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50.0 mL of 1.00 M of a solution of HCl at 18.0 ºC is combined with...

50.0 mL of 1.00 M of a solution of HCl at 18.0 ºC is combined with 50.0 mL of 1.00 M solution of NaOH at 18.0 ºC. The final temperature of the reaction mixture is 24.6 ºC. If the heat capacities and densities of these solutions are approximately the same as water (4.184 J/g · ºC, 1.00 g/mL), what is the molar heat of neutralization (∆Hneut.)? Is this reaction endothermic or exothermic? [2 pt.]

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

Solutions

Expert Solution

No of mole of HCl = 50*1/1000 = 0.05 mole

No of mole of NaOH = 50*1/1000 = 0.05 mole

total mass of mixture = d*V

                        = (50+50)*1 = 100 g

heat released in the reaction(q) = m*s*DT

                                   = 100*4.184*(24.6-18)

                                   = 2.7614 kj

enthalpy of neutralisation(DH0neu) = 2.7614 kj/0.05 mole

                                     = 55.23 kj/mol

as temperature increases, it is exothermic process

queen_honey_blossom answered 1 year ago
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