Enthalpy and Entropy in Reactions and Solutions
Introduction
In the course of most physical processes and chemical reactions there is a change in energy. In chemistry what is normally measured is H (enthalpy change), the change in heat at constant pressure and ignoring any work done by the reacting system. If the reaction is exothermic, heat is given off and H has a negative value. When heat is absorbed from the surroundings during the reaction, the reaction is endothermic and H has a positive value. The change in enthalpy is a measure of the difference in energy between the bonds in the products and the bonds in the reactants. The absolute energy of a compound cannot be measured directly, but the change in enthalpy that occurs during a chemical reaction can be measured. We will do this in the first part of today's lab.
Reactions and physical processes (like phase changes and formation of solutions) are also influenced by changes in entropy. Recall that reactions are spontaneous (i.e. thermodynamically favored) when the Gibbs free energy change, G, is negative. G depends on both the enthalpy change H and the entropy change, S. In the second part of the lab we will see how enthalpy and entropy explain why water dissolves polar substances and does not dissolve non-polar organic substances. We will use a special "Magic Sand" to model "hydrophobic" reactions between water and non-polar substances, a type of reaction that is important in biological systems.
I. Enthalpy of Neutralization
The reaction in today’s experiment is acid / base neutralization which is an exothermic process.
HCl (aq) + NaOH(aq) H2O(l) + NaCl(aq) + heat
The heat released by the reaction will be absorbed by the products (salt water solution). Coffee Cup Calorimetry will be employed to determine the amount of heat lost by the reaction and gained by the salt water solution. A calorimeter is simply a container used to measure the heat change. Coffee Cup Calorimetry just means that we will be measuring heat at constant pressure, H. The heat lost by the reaction will actually be transferred to both the salt water and its surroundings (the calorimeter.) The heat capacity of the calorimeter (in this case, two Styrofoam cups) usually would be calculated first; however, we have found that the heat capacity of the cups is so small that it can be neglected.
As the First Law of Thermodynamics applies,
Heat (q) lost by the reaction + heat (q) gained by the salt water solution = 0
q = (s) (m) (ΔT)
where: s = specific heat capacity of the solution (3.87 J/gC)
m = mass of the solution (The density of the solution is 1.04 g/mL)
ΔT= change in temperature of the solution (Tfinal – Tinitial)
Equipment
2 Styrofoam cups (2) Thermometers
50 or 100 mL graduated cylinder Stirring rod
150 mL beaker
Chemicals
2.00 M HCl, 2.00 M NaOH
Spill/Disposal
Acid/Base Spill/Disposal: B1 Neutralization Product: Spill/Disposal A
Procedure
1. Rinse and dry the Styrofoam cups (calorimeter). Place one cup inside the other. Measure out 25.0 mL of 2.00 M NaOH and pour it into the calorimeter. In a clean dry 150 mL beaker, take 25.0 mL of 2.00 M HCl.
2. Determine the temperature of both the acid and the base to the nearest 0.1C. Average the temperatures. Record this average as the initial temperature Tinitial.
3. Pour the acid into the base quickly and carefully with gentle stirring. Start monitoring the temperature as soon as the two are mixed. Continue to stir and monitor the temperature. Record the maximum temperature that the solution reaches. This is Tfinal .
4. Repeat Steps 1 - 3 (above) for Run 2.
Disposal
All reactants and products may be disposed of into the sink.
Heat of Neutralization: Pre Lab Name: ________________________________
In a coffee cup calorimeter, 25.0 mL of 2.00 M NaOH and 250.0 mL of 2.00 M HCl are mixed. Both solutions were originally at 24.6 C. After the reaction, the final temperature is 38.7 C.
HCl (aq) + NaOH(aq) H2O(l) + NaCl(aq) + heat.
Heat lost by the reaction + heat gained by the NaCl solution = 0
q lost + q gained = 0 q gained = (s) (m) (ΔT)
(A) Calculate the volume of NaCl(aq) (this will be the solution in the calorimeter).
Given that the density of 1.00 M NaCl is 1.02 g/mL, calculate the mass of the NaCl(aq).
Mass = _______________
(B) Calculate the ΔT for the NaCl(aq).
ΔT = __________________
(C) Given that the specific heat capacity (s) of 1.00 M NaCl(aq) is 3.87 J/gC, calculate the heat gained by the solution (q gained)
q gained = _______________
(D) The heat gained by the solution is equal but opposite in sign of the heat lost by the reaction (q lost). Calculate q of the reaction (q lost).
q lost = ___________________
(E) Given that 25.0 mL of 2.00 M NaOH and 25.0 mL of 2.00 M HCl reacted. Calculate the number of moles of either NaOH or HCl that reacted. Hneutralization is the heat for the reaction in units of kJ/mole.
Calculate Hneutralization. (Pay close attention to the sign)
Hneutralization= _______________
Maybe you should actually do the experiment before trying to answer the questions?
To calculate the volume of NaCl(aq) (solution in the calorimeter):
The volume of NaOH(aq) used is 25.0 mL and the volume of HCl used is also 25.0 mL, so the total volume of the solution in the calorimeter is:
VolumeNaCl(aq) = VolumeNaOH(aq) + VolumeHCl(aq)
= 25.0 mL + 25.0 mL
= 50.0 mL
Therefore, the volume of NaCl(aq) in the calorimeter is 50.0 mL.
To calculate the mass of NaCl(aq):
Given that the density of 1.00 M NaCl(aq) is 1.02 g/mL, we can use the density formula:
Density = Mass / Volume
Rearranging the formula for mass:
Mass = Density * Volume
Plugging in the values:
Mass = 1.02 g/mL * 50.0 mL
Mass = 51.0 g
Therefore, the mass of NaCl(aq) in the calorimeter is 51.0 g.
To calculate the ΔT for the NaCl(aq):
ΔT = Tfinal - Tinitial
Plugging in the values:
ΔT = 38.7°C - 24.6°C
ΔT = 14.1°C
Therefore, the ΔT for the NaCl(aq) is 14.1°C.
To calculate the heat gained by the solution (q gained):
q gained = (s) (m) (ΔT)
Plugging in the values:
q gained = (3.87 J/g°C) * (51.0 g) * (14.1°C)
q gained = 2820.57 J
Therefore, the heat gained by the solution is 2820.57 J.
To calculate the heat lost by the reaction (q lost):
Since q lost = -q gained (opposite in sign), we can simply take the negative of q gained:
q lost = -2820.57 J
Therefore, the heat lost by the reaction is -2820.57 J.
To calculate the number of moles of NaOH or HCl that reacted:
Given that 25.0 mL of 2.00 M NaOH and 25.0 mL of 2.00 M HCl reacted, we can use the equation:
Molarity = Moles / Volume
Rearranging the equation for moles:
Moles = Molarity * Volume
Plugging in the values:
Moles = (2.00 mol/L) * (0.025 L)
Moles = 0.05 mol
Therefore, the number of moles of either NaOH or HCl that reacted is 0.05 mol.
To calculate the heat of neutralization (Hneutralization):
Hneutralization = q lost / Moles
Plugging in the values:
Hneutralization = -2820.57 J / 0.05 mol
Hneutralization = -56411.4 J/mol
Therefore, the heat of neutralization is -56411.4 J/mol.
To calculate Hneutralization (the heat of neutralization) in this experiment, you will need to follow these steps:
Step A: Calculate the volume of NaCl(aq) (the solution in the calorimeter).
Given that the density of 1.00 M NaCl is 1.02 g/mL, calculate the mass of the NaCl(aq).
Mass = Volume x Density
Mass = [Concentration (M) x Volume (L)] x Density (g/mL)
Mass = [2.00 M x (25.0 mL / 1000 mL)] x 1.02 g/mL
Mass = 0.051 g
Step B: Calculate the ΔT for the NaCl(aq).
ΔT = Tfinal - Tinitial
ΔT = 38.7°C - 24.6°C
ΔT = 14.1°C
Step C: Given that the specific heat capacity (s) of 1.00 M NaCl(aq) is 3.87 J/g°C, calculate the heat gained by the solution (q gained).
q gained = (s) (m) (ΔT)
q gained = (3.87 J/g°C) x (0.051 g) x (14.1°C)
q gained = 2.80 J
Step D: The heat gained by the solution is equal but opposite in sign to the heat lost by the reaction (q lost). Calculate q of the reaction (q lost).
q lost = -q gained
q lost = -2.80 J
Step E: Given that 25.0 mL of 2.00 M NaOH and 25.0 mL of 2.00 M HCl reacted, calculate the number of moles of either NaOH or HCl that reacted. Hneutralization is the heat for the reaction in units of kJ/mol.
Because the reaction is 1:1 between HCl and NaOH, the number of moles of HCl is equal to the number of moles of NaOH.
Moles HCl = Concentration (M) x Volume (L)
Moles HCl = 2.00 M x (25.0 mL / 1000 mL)
Moles HCl = 0.050 mol
Moles NaOH = Moles HCl = 0.050 mol
Finally, calculate the Hneutralization (heat of neutralization) using the equation:
Hneutralization = q lost / Moles HCl (or Moles NaOH)
Hneutralization = (-2.80 J) / (0.050 mol)
Hneutralization = -56.0 J/mol
Note: Remember to convert J to kJ to obtain the final answer in units of kJ/mol.