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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