to understand the concept of neutralization and how to calculate the enthalpy change associated with it.

Understanding Neutralization

Neutralization is a chemical reaction between an acid and a base, resulting in the formation of water and a salt. In this case, sulfuric acid (H₂SO₄) reacts with sodium hydroxide (NaOH). The balanced chemical equation for this reaction is:

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

Calculating Moles of Reactants

Next, we need to calculate the number of moles of each reactant. The normality (N) of a solution indicates the number of equivalents of solute per liter of solution. For sulfuric acid, which is a diprotic acid, 1 N corresponds to 1 mole of H⁺ ions. Therefore:

• For H₂SO₄: 0.1 N means 0.1 equivalents per liter. In 100 ml (0.1 L), the moles of H₂SO₄ are:
• 0.1 N × 0.1 L = 0.01 equivalents of H₂SO₄.
• Since H₂SO₄ provides 2 moles of H⁺, the moles of H₂SO₄ are:
• 0.01 equivalents / 2 = 0.005 moles of H₂SO₄.

• For NaOH: 0.15 N means 0.15 equivalents per liter. In 100 ml (0.1 L), the moles of NaOH are:
• 0.15 N × 0.1 L = 0.015 equivalents of NaOH.
• Since NaOH provides 1 mole of OH⁻, the moles of NaOH are:
• 0.015 moles of NaOH.

Determining the Limiting Reactant

In this reaction, we have:

• 0.005 moles of H₂SO₄
• 0.015 moles of NaOH

According to the stoichiometry of the reaction, 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore, to fully react with 0.005 moles of H₂SO₄, we need:

0.005 moles H₂SO₄ × 2 = 0.01 moles of NaOH.

Since we have 0.015 moles of NaOH available, H₂SO₄ is the limiting reactant.

Calculating Enthalpy of Neutralization

The enthalpy of neutralization is typically measured in joules per mole of water formed. The reaction produces 2 moles of water for every mole of H₂SO₄ that reacts. The standard enthalpy change for the neutralization of a strong acid with a strong base is approximately -57 kJ/mol.

Since we have 0.005 moles of H₂SO₄ reacting, the total enthalpy change can be calculated as:

Enthalpy change = moles of H₂SO₄ × standard enthalpy of neutralization

Enthalpy change = 0.005 moles × (-57 kJ/mol) = -0.285 kJ

Final Thoughts

The enthalpy of neutralization for the reaction of 100 ml of 0.1 N H₂SO₄ with 100 ml of 0.15 N NaOH is approximately -0.285 kJ. This negative value indicates that the reaction is exothermic, releasing heat as the acid and base neutralize each other.