General Chemistry Solution Calorimetry Lab Reports

Solution Calorimetry Theory/Principles A thermodynamic system is a system that requires variables such as temperature, entropy, pressure, surroundings, pressure and internal energy to exist. Enthalpy1 is one of the properties of thermodynamic systems, it is the sum of the internal energy and the product of pressure with volume in a thermodynamic system. Enthalpy, when heat is released to the surroundings it is negative, while when the enthalpy change is absorbed by the system then it will be positive. In order to find the enthalpy of a system, we will need to use the following equation: (1) q = m × C s × ΔT (2) q = m × C s × (T f − T i ) As mentioned before to find the enthalpy of a system; internal energy ( ΔE ), pressure (P) and volume (V) need to be known to calculate. It is mentioned in the manual that the process taking place in a calorimeter that is opened to the atmosphere, the enthalpy, ΔH, change is equal to qp meaning that other equations given in manual2 can be replaced with the following equation for a shortcut: (3) ΔH = q p =− q c = mC H 2O (T i − T f ) In part A of the lab, three acids are mixed with based NaOH and their balanced equations are as followed: (4) N aOH(aq) + H Cl(aq) → N aCl(aq) + H 2O(l) Base sodium hydroxide was added to strong acid hydrogen chloride into a calorimeter to create a reaction that would allow calculating the change in enthalpy, same goes for the following balanced equations, all were acid-base reactions: (5) N aOH(aq) + H N O 3 (aq) → N aN O 3 (aq) + H 2 O(l) (6) N aOH(aq) + C H 3 COOH(aq) → C H 3 COON a(aq) + H 2 O(l) In part B of the lab, the solutions of N a2 S 2 O3 and N a2 S 2 O3 • H 2 O had data recorded such as the mass of solid, the change of temperature and number of moles in order to calculate the enthalpy of solution. The balance equations of the solutions are found in the following: (7) N a2 S 2 O3 (s) + H 2 O(l) → H 2 S 2 O3 (s) + N aOH(l) (8) N a2 S 2 O 3 • 5H 2 O(aq) → N a2 S 2 O 3 (s) + 5H 2 O(l) In part C of the lab, solids such as magnesium metal and magnesium oxide were used to calculate the enthalpies of dissolutions. The same calculation process was used as finding the enthalpy of solution. The following are the balanced equations of magnesium metal and magnesium oxide: (9) M g(s) + 2HCl(aq) → M gCl2 (aq) + H 2 (g) (10) M gO(s) + 2HCl(aq) → M gCl2 (aq) + H 2 O(l) HCl(aq) Raw Data Part (A) Table 1. Strong and Weak Acids Data Class Data for Temperature of Strong and Weak Acids (℃) HCl HNO3 Acetic acid Ti Tf Ti Tf Ti Tf 1. 23.9 30.6 22.4 28.9 23.3 27.7 2. 22.5 27.6 22.5 27.4 22.0 27.6 3. 22.9 28.3 22.1 29.2 22.7 28.0 22.33 28.5 22.67 27.76 4. 22.4 29.7 Average: 22.55 29.05 Part (B) 1. N a2 S 2 O3 Mass: 2.981 g Ti=23.1℃ Tf=26.1℃ 2. N a2 S 2 O3 • H 2 O Mass: 4.932 g Ti =23.1℃ Tf=18.3℃ Part (C) 1. Magnesium metal Mass: 0.125 g Ti= 22.5℃ Tf= 27.1℃ 2. Magnesium oxide Mass: 0.795 g Ti= 22.5℃ Tf= 22.7℃ Experimental Found in manual 2 *Changes: part a was split between three benches, each bench was assigned an acid to work with. *Error in part c, magnesium oxide provided was not pure. Results Table 2. Enthalpy of Neutralization Acids Enthalpies of Neutralization of Strong and Weak Acids (J/mol) HCl HNO3 Acetic Acid 54.393 J/mol 51.630 J/mol 42.593 J/mol Table 3. Enthalpy of Solution Enthalpy of Solution (J/mol) N a2 S 2 O 3 N a2 S 2 O 3 • H 2 O 68.598 J/mol -106.052 J/mol Table 4. Enthalpy of Dissolution Enthalpies of Solution of Magnesium Metal and Magnesium Oxide (J/mol) Magnesium metal Magnesium oxide 374.694 J/mol 45.326 J/mol Sample Calculations: ● q = m × C s × (T f − T i ) J ×( q = 100g × 4.184 g.K 29.05℃ − 22.55℃) = 2719.6 J q(J) # mol acid ● = 2719.6 J 0.0500 mol of HCl 1 × 1000 = 5 4.393 J/mol = Δ H q = (m + solid) × C s × (T f − T i ) q = (100g + 0.125g) × 4.184 1927.0458 J 0.00514 mol × 1 1000 J g.K × (27.1℃ − 22.5℃) = 1927.04 J = 3 74.694 J/mol Discussion Questions: 1. Compare the values of the enthalpy of neutralization for the three acids. Explain any similarities and/or differences. The similarity between the two of the acids: HCl and HNO3 , they both have around the same amount of enthalpy, while acetic acid has a lower amount of enthalpy. Both HCl and HNO3 are considered strong acids but acetic acid is considered a weak acid, that may be the factor that contributes to the difference in enthalpy between the three acids. 2. Calculate the enthalpy change for the reaction: N a2 S 2 O3 (s) + 5H 2 O(l) → N a2 S 2 O3 • 5H 2 O(s) ΔH rxn =− 106.052 J /mol − (68.598 J /mol + (5 ×− 285.83 kJ /mol) =− 1603 kJ 3. Use the data from your work to calculate the enthalpy change of the following reaction: H 3 CCOOH (aq) → H +(aq ) + H 3 CCOO −(aq) 4. Use your data and the known value of ΔH° , for liquid water to calculate ΔH° , (MgO(s)). q = (100.125g) × 4.184 J g.K × (27.1℃) − (22.5℃) = 1927.04 J 1927.0458 J 0.00514 mol × 1 1000 = 3 74.694 J/mol 5. Use your data in Part C and the known value for ΔH° , (H+ (aq)) to calculate ΔH° , (Mg2+(aq)). 6. Calculate the uncertainty in the enthalpy of neutralization for two of the acids in today’s experiment. (Make sure you list and identify the uncertainties for each measurement.) Does your result agree with literature results within your experiment uncertainty? ● N aOH(aq) + H N O 3 (aq) → N aN O 3 (aq) + H 2 O(l) q = (100 + 0.05 mL)(4.184 J/g .K)(28.5 + 0.05) − (22.33 + 0.05) = 2582.81 (2582.81/0.0500 mol)/1000 = 51.65 J/mol uncertainty ● N aOH (aq) + H Cl(aq ) → N aCl(aq) + H 2O(l) q = (100 + 0.05 mL)(4.184 J/g .K)(29.05 + 0.05) − (22.55 + 0.05) = 2720.95 (2720.95/0.0500 mol)/1000 = 54.41 J/mol uncertainty Conclusion Understanding how a calorimeter works as a thermodynamic system with exothermic and endothermic reactions, it is possible to find the enthalpy of solids and solutions. In this lab, it was possible to find the enthalpy of neautralization of three acids, two solutions of salt, and two metals with an acid.