Thermodynamics Lab
Thermodynamics-Enthalpy of Reaction and Hess’s Law
Purpose To demonstrate the principle of Hess’s Law and to find the heat capacity of the coffee cup calorimeter using three different reactions.
Data
Tave = (46.4-45.2)/2 = 45.8 qwater = -(100g)(4.184)(46.56-45.8) = -318 J Ccal = 318J/(46.56-21.2) = -12.53J/g*C
Tinitial = (27.1+23.8)/2 = 25.45 qrxn = -(100g)(4.18)(38.43-24.45)+(-12.53x12.98)
=-5400J/.1mol(1J/1000kJ) = -54.0 kJ/mol
Tinitial = (26.0-24.5)/2 qrxn =-(100g)(4.18)(26.26-25.25)+(-12.53x1.01)= -40.9J/.1mol(1J/1000kJ) = -4.09kJ/mol
Tinitial = (23.9+22.9)/2 =23.4 qrxn =-(100g)(4.18)(29.48-23.4)+(-12.53x 6.08)=
-2620J/.1mol(1/1000kJ)= -26.2 kJ/mol
Verify Hess’s Law 1. HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) H+ + Cl- + Na+ + OH- → Na+ +Cl- + H2O (l) H+ + OH- → H2O (l)
NH4Cl (aq) + NaOH (aq) → NH3 (aq) + NaCl (aq) + H2O (l)
NH4+ + Cl- + Na+ + OH- → NH3+ + Na+ +Cl- + H2O (l)
NH4++ OH- → NH3+ + H2O (l)
NH3 (aq) + HCl (aq) → NH4Cl (aq)
NH3+ + H+ + Cl- →NH4+ + Cl-
NH3+ + H+ → NH4+
H+ + OH- → H2O (l) ∆H= -54.0 kJ/mol
NH4++ OH- → NH3+ + H2O (l) (x-1) ∆H=-4.09 kJ/mol (x-1)
-54.0+4.09=-49.9
NH3+ + H+ → NH4+ =- 49.9 kJ /mol
(-49.9- -26.2)/-49.9 x 100 =47.3%
Lab Questions:
Pre:
1. The change in thermal content in a reaction
2. The amount of energy needed to change one gram of a substance 1 degree C
3. (50 mL)(1.02 g/mL)(4.18 J/g C)(25.3-21.4) = 830 J
4. 830J+8.20J/g C(25.3-21.4)= 860J
5. .25L(.6mol/1L)+.25L(.L/1mol)= 0.3mol AB 830J/.3mol(1kJ/1000J)= 2.7 kj/mol
Post
1. Calorimetry is the measure of heat flow. Heat that is measured between the reactants and products.
2. By tracking how close the data is to the best-fit line. Using the best-fit line you can find the initial temperature of the mixture
3. To show the heat that was absorbed by the calorimeter. Showing that the number is the same value of the reaction but has the opposite sign.
4. The lab
Purpose To demonstrate the principle of Hess’s Law and to find the heat capacity of the coffee cup calorimeter using three different reactions.
Data
Tave = (46.4-45.2)/2 = 45.8 qwater = -(100g)(4.184)(46.56-45.8) = -318 J Ccal = 318J/(46.56-21.2) = -12.53J/g*C
Tinitial = (27.1+23.8)/2 = 25.45 qrxn = -(100g)(4.18)(38.43-24.45)+(-12.53x12.98)
=-5400J/.1mol(1J/1000kJ) = -54.0 kJ/mol
Tinitial = (26.0-24.5)/2 qrxn =-(100g)(4.18)(26.26-25.25)+(-12.53x1.01)= -40.9J/.1mol(1J/1000kJ) = -4.09kJ/mol
Tinitial = (23.9+22.9)/2 =23.4 qrxn =-(100g)(4.18)(29.48-23.4)+(-12.53x 6.08)=
-2620J/.1mol(1/1000kJ)= -26.2 kJ/mol
Verify Hess’s Law 1. HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) H+ + Cl- + Na+ + OH- → Na+ +Cl- + H2O (l) H+ + OH- → H2O (l)
NH4Cl (aq) + NaOH (aq) → NH3 (aq) + NaCl (aq) + H2O (l)
NH4+ + Cl- + Na+ + OH- → NH3+ + Na+ +Cl- + H2O (l)
NH4++ OH- → NH3+ + H2O (l)
NH3 (aq) + HCl (aq) → NH4Cl (aq)
NH3+ + H+ + Cl- →NH4+ + Cl-
NH3+ + H+ → NH4+
H+ + OH- → H2O (l) ∆H= -54.0 kJ/mol
NH4++ OH- → NH3+ + H2O (l) (x-1) ∆H=-4.09 kJ/mol (x-1)
-54.0+4.09=-49.9
NH3+ + H+ → NH4+ =- 49.9 kJ /mol
(-49.9- -26.2)/-49.9 x 100 =47.3%
Lab Questions:
Pre:
1. The change in thermal content in a reaction
2. The amount of energy needed to change one gram of a substance 1 degree C
3. (50 mL)(1.02 g/mL)(4.18 J/g C)(25.3-21.4) = 830 J
4. 830J+8.20J/g C(25.3-21.4)= 860J
5. .25L(.6mol/1L)+.25L(.L/1mol)= 0.3mol AB 830J/.3mol(1kJ/1000J)= 2.7 kj/mol
Post
1. Calorimetry is the measure of heat flow. Heat that is measured between the reactants and products.
2. By tracking how close the data is to the best-fit line. Using the best-fit line you can find the initial temperature of the mixture
3. To show the heat that was absorbed by the calorimeter. Showing that the number is the same value of the reaction but has the opposite sign.
4. The lab