What is the change in enthalpy of the following reaction?

2NaBr (aq) + PbCl2 (aq) 2NaCl (aq) + PbBr2 (aq)

Given:

NaCl: ∆H = - 411 kJ

NaBr: ∆H = - 360 kJ

PbCl2: ∆H = - 359 kJ

PbBr2: ∆H = - 277 kJ

We are given with the balanced chemical reaction and the individual enthalpy of formation of the species involved in the reaction. To get the enthalpy of the reaction, the difference between the sum of the enthalpy of formations of the products and the reactants is calculated taking in mind the stoichiometry of the reaction.

∆Hrxn = [2 (-411) + (-277) ] - [2 (-360) + (-359) ]

∆Hrxn = - 20 kJ