Concept Explanation: Enthalpy of Neutralization

The enthalpy of neutralization is the heat change when one mole of water is formed from the reaction of an acid and a base. For strong acids and bases like HCl and NaOH, it is typically around -57.1 kJ/mol. In this problem, we are given experimental data to calculate it.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation

The reaction is: $\mathrm{HCl}+\mathrm{NaOH}\to \mathrm{NaCl}+H_{2}O$

This shows that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of water.

Step 2: Determine the limiting reactant and moles of water produced

Moles of HCl = Molarity × Volume (in L) = 0.10 mol/L × 0.050 L = 0.005 mol

Moles of NaOH = 0.10 mol/L × 0.050 L = 0.005 mol

Since they are in a 1:1 ratio and equal moles, neither is limiting. Moles of water produced = 0.005 mol.

Step 3: Calculate the heat released (q)

The total volume of solution = 50.0 mL + 50.0 mL = 100.0 mL. Assuming the density is similar to water (1 g/mL), the mass (m) is 100.0 g.

The specific heat capacity (c) for dilute aqueous solutions is approximately 4.18 J/g°C.

The temperature change (ΔT) = 3.0 °C.

Using the formula: $q=m\times c\times \mathrm{\Delta }T$

$q=100.0\mathrm{g}\times 4.18\frac{\mathrm{J}}{\mathrm{g}°\mathrm{C}}\times 3.0°\mathrm{C}=1254\mathrm{J}$

This heat (q) is released, so it is exothermic: q = -1254 J

Step 4: Calculate enthalpy change per mole of HCl (ΔH)

This heat was released for 0.005 moles of HCl.

Therefore, ΔH = q / moles of HCl = -1254 J / 0.005 mol = -250800 J/mol

Convert to kJ/mol: -250800 J/mol ÷ 1000 = -250.8 kJ/mol ≈ -2.5 × 10² kJ/mol

Final Answer: $-2.5\times {10}^2\frac{\mathrm{kJ}}{\mathrm{mol}}$

Related Topics:

• Calorimetry: The science of measuring the heat of chemical reactions.
• Thermochemistry: The study of the energy and heat associated with chemical reactions.
• Acid-Base Reactions: Reactions between acids and bases, often leading to salt and water formation.
• Limiting Reactant: The reactant that is completely consumed first and limits the amount of product formed.

Key Formulae and Theory:

Heat Transfer (q): $q=m\times c\times \mathrm{\Delta }T$

Where:
q = heat (J)
m = mass (g)
c = specific heat capacity (J/g°C)
ΔT = change in temperature (°C)

Enthalpy Change (ΔH): $\mathrm{\Delta }H=\frac{q}{n}$

Where:
ΔH = enthalpy change (J/mol or kJ/mol)
q = heat (J)
n = number of moles of the limiting reactant

Moles from Concentration: Moles = Molarity (M) × Volume (L)

Theory: For strong acid-strong base neutralization, the net ionic reaction is always $H^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to H_{2}O\left(l\right)$, and the standard enthalpy change is approximately -57.1 kJ/mol. Experimental values may differ slightly due to specific heat capacity assumptions or heat loss.