Question

The following data was collected by a student performing the DETERMINING THE MOLAR ENTHALPY OF NEUTRALIZATION portion of the experiment. 50.00 mL of a 0.250 M acid is combined with 50.00 mL of 0.255 M NaOH. Before the reaction, the acid and base are at a temperature of 24.92 °C. After mixing, the neutralized solution reaches a maximum temperature of 26.50 °C in a calorimeter (Ccal=58.4 J/°C). The neutralized solution has a specific heat of 3.89 J/g°C and a density of 1.04 g/mL. What is the molar enthalpy of neutralization in kJ/mol?

Answer 1

moles of acid = 50 x 0.250 / 1000 = 0.0125

moles of NaOH = 50 x 0.255 / 1000 = 0.01275

limiting reagent is acid

total volume = 100 mL

mass of solution = volume x density

                          = 100 x 1.04

                         = 104 g

dT = 26.5 -24.92 = 1.58 oC

Q = m Cp dT + Cp dT

Q = 104 x 3.89 x 1.58 + 58.4 x 1.58

Q = 639 + 92.3

Q = 731 J

molar enthalpy of neutralization = - Q / n

                                                    = -0.731 / 0.0125

                                                   = - 58.5 kJ / mol