Given the reactions: 1) HF ⇔ H⁺ + F⁻ K= 6.8 x 10⁻⁴
2) 2HF + C₂O₄²⁻ ⇔ 2F⁻ + H₂C₂O₄ K= 0.24
Determine the value of the equilibrium constant for the reaction:
3) H₂C₂O₄ ⇔ 2H⁺ + C₂O₄²⁻

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

i dont know

explanation: