While performing a thermodynamics experiment, a student made the following observations, HCl + NaOH NaCl + H O H = –57.3 kJ mol CH COOH + NaOH CH COONa + H O H = –55.3 kJ mol . The enthalpy of ionizat

Question

While performing a thermodynamics experiment, a student made the following observations, HCl + NaOH NaCl + H O H = –57.3 kJ mol CH COOH + NaOH CH COONa + H O H = –55.3 kJ mol . The enthalpy of ionization of CH COOH as calculated by the student is ______ kJ mol . (nearest integer) 2 –1 3 3 2 –1 3 –1

Prashant Kotian · Accepted Answer

Correct answer: 2.00. **Step 1: Identify the relevant reactions** Strong acid (HCl) neutralization: HCl + NaOH → NaCl + H₂O, ΔH = –57.3 kJ/mol Weak acid (CH₃COOH) neutralization: CH₃COOH + NaOH → CH₃COONa + H₂O, ΔH = –55.3 kJ/mol **Step 2: Write the ionization equation for acetic acid** CH₃COOH → CH₃COO⁻ + H⁺, ΔH_ionization = ? **Step 3: Apply Hess's Law** The neutralization of CH₃COOH can be broken into two steps: - Step A: CH₃COOH → CH₃COO⁻ + H⁺ (ionization) - Step B: H⁺ + OH⁻ → H₂O (neutralization of H⁺) The strong acid neutralization shows that H⁺ + OH⁻ → H₂O releases 57.3 kJ/mol **Step 4: Use the relationship** For weak acid neutralization: ΔH(CH₃COOH + NaOH) = ΔH_ionization + ΔH(H⁺ + OH⁻) –55.3 = ΔH_ionization + (–57.3) **Step 5: Solve for ionization enthalpy** ΔH_ionization = –55.3 + 57.3 = 2.0 kJ/mol Therefore, the answer is 2.00.

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