Question

The standard enthalpy of formation of the fumarate ion is ∆_fH^o = -777.4 kJ mol^-1. If the standard enthalpy change of the reaction,

fumarate^2- (aq) + H₂(g) → succinate^2- (aq), is 131.4 kJ mol^-1, what is the enthalpy of formation of the succinate ion?

Answer 1

The given reaction is

Fumerate^2- (aq) + H₂ (g) ---------> Succinate^2- (aq)

The enthalpy change for the reaction is given as

ΔH⁰_rxn = ΣΔH⁰_f(products) – ΣΔH⁰_f(reactants) = [ΔH⁰_f (Succinate^2-, aq)] – [ΔH⁰_f (Fumerate^2-, aq) + ΔH⁰_f (H₂, g)]

===> 131.4 kJ/mol = [ΔH⁰_f (Succinate^2-, aq)] – [(-777.4 kJ/mol) + (0.0 kJ/mol)] (enthalpy of formation of homoatomic compounds is zero).

===> 131.4 kJ/mol = [ΔH⁰_f (Succinate^2-, aq)] + 777.4 kJ/mol

===> [ΔH⁰_f (Succinate^2-, aq)] = (131.4 – 777.4) kJ/mol = -646.0 kJ/mol.

The standard enthalpy of formation of Succinate is -646.0 kJ/mol (ans).