Thermodynamics Enthalpy OF Formation Magnesium Oxide PDF

Title Thermodynamics Enthalpy OF Formation Magnesium Oxide
Author mudasir elahii
Course Chemistry
Institution Quaid-i-Azam University
Pages 10
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its lab report about the chemistry and will required criteria. It is all about the chemistry 101 course...


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LAB REPORT EXPERIMENT THERMODYNAMICS: ENTHALPY OF FORMATION MAGNESIUM OXIDE

TITLE: To determine the enthalpy of formation of magnesium oxide.

ABSTRACT: The objective of this experiment was to determine the enthalpy of formation of MgO. Two activities were performed. In first activity I reacted MgO with HCl and in second activity I reacted Mg with HCl. By measuring the temperature change I was able to calculate the heat generated during the reactions and was able to find the H total. It was difficult to determine the enthalpy of formation directly. However, Hess’ Law was used to find the enthalpy of reaction indirectly. Hess’ Law states that if a reaction is carried out in a series of steps, ΔH for this reaction will be equal to the sum of the enthalpy changes for the individual steps. INTODUCTION: The goal of this exercise is to measure the enthalpies of formation of Mg2+(aq) and MgO(s). The enthalpy of formation of Mg2+(aq) can be determined from the enthalpy of dissolution of 1 mol of Mg metal in a very large amount of very dilute acid (eq 1). Mg(S) + 2 H+(aq) → Mg2+(aq) + H2 (g)

∆H = ∆H˚f (Mg2+, aq)

(1)

The ∆H for this reaction is the ∆H˚f for Mg2+, because the ∆H˚f for Mg(s), H+(aq), and H2 (aq) are zero by definition. The only species in equation 1 with a nonzero ∆H˚f is Mg2+(aq), so that its enthalpy of formation is simply the enthalpy of this reaction. The enthalpy of formation of MgO is more difficult to measure directly. Mg(s) + 1/2 O2(g) → MgO(s)

∆H = ∆H˚f (MgO)

(2)

It is more convenient to use the first law of thermodynamics in the form of Hess's law to simplify the measurement of ∆H˚f (MgO). The enthalpy changes for Eq. (2) can be obtained by combining equations (3), (4), and (5) in the correct manner. Consider the following reaction equations: MgO(s) + 2 HCl(aq) → MgCl2(aq) + H2O (l)

∆H1

(3)

Mg(s) +2 HCl(aq) → MgCl2(aq) + H2(g) H2 (g) + 1/2 O2(g) → H2O(l)

∆H2 ∆H3

Mg(s) + 1/2 O2(g) → MgO(s)

(4) (5)

∆Hf =?

The heat effect for a chemical reaction run at constant pressure (such as those run on the bench top in open vessels) is the enthalpy change, qrxn = ∆H. The heat evolved for a chemical reaction can be determined by running the reaction in a calorimeter and measuring the temperature change. q= m x CP x ∆T

(6)

where CP is the heat capacity of the system. The heat capacity of the system is sum of the heat capacity of the solution in the calorimeter, Csoln, and the heat capacity of the calorimeter, Csys = Csoln + Ccp

(7)

The heat capacity of the calorimeter includes the Styrofoam cups and the thermometer. The heat capacity of the solution is given by the specific heat of the solution, Cs, multiplied by the mass of the solution: Csoln = Cs x mass of the solution

(8)

The specific heat of 0.4 M HCl, which is the average molarity of the acid used during today’s reaction, is 4.07 J g-1 ˚C-1 and its density is 1.01 g/cm3. The heat capacity of the calorimeter must be measured in a separate experiment. The heat effect for the chemical reaction is then ∆H = qrxn = -qcal The equation was used to find the heat that was lost by the reaction. Which equals to ΔH. (ΔH = qrxn) ΔHKJ/mol = ΔH/mole Percent Error, % = [Theoretical Value - experimental value/Theoretical Value]x100 Used to calculate how accurate was the value that was obtained, compared to the

actual value.

EXPERIMENTAL: MATERIALS REQUIRED: 

Computer



Vernier computer interface



LoggerPro software



Vernier temperature probe



Ring stand



Utility clamp



Slit stopper



100 mL graduated cylinder



250 mL beaker



Styrofoam cup



1.00 M HCl



Magnesium oxide (MgO)



Magnesium ribbon (Mg)



Stirring rod



Balance



Weighing paper

PROCEDURE:

o Firstly, I made some safety precautions and wear a lab coat and safety glasses (to protect my eyes).

o First Reaction MgO(s) + 2 HCl(aq) → MgCl2(aq) + H2O (l) o I connected the probe to the computer interface and prepared the computer for data collection by opening the file “19 Heat of Combustion” from the Chemistry with Vernier folder in LoggerPro. The vertical axis has temperature scaled from 0 to 500 oC. The horizontal axis has time scaled from 0 to 500 seconds.

o I placed a Styrofoam cup inside a 250 mL beaker and measured out 100.0 mL of 1.00M HCl into the Styrofoam cup. o I used a utility clamp and a slit rubber stopper to suspend the temperature probe from a ring stand. Lower the temperature probe into the solution in the Styrofoam cup. o I weighed out about 1.0 g of magnesium oxide, MgO, on a piece of weighing paper. o I recorded the exact mass used on the data table. o I collected the data and obtained the initial temperature, T1, that was 17.1 oC. After three or four readings at the same temperature I added the white magnesium oxide powder to the solution. o I used a stirring rod to stir the cup contents until a maximum temperature, T 2 that was 25.2 oC reached and the temperature starts to drop. o I collected the data and made calculations.

o Second Titration Mg(s) +2 HCl(aq) → MgCl2(aq) + H2(g) o I connected the probe to the computer interface and prepared the computer for data collection by opening the file “19 Heat of Combustion” from the Chemistry with Vernier folder in LoggerPro. The vertical axis has temperature scaled from 0 to 500 oC. The horizontal axis has time scaled from 0 to 500 seconds. o I placed a Styrofoam cup inside a 250 mL beaker and measured out 100.0 mL of 1.00M HCl into the Styrofoam cup. o I used a utility clamp and a slit rubber stopper to suspend the temperature probe from a ring stand. Lower the temperature probe into the solution in the Styrofoam cup. o I weighed out about 0.5 g of magnesium ribbon on a piece of weighing paper. o I recorded the exact mass used on the data table. o I collected the data and obtained the initial temperature, T1, that was 17.1 oC. After three or four readings at the same temperature I added the white magnesium oxide powder to the solution. o I used a stirring rod to stir the cup contents until a maximum temperature, T 2 that was 39.6 oC reached and the temperature starts to drop. o I collected the data and made calculations.

DATA AND RESULTS

Experimental Data

MgO (s) + 2HCl (aq) → MgCl2 (aq) + H2O (l)

Data 1.Volume of 1.00 M HCl (aq)

Mg(s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)

100.0mL = 100.0g

100.0mL = 100.0g

(C°)

25.2 oC

39.6 oC

3. Initial temperature, T1 (C°)

17.1 oC

17.1 oC

(#2 – #3)

(#2 – #3)

m = (#1+#5),

________________ J

________________J

Cp = 4.18J/g-C,

________________ kJ

________________ kJ

________________ Kj

________________ kJ

2. Final temperature, T2

4. Change in temperature, ΔT C° 5. Mass of solid, g Calculations 6. q = mCp ΔT

ΔT = #4 7. ΔH = -q ΔH = –#6 8. Moles of solid

(#5÷40.311) =

(#5÷24.312) =

9. ΔH/mole #7÷#8 = ___ kJ/mol

______________ kJ/mol

______________ kJ/mol

 CALCULATIONS Results and calculations Result table MgO (s) + 2HCl (aq) → MgCl2 (aq) + H2O (l)

Data 1.Volume of 1.00 M HCl (aq) 2. Final temperature, T2

(C°)

Mg(s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)

100.0mL = 100.0g

100.0mL = 100.0g

25.2 oC

39.6 oC

3. Initial temperature, T1 (C°)

17.1 oC

17.1 oC

4. Change in temperature, ΔT C°

8.1 oC

22.5 oC

5. Mass of solid, g

1.0 g

0.5 g

m = (#1+#5),

3419.658 J

9452.025 J

Cp = 4.18J/g-C,

3.419658kJ

9.452025kJ

ΔH = –#6

-3.419658kJ

-9.452025kJ

8. Moles of solid

0.025 moles

0.021 moles

-136.79 kJ/mol

-450.10 kJ/mol

Calculations 6. q = mCp ΔT

ΔT = #4 7. ΔH = -q

9. ΔH/mole #7÷#8 = ___ kj/mol

Enthalpy of formation of MgO = -572.0 kJ/mol Percentage error = 4.9 %

Sample Calculations: 1. For MgO A. Heat released by reaction q= m x Cp x ΔT m = 1.0 g Cp= 4.18 J/ g. oC ΔT= T2 – T1 = 25.2 oC – 17.1 oC = 8.1 oC

Now we put the value of m, Cp and ΔT I above equation q= 1.0 g x 4.18 J/ g. oC x 8.1 oC q= 3419.658 J

,

q=3.419658Kj

B. Enthalpy change for reaction ΔH = -q ΔH = -3.419658kJ C. Moles of Solid Moles of solid = mass of solid /molar mass of solid = 1.0 g / 40.311 g.mol-1 = 0.025 moles D. Enthalpy change of reaction per mole ΔH/mole = -3.419658kJ / 0.025 moles = -136.79 kJ/mol

2. For Mg A. Heat released by reaction q= m x Cp x ΔT m = 0.5 g Cp= 4.18 J/ g. oC ΔT= T2 – T1 = 39.6 oC – 17.1 oC = 22.5 oC Now we put the value of m, Cp and ΔT I above equation q= 0.5 g x 4.18 J/ g. oC x 22.5 oC q=9452.025 J,

q=9.452025kJ

B. Enthalpy change for reaction

ΔH = -q ΔH = -9.452025kJ C. Moles of Solid Moles of solid = mass of solid /molar mass of solid = 0.5 g / 24.312 g.mol-1 = 0.021 moles D. Enthalpy change of reaction per mole ΔH/mole = -9.452025kJ / 0.021 moles = -450.10 kJ/mol Enthalpy of formation of MgO = - 450.10Kj/mol – (-136.79Kj/mol) + (-258.8KJ/mol) = -572.0 kJ/mol Percentage error = experimental ΔHf – actual ΔHf / actual ΔHf x 100

= -572.0Kj/mol – (-601.7 KJ/mol)/ -601.7 kJ/mol x 100

= - 4.9 %

DISCUSSION: An incorrect weighing of the reactants or incorrect calculations could have been potential causes of error that caused the experimental value to differ from the actual value. Another source of potential inaccuracy could have been that the calorimeter was not tightly sealed, allowing heat to escape and cause a change in the final temperature data. The temperature

of the solution remained constant before adding the MgO. After it was added, the temperature of the solution increased and then decreased slowly as the solution was stirred. The solution became cloudy. After the temperature stabilized, the solution became clear again

CONCLUSION: The actual value for the enthalpy of formation of MgO is -601.7 kJ/mole but the value obtained in this experiment was -572.0 kJ/mole. One reason for this difference could be that not all of the MgO and Mg that were calculated on the weighting paper, were put inside of the solution. Another reason could be that the volume of HCl wasn’t accurately measured. Percentage error that was calculated in this experiment was approximately 4.9%.

REFERENCES  https://www.colby.edu/chemistry.  https://socratic.org/questions.  https://chemistry.stackexchange.com/questions/10146/finding the enthalpy of formation of magnesium oxide using hess law.  https://phdessay.com/heat-of-formation-of-magnesium-oxide.  https://everettcc.instructure.com/courses/1232740/files/59680724/download....


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