Question

Insoluble PbBr2(s) precipitates when solutions of Pb(NO3)2(aq) and NaBr(aq) are mixed.

Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq) ΔrH° = ?

To measure the enthalpy change, 200. mL of 0.75 M Pb(NO3)2(aq) and 200. mL of 1.5 M NaBr(aq) are mixed in a coffee-cup calorimeter. The temperature of the mixture rises by 2.44 °C. Calculate the enthalpy change for the precipitation of PbBr2(s), in kJ/mol. (Assume the density of the solution is 1.0 g/mL, and its specific heat capacity is 4.2 J/g ∙ K.)

Answer 1

Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq)

no of moles Pb(NO3)2 = molarity * volume in L

                                    = 0.75*0.2 = 0.15 moles

no of moles of NaBr    = molarity * volume in L

                                     = 1.5*0.2 = = 0.3

0.15 moles of Pb(NO3)2 react with 0.3 moles of NaBr to gives 0.15 moles of PbBr2

total volume = 200+ 200 = 400ml

mass of solution = volume * density

                              = 400*1 = 400g

q = mC deltaT

      = 400*4.2*2.44   = 4099.2 J

     J/mole = 4099.2/0.15   = 27328J/mole = 27.328KJ/mole