Question

AH and Calorimetry: 8. Suppose 33mL of 1.20M HCl is added to 42mL of a solution containing excess sodium 0 hydroxide in a coffee-cup calorimeter. The solution temperature, originally at 20. 0 31.8°C. COMPARE ANSWER WITH PROBLEM #9. 2pts Provide a thermochemical equation, including AH in kJ. (include sign) What is the energy (AH) evolved in kJ per mole of HCI? (include sign) For simplicity, assume the heat capacity and the density of the final solution are those of H2O. (Note: in more accurate work, these values are not the same and must be calculated). Also assume that the total volume of the solution equals the sum of the Cvolumes of HCl(aq) and NaOH(aq). work space Provide a thermochemical equation. HCl + NaOH Nad & Ho 199) (a) (9) (1) b. Must include sign b) 9. (Basically the same reaction as #8 above, without some of the assumptions) In one experiment, a student placed 50.0mL of 1.00M HCl at 25.5°C in a coffee-cup calorimeter. To this was added 50.0mL of 1.00M NaOH solution also at 25.5°C. The mixture was stirred, and the temperature quickly increased to a maximum of 32.2°C. 2pts a. Provide a thermochemical reaction, including AH in kJ. (include sign) What is the energy (AH) evolved in kJ per mole of HCI? (include sign) Because the solutions are relatively dilute, we can assume their specific heats are close to that of H2O, 4.184Jg*°C. However, do not assume the density of the solutions is that of water. The density of 1.00M HCl is 1.20g/mL and 1.00M NaOH is 1.40g/mL. work space: Provide a thermochemical equation.

Answer 1

8. Total volume of solution=Volume of HCl solution +Volume of NaOH solution

=33 mL+42 mL=75 mL

Density of solution=1 g/mL (=Density of water)

Mass of solution =Density x volume=1 g/mL x 75 mL=75 g

Rise in temperature of solution=Final temperature-initial temperature

=31.8°C-25.0°C=6.8°C

Specific heat of solution=4.184 J/g°C(=specific heat of water)

As the temperature of the solution increases, it absorbs the heat released by the reaction

So heat absorbed by the solution=Mass of solution x specific heat of solution x rise in temperature

=75 g x 4.184 J/g°C x 6.8°C=2133.84 J

=2133.84 J/1000 J/kJ=2.133 kJ

The energy which is absorbed by the solution and raises the temperature of the solution is actually the energy released during the neutralisation of HCl with NaOH

So energy released during neutralisation of HCl and NaOH=-2.133 kJ

So the thermochemical equation for the reaction is

NaOH(aq) + HCl(aq) + NaCl(aq) + H2O(1) AH = -2.133kJ

Number of moles of HCl=Molarity x volume (L)

=1.20 M x 33 mL/1000 mL/L (1 L=1000 mL)

=0.0396 mol

So energy involved in reaction in kJ per mol of HCl

=Energy released during neutralisation of HCl and NaOH/number of moles of HCl

=-2.133 kJ/0.0396 mol=-53.88 kJ/mol

9. Volume of HCl solution=50.0 mL

Density of HCl solution=1.20 g/mL

Mass of HCl solution=density x volume

=1.20 g/mL x 50.0 mL=60.0 g

Volume of NaOH solution=50.0 mL

Density of NaOH solution=1.40 g/mL

Mass of NaOH solution=Density x volume

=1.40 g/mL x 50.0 mL=70.0 g

Mass of solution =Mass of HCl solution+ Mass of NaOH solution=60.0 g+70.0 g=130.0 g

Rise in temperature of solution=Final temperature-initial temperature

=32.2°C-25.5°C=6.7°C

Specific heat of solution=4.184 J/g°C(=specific heat of water)

As the temperature of the solution increases, it absorbs heat released by the reaction

So heat absorbed by the solution=Mass of solution x specific heat of solution x rise in temperature

=130 g x 4.184 J/g°C x 6.7°C= 3644.26 J

=3644.26 J/1000 J/kJ=3.644 kJ

The energy which is absorbed by the solution and raises the temperature of the solution is actually the energy released during the neutralisation of HCl with NaOH

So energy released during neutralisation of HCl and NaOH=-3.644 kJ

So the thermochemical equation for the reaction is

NaOH(aq) + HCl(aq) + NaCl(aq) + H2O(1) AH = -3.644kJ

Number of moles of HCl=Molarity x volume (L)

=1.00 M x 50 mL/1000 mL/L (1 L=1000 mL)

=0.05 mol

So energy involved in reaction in kJ per mol of HCl

=Energy released during neutralisation of HCl and NaOH/number of moles of HCl

=-3.644 kJ/0.05 mol=-72.88 kJ/mol

so we see that the value for enthalpy ($\Delta H$) and molar enthalpy is higher in problem 9 for the same reaction as in problem 8 as we take into account densities of individual solutions to calculate the mass of solution.