Question

A quantity of 12.0 mL of 6.00 M HBr is mixed with 12.0 mL of 6.00 M NaOH in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 21.30°C. The heat of neutralization when 1.00 mol of HBr reacts with 1.00 mol NaOH is -56.1 kJ/mol. Assume that the densities and specific heats of the solution are the same as for water (1,00 g/mL and 4.184 °C. respectively). What is the final temperature of the solution? (15 points)

Answer 1

Initial moles of H+ = 120mLx_1L ) 6.00 mol HBr 1molH* * * 1000mL). 1L 1 m ol HBr = 0.0720 molt 1L 6.00 mol NaOH Initial moles of OH = 1mol OH 1000mL) 1mol NaOH = 0.0720 mol OH 2.0 mLx 1L The two reactants are in equal amounts that are in stoichiometric proportions as per reaction. So the moles of water produced from either of initial moles will be: 0.0720 mol H+xlmol 1,0 H imolH+ = = 0.0720mol H, 0.0720 mol OH x mol H2O = 007 - 1 mol H2O =0.0720 moiH, Imol OH Thus, 0.0720 moles of H, produced - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Enthalpy of neutralization = 0.0720 mol H.Ox-30.1 kJ Enthalpy of neutralization = 0. 0 0 m 2 mol H,O = - 4.0392 kJ heat absorbed , calain = -freute = -(-4.0392 kJ ) = 4.0392 kJ = 4039.2J -------------- Volume of water = (12.0+12.0) mL = 24.0 mL The heat absorbed by calorimeter contents (water is: - - - - - - - - - - - - - - - - - - - Posen = (24.0 mix 2028) 1949 (3,-21.30) Carta = (2403)*41847 < (33-21.3°c) 40392 = 100.4165«(7, -21.30°C) 4039.2J T; - 21.30°C = 10416 1 = 40.22°C °C 7 = 21.3°C+40.22°C = 61.52°C