Question

The enthalpy change of neutralization for the reaction of a strong acid (Hal) with a strong base (NaOH) is -56 KJ/ mol water produced. When CH3COOH reacts with strong base (NaOH) the enthalpy change is -55 KJ/mol water produced. What is the enthalpy change for the following reaction: CH3COOH (aq) --> H+(aq) + CH3COO-(aq)

Answer 1

NaOH + H⁺----------------> Na⁺ + H₂O    ΔH = -56 kj/mol   ------------(1)

CH₃COOH (aq)+ NaOH --> H₂O(aq) + CH3COO^-(aq) + Na⁺                ΔH = -55 kj/mol -----------(2)

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Subtract the second equation from first (Hess Law) gives

CH₃COOH---------------------->CH₃COO- + H⁺ ΔH = -56 kj/mol -(- ΔH = -55 kj/mol ) = -1 kj/mol

so the enthalpy change for the reaction CH₃COOH (aq) --> H+(aq) + CH₃COO-(aq) is -1 kj/mol