Question

Experiment - Enthalpy of Neutralization

Use Hess' Law to Calculate the enthalpy of dissociation for chloroacetic acid.

H (aq) + OH -> H₂O (l) ΔH_n(HCl)

HC₂H₂ClO₂(aq) + OH (aq) -> C₂H₂ClO₂ + H2O (l) ΔH_n(HC₂H₂ClO₂)

We used a strong acid, HCl as the limiting reactant and NaOH as the the excess.

Weak acid: HC₂H₂ClO₂ (aq) + H (aq) + C₂H₂ClO₂ (aq) ΔH_d(HC₂H₂ClO₂)

HCl: Average ΔH_n = -59 KJ

HC₂H₂ClO_2: Average ΔH_n = -61 KJ

Answer 1

Eq.1; H⁺ (aq) + OH^-(aq) H₂O (l) ; ₁ = -59 KJ

Eq.2 ; HC₂H₂ClO₂ (aq) + OH^- (aq) C₂H₂ClO₂ + H₂O ; ₂= -61 KJ

Eq. 3 ; HC₂H₂ClO₂ (aq) H⁺ (aq) + C₂H₂ClO₂ (aq) .

Now,

Eq.3 = Eq.2 - Eq.1

So, by Hess law

= ₂ - ₁

= {- 61 - ( -59) } KJ

= - 2 KJ.

So, enthalpy of dissociation of chloroacetic acid = -2 KJ.