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CHEM 1211 Lab Manal- Revised 052017 Standard Enthalpy of Formation Thermochemistry II Learning Outcomes 1. Demonstrate your understanding of chemical equations representing formation reactions 2. Use data from calorimetry experiments and Hess's Law to determine the enthalpy of formation for MgO. 3: Demonstrate your ability to mathematically determine limiting reactants. Introduction As in the previous experiment, this experiment will investigate Hess's Law of the transfer of heat, however, this time the calculated value will be the standard enthalpy of formation, AHs. Remember that Hess's Law states that "if a reaction can be expressed as the sum two or more reactions, then the heat flow for the overall reaction is equal to the sum of the heat flows." At constant pressure, the flow of heat is called enthalpy, and this experiment is done as constant pressure in the lab. The standard enthalpy of formation is defined as the enthalpy involved in the formation of le of a compound from the corresponding elements as they are found usually at 25°C and I atmosphere pressure. Note the examples in Eq. I through Eq.4 (g) CHs (g)+2 0: (g)2 H:O (g) + CO C dano ed (s) + O2(g) + CO2 (g) C gnphite (s)+ O (g) Eq. 2 Eq. 3 The first three reactions all form carbon dioxide and can occur at 25°C and 1 pressure. However, in equation 1, the starting materials are not all elements since methane is a compound, and two products are formed. Hence equation 1 is exel definition of a standard formation reaction. The natural state of carbon i graphite, not diamond, and thus equation 2 does reaction. Finally equation 3 represents "the formation of the compound CO2 from the elements as they are found at standard conditions." Equation 4 does the same for li water s accepted as not fit the definition of a standard formation When magnesium metal is burned at constant pressure in the presence of oxygen, the heat evolved per mole of magnesium is the ??? for MgO(s), (Eq, 5). This process will be demonstrated for you in the lab. The o in ??? indicates standard conditions. Mg(s) + ½Odg) + MgO(s) 73

Part 1!

Answer 1

Trial - 1 Mg + 2HCl Rightarrow H_2 + Mg Cl_2 Mass of Mg = 0.549 g Volume of Hcl = 94 mL Molarity of Hcl = 3.00 M 0.549 g Mg/24.305 g/mol = 0.02258 mol mg 0.094 L Hcl times 3.00 M = 0.282 mol HCL 0.0258 mol mg times 2 mol HCL/1 mol Mg = 0.04516 mol HCL This means that you need 0.04516 mol of HCl to react 0.02258 mol mg, 0.02258 mol of mg, the limiting reagent. Limiting reagent is Mg Mole product of limiting reagent is 0.02258 mol. T_1 degree = 21.0 degree c T_2 degree = 46.6 degree c q = mc Delta T q = 0.549 g times 4.18 J/g degree c times (46 - 21) degree c q = 57.3705 J Delta H = -q/12 Delta H = -57.3705/0.02258 = Delta H = -2540.766 J/mol Delta H = -2.540 kJ/mol