Mg O lab report PDF

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Chemical Thermodynamics: Heat of Formation of MgO(s) Purpose The purpose of this lab is to determine the enthalpy of formation of MgO (s). However, because 79% of air is made up of N2(g), without a pure oxygen environment, it is impossible to prevent the formation of Mg3N2 when Mg(s) burns in air. Therefore, in situations like this, Hess’s law is useful. This lab will also teach how to set up a calorimeter and apply the principle of Hess’s Law. The heat capacity, or ability of a substance to absorb heat, of an unknown metal will also be found. Introduction The lab has a practical use as magnesium is used in many alloys or mixtures of two metals to form a solid solution. Additionally, water has one of the highest heat capacities known. This means that it takes a large amount of energy to change the temperature of a body of water. In this lab, the Styrofoam cup calorimeter is a model for the heat generated and lost in the universe. The theory used in the experiment was Hess’s Law. According to Hess’s law, if a reaction is carried out in a series of steps, the enthalpy of formation for the reaction will equal the sum of the enthalpy changes for individual steps. The enthalpy of formation, ∆Hf, is defined as the heat given off when one mole of a substance is prepared from its constituent elements in their standard states. Therefore, the following equations can be used to find the enthalpy of MgO (s): A). H2 (g) + ½ O2 (g)  H2O (l) B). Mg(s) + 2HCl(aq)  MgCl2 (aq) + H2 (g) C). MgO(s) + 2HCl(aq)  MgCl2 (aq) + H2O(l)

∆Hf = -68.3 kcal/mole

According to Hess’s law ∆Hf(MgO) = ∆HB - ∆HC + ∆HA = ∆HB - ∆HC + (-68.3 kcal/mole) = ∆HD

H 2 (g)

+½O 2 (g)

H

2

O (l)

H f

= -68.3 kcal/mole Mg (s)

+ 2 HCl (aq)

 MgCl 2 (aq)

+H 2 (g)

MgO (s)

+ 2 HCl (aq)

 MgCl

2 (aq)

+H 2

O (l)

Mg (s)

+½O 2 (g)

 MgO

(s

H 2 (g)

+½O 2 (g)

H

2

O (l)

f

H

= -68.3 kcal/mole Mg (s)

+ 2 HCl (aq)

 MgCl 2 (aq)

+H 2 (g)

MgO (s)

+ 2 HCl (aq)

 MgCl 2 (aq)

+H 2

O (l)

Mg (s)

+½O 2 (g)

 MgO (s D). Mg(s) + ½O2 (g)  MgO(s) The formula: ∆H = mC∆T will be used to calculate the heat capacity of the unknown metal. Procedure Determination of the Heat of Formation of MgO(s) 1. Set up calorimeter 2. Prepare 250 mL of 1 M HCl from the 3 M stock solution. 3. Use a graduated cylinder to transfer exactly 100 mL of the dilute HCl solution into the calorimeter Styrofoam cup. 4. Clean a piece of Mg ribbon with steel wool and weigh it. a. record the weight to the nearest 0.0001 g. 5. Wind the ribbon around the loop of the stirrer. 6. Record the initial temperature of the HCl solution. 7. Immerse the stirrer in the HCl and stir gently. 8. Monitor the temperature rise and record the highest value reached. 9. Analyze the data with the appropriate formula to get ∆Hrxn. 10. Rinse and dry the calorimeter Styrofoam cup. 11. Place 100 mL of fresh acid solution into the Styrofoam cup. 12. Weigh approximately 0.4 g of MgO and record the exact weight. 13. Repeat the above procedure (1-8) with MgO in pace of Mg. a. Stir until all of the MgO is dissolved. Specific Heat of a Metal

1. 2. 3. 4. 5.

Obtain metal rod from instructor and record the number stamped on it Mass the rod and record the weight. Fill a beaker with water, and place on a hot plate to bring the water to boil. Use a copper wire to suspend the unknown rod in a beaker of boiling water. In the meantime, transfer 50.0mL of room temperature water to a clean dry Styrofoam cup set inside another Styrofoam cup. a. Record the temperature of the water to the nearest .2 °C. 6. After 10 minutes, transfer the metal rod from the boiling water to the water in the Styrofoam cup. 7. Record the highest temperature reached to the nearest 2 °C. 8. Repeat the experiment and determine the average specific heat of the unknown.

Data/Observations Determination of the Heat of Formation of MgO(s) Mg Initial Temperature (°C) 22.0 Final Temperature (°C) 32.6 Temperature Change (°C) -10.6 Weight (g) 0.256 Atomic / Mole Weight (g) 24.32 # of Moles 1.053 x 10-2 Heat given off (kcal) 1.06 ∆Hrxn (kcal/mole) -100.66

MgO 21.0 24.2 -3.2 0.368 40.31 9.129 x 10-3 0.32 -35.05

Specific Heat of a Metal Weight of H2O (v x d) Initial H2O Temperature Hot Metal Temperature Equilibrium Temperature ∆T of metal ∆T of H2O Sp. Heat of H2O Sp. Heat of Metal

50 g 21.2 °C 100 °C 26.2 °C -73.8 °C +5 °C 1.0 cal/g°C 0.254 cal/g°C

Calculations Determination of the Heat of Formation of MgO(s) Mg: q = mCs∆T q = (100g)(1.0 cal/g°C)(-10.6°C) q = -1060 cal = -1.06 kcal ∆Hrxn = -q/moles ∆Hrxn = -(-1.06)/( 1.053 x 10-2) ∆Hrxn = -100.66 kcal/moles MgO: q = mCs∆T q = (100g)(1.0 cal/g°C)(-3.2°C) q = -320 cal = -0.32 kcal ∆Hrxn = -q/moles ∆Hrxn = -(-0.32 kcal)/( 9.129 x 10-3) ∆Hrxn = -35.05 kcal/moles Calculated enthalpy of formation of MgO (kcal/mol): (-100.66)-( -35.05) + (-68.3) =-133.91 kcal/moles Enthalpy of formation of MgO (kJ/mol): 4.184 kJ −133.91 kcal x = -560.28 kJ/mol moles 1 kcal Specific Heat of a Metal: (heat lost) = -(heat gained) mCs∆Tmetal = -mC∆Twater (13.333g)(Cs)(-26.2°C-100°C) = -[(50g)(1 cal/g°C)(26.2 °C-21.2°C)] (13.333g)(Cs)(-73.8°C) = -[(50g)(1 cal/g°C)(5 °C)] -983.9754 Cs =250 cal Cs = 0.254 cal/g°C

Discussion A calorimeter was used to find the enthalpy of formation of MgO, which was calculated to be -560.28 kJ/mol. The specific heat capacity of the unknown metal rod was also found using a calorimeter. The unknown metal was Aluminum. This was concluded as the specific heat of Al is .24 cal/g°C, making it the closest to the found value out of all of the other metal’s specific heats. The difference between the calculated specific heat and actual specific heat was only 0.14 cal/g°C difference, suggesting that the data is important, as it is usable. An error may have occurred if the Mg strip was not cleaned thoroughly with the wool. This could cause the weight to be higher, making the # moles higher, and thus producing a lower ∆Hrxn value. This can be prevented in the future by making sure to clean both sides of the Mg strip, and to not rush this step. Another source of error could have occurred if the 1M HCl was not diluted properly, or if the HCl was not measured precisely. This could have occurred if the bottom of the meniscus curve was not read at eye level. To prevent this from happening in the future, it is important to carefully read the graduated cylinder, and bend down to read the volume correctly. Finally, a third source of error was that some of the MgO was spilled while transporting it from the scale back to the table. The amount lost was miniscule, but it could have slightly caused the ∆Hrxn calculated to be slightly lower, as less of the MgO was reacting than was recorded. To prevent this in the future, it is important to move slowly and carefully in the laboratory....