Question

How can i calculate the enthalpy of reaction for both neutralization reactions (HCL + NaOH) and (Citric acid + NaHCO3) with units of Kj/mol?

neutralization reactions that occur when an acid reacts with a base. HCI (aq) + NaOH (aq) → H20 (1) + NaCl (aq) HzC6H50, (citric acid) (aq) + 3 NaHCO3 (s) → Na2C8H:07 (aq) + 3 CO2 (g) + 3 H20 (1)

Enthalpy of formation:

Hcl (aq): -167.2 kj/mol

NaOH (aq): -469.6 kj/mol

citric acid (aq): not sure

NaHCO3 (s): -947.7 kj/mol

Answer 1

1.Ans :-

Neutralization reaction is :

HCl (aq) + NaOH (aq) ----------------> NaCl (aq) + H₂O (l) , ΔH_rxn⁰ = ?

Given Enthalpy of formation :

ΔH_f⁰ of HCl (aq) = - 167.2 KJ/mol

ΔH_f⁰ of NaOH (aq) = - 469.6 KJ/mol

ΔH_f⁰ of H₂O (l) = - 285.8 KJ/mol and

ΔH_f⁰ of Na⁺ (aq) = - 240.1 KJ/mol and

ΔH_f⁰ of Cl^- (aq) = - 167.2 KJ/mol

Enthalpy of reaction i.e. ΔH_rxn⁰ = Sum of enthalpy of formation of products - Sum of enthalpy of formation of reactants

ΔH_rxn⁰ = [ΔH_f⁰ of Na⁺ (aq) + ΔH_f⁰ of Cl^- (aq) + ΔH_f⁰ of H₂O (l)] - [ ΔH_f⁰ of HCl (aq) + ΔH_f⁰ of NaOH (aq) ]

= [ -240.1 KJ/mol -167.2 KJ/mol -285.8 KJ/mol] - [ - 167.2 KJ/mol - 469.6 KJ/mol ]

= - 693.1 KJ/mol + 636.8 KJ/mol

= - 56.3 KJ/mol

Hence, Enthapy of reaction = ΔHrxn0 = - 56.3 KJ/mol