Question

5. 0.20 L of a 0.40 M HCI is totally reacted with 0.20 L of a NaOH solution in a Styrofoam coffee cup. The temperature of both solutions before mixing is 25.1 °C and 26.6 °C after. Assume that both the HCI and the NaOH solutions have a density of 1.00 g/mL, the heat absorbed by the cup is negligible, and the specific heat capacity of the solutions is 4.18 J/g C. Calculate the heat released per mol HCl in this neutralization reaction. (Hint: How many moles of HCI was reacted in the experiment?)

Answer 1

dT = 26.6 - 25.1

dT = 1.5 oC

moles of HCl = 0.20 x 0.40

                    = 0.08

total volume of the solution = 0.2 + 0.2 = 0.4 L = 400 mL

mass of solution = density x volume

                           = (1.00 g / mL ) x 400 mL

                          = 400 g

Q = m Cp dT

Q = 400 x 4.184 x 1.5

Q = 2510.4 J

heat = 2510.4 J

Q = 2.510 kJ

heat released per mole of HCl = - Q / n

                                                = -2.510 / 0.08

                                                = -31.4 kJ / mol