Question

How would i determine the Enthalpy of Dissociation for CH3COOH --> CH3COO- + H+ from the data of

NaOH + HCl --> NaCl + H2O Delta H = -59.4 Kj/mol

NaOH + CH3COOH --> NaCH3COO + H2O Delta H = -51.6 Kj/mol

Answer 1

Hess's law of constant heat summation is to find the enthalpy of the desired reaction by combining enthalpies of several reactions.The sign of enthalpy change when the reaction is reversed.Enthalpy is a state function so whether the reaction goes in many steps there is no effect.

If ΔH is positive, the reaction is endothermic

If ΔH is negative, the reaction is exothermic

(a) NaOH + HCl --> NaCl + H2O Delta H = -59.4 Kj/mol

(b) NaOH + CH3COOH --> NaCH3COO + H2O Delta H = -51.6 Kj/mol

Flip the reaction (a) Delta H value will change from negative to positive.

(a) NaCl + H2O --> NaOH + HCl Delta H = + 59.4 Kj/mol

Add reaction (b) to the above reaction we get,

(a) Na⁺Cl^- + H2O --> NaOH + H⁺Cl^- Delta H = + 59.4 Kj/mol

(b) NaOH + CH3COOH --> Na⁺CH3COO^- + H2O Delta H = -51.6 Kj/mol

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CH3COOH --> CH3COO- + H⁺ Delta H = + 59.4 Kj/mol -51.6 Kj/mol = + 7.8 Kj/mol

as we know dissociation of a weak acid absorbed energy so ΔH is positive.

N.B:-  [all the bold, italicized and underlined atoms cancel each other in the reaction]

Enjoy :-)