Question

A quantity of 2.30 × 102 mL of 0.900 M HNO3 is mixed with 2.30 × 102 mL of 0.450 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 18.46°C. The heat of neutralization when 1.00 mol of HNO3 reacts with 0.500 mol Ba(OH)2 is −56.2 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 J/g · °C, respectively). What is the final temperature of the solution? (in C)

Answer 1

Final temperature of the solution = 24.50 ^oC

Explanation

molarity HNO₃ = 0.900 M

volume HNO₃ = 230 mL = 0.230 L

moles HNO₃ = (molarity HNO₃) * (volume HNO₃ in Liter)

moles HNO₃ = (0.900 M) * (0.230 L)

moles HNO₃ = 0.207 mol

Similarly, moles Ba(OH)₂ = 0.1035 mol

Heat released = (heat of neutralization) * (moles HNO₃)

Heat released = (-56.2 kJ/mol) * (0.207 mol)

Heat released = -11.6334 kJ

Heat released = -11633.4 J

heat absorbed by solution = -(Heat released)

heat absorbed by solution = -(-11633.4 J)

heat absorbed by solution = 11633.4 J

Temperature change of solution = (heat absorbed by solution) / [(mass of solution) * (specific heat of solution)]

Temperature change of solution = (11633.4 J) / [(230 g + 230 g) * (4.184 J/g.^oC)]

Temperature change of solution = 6.0 ^oC

Final temperature = (initial temperature) + (Temperature change of solution)

Final temperature = (18.46 ^oC) + (6.0 ^oC)

Final temperature = 24.50 ^oC