LAB REPORT CHM 432 HESS LAW CALORIMETER PDF

Title LAB REPORT CHM 432 HESS LAW CALORIMETER
Author gempoi cun
Course physical chemistry
Institution Universiti Teknologi MARA
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Summary

FACULTY OF APPLIED SCIENCESLABRORATORY REPORTPHYSICAL CHEMISTRY (CHM 432)Title of Experiment Calorimetry: Hess`s LawNo. Experiment 1Name of StudentStudent ID numberProgramme Code ASDate of experiment is done 18/10/Date of lab report is submitted 29/10/Lecturer`s name NOR AZIRA IRMA BINTI MUHAMMADOBJ...


Description

FACULTY OF APPLIED SCIENCES LABRORATORY REPORT PHYSICAL CHEMISTRY (CHM 432) Title of Experiment

Calorimetry: Hess`s Law

No. Experiment

1

Name of Student Student ID number Programme Code

AS245

Date of experiment is done

18/10/21

Date of lab report is submitted

29/10/21

Lecturer`s name

NOR AZIRA IRMA BINTI MUHAMMAD

OBJECTIVES (a) To compare the heat capacities of a coffee cup calorimeter and a copper calorimeter. (b) To determine the standard enthalpy of formation of magnesium oxide, ∆Hof INTRODUCTION The heat of reaction for a given reaction,can be calculated from the values for other reactions. This is done by making use of Hess’s Law. In this experiment you will determine the standard enthalpy (heat) of formation of magnesium oxide. The equation for the formation of magnesium oxide is: 1 Mg( s) + O 2 ( g ) ⟶ MgO(s) … … .. ( 1) 2 Mg( s) +2 HCl(aq ) ⟶ MgCl 2 (aq)+ H 2 (g) … … . (2)

Mg( s) +2 HCl(aq ) ⟶ MgCl( aq ) + H 2 O(l) … … .(3) 1 H 2 ( g) + O 2 ( g ) ⟶ H 2 O(l) … … .(4) 2 The heat of reactions for Reactions (2) and (3) will be obtained experimentally and for the Reaction (4), the heat of formation can be obtained from the literature.The heat capacity of a calorimeter can be obtained by measuring the temperature change when a known weight of hot water is added to a known amount of cold water in the calorimeter. APPARATUS 

Thermometer (2)



Styrofoam cup with cardboard cover lid (2)



Copper calorimeter with insulating cover (1)



100 cm3 beaker



100 cm3 graduated cylinder

CHEMICALS 

2.0 M hydrochloric acid



Magnesium oxide powder



Magnesium powder

PROCEDURE These experiments were performed at standard conditions (atmospheric pressure and 25C). All experiments were performed in styrofoam cups (coffee cups) ‘double’ and calorimeters copper calorimeter. A. Heat capacity of calorimeter Burettes was used to deliver exactly 50 cm3 tap water into the calorimete

Every 15 seconds, the temperature was recorded for next 3 minutes.

The cover and thermometer have been replaced. Water temperature for four minutes at one -minute intervals were recorded.

The lid was replaced and the water was stirred carefully with the thermometer.

50 cm3 hot water was measured (40 - 50 C) using graduated cylinder and pour into a beaker.

The temperature of hot water was recorderd quickly using another thermometerand poured it completely into calorimeter (containing cold water) in the fifth minute.

Steps (1) and (2) were repeated using a copper calorimeter.

B. Reaction 2: Magnesium oxide with Hydrochloric Acid 1.0- 1.1 g of magnesium powder were weighed. The actual weight used was recorded.

Temperature was recorded for the next 3 minutes at 15 seconds time intervals.

50 cm3 2 M HCl was drained from the burette to the calorimeter.

The lid is replacedand the contents of the calorimeter are carefully stirred with thermometer.

The cover and thermometer has been replaced.

At the fifth minute,the magnesium powder was poured quickly into the HCl.

C. Reaction 3: Magnesium oxide with Hydrochloric Acid

1.6 - 1.8 g of magnesium oxide was weighed

The actual weight used was recorded.

Step (2) in (B) has been repeated .

D. Reaction 4: Hydrogen gas with Oxygen gas

For safety reasons, this value will not be determined experimentally in laboratory. It has been professionally determined and verified. The value is already listed in standard reference sources

RESULT

Mass, m (g) Initial temperature (°c) (HCl) Final temperature (°c) (HCl + Mg or MgO) Temperature change (ΔT)

Magnesium (Mg) 1.0265 24.5

Magnesium oxide (MgO) 1.6598 24.5

62.5

43.0

38.0

18.5

GRAPH 1.Graph of time versus temperature of coffea cup calorimeter

2.Graph of time versus temperature of copper calorimeter

3.Graph of time versus temperature of Mg and HCl

4. Graph of time versus temperature of MgO and HCl

DISCUSSION Heat capacity of coffee cup q hot = (50)(4.18)(42-32.5)

= 1985.5 J q cold = (50)(4.18)(32.5-24.5) = 1672 J q cal = 1985.5 – 1672

= 313.5 J c cal =

313.5 (32.5−24.5)

= 39.19 J/˚C Heat capacity of copper cup q hot = (50)(4.18)(45-33.5) = 2403.5 J q cold = (50)(4.18)(33.5-24.5)

= 1881 J q cal = 2403.5 – 1881 = 522.5 J c cal =

522.5 (33.5−24.5)

= 58.06 J/˚C

From the experiment, I calculated the value of heat capacity of both calorimeter which are 39.19 J/˚C for coffee cup calorimeter and 58.06 J/˚C for copper calorimeter. The value of both calorimeters is different as the temperature changes are also different. This is because copper is an excellent conductor of heat so it can conduct heat more effective than coffee cup. The copper calorimeter can carry out heat to surrounding faster than the coffee cup which made the value of heat capacity of copper calorimeter is bigger than the value of heat capacity of coffee cup calorimeter.

Standard enthalpy of formation of MgO Mg(s )+2 HCl( aq ) ⟶ MgCl2 ( aq) +H 2 ( g ) 1.00 g 1000 ml ×0.05 l=50 g × 1l ml 1.0265 g 24.305 g/mol

n Mg =

= 0.0422 mol q rxn =

q HCl +

∆ H rxn =

q cal

(50 )(4.18 )(54.5−24.5)+(39.23 )(54.5−24.5) 0.0422

= 176.4668 kJ/mol

Mg( s) +2 HCl(aq ) ⟶ MgCl( aq ) + H 2 O(l) 1.00 g 1000 ml ×0.05 l=50 g × 1l ml 1.6589 g 40.31 g /mol

n MgO =

= 0.0411 mol q rxn =

q HCl +

∆ H rxn =

q cal

(50 )(4.18 )( 41.5−24.5 )+(39.23 )( 41.5−24.5) 0.0411

= 102.6742 kJ/mol

Mg( s) +2 HCl(aq ) ⟶ MgCl2 (aq ) +H 2( g) ∆ H =176.4668 kJ / mol Mg( s) +2 HCl(aq ) ⟶ MgCl( aq )+H 2 O( l ) ∆ H = 102.6742 kJ/mol 1 H 2 ( g) + O 2 ( g ) ⟶ H 2 O( l) ∆ H = -285.820 kJ/mol 2

Mg( s) +2 HCl(aq ) ⟶ MgCl2 (aq ) +H 2( g) ∆ H =176.4668 kJ / mol MgCl(aq) + H 2 O(l) ⟶ MgO(s )+ 2 HCl (aq ) ∆ H = -102.6742 kJ/mol

1 H 2 ( g) + O 2 ( g ) ⟶ H 2 O( l) ∆ H = -285.820 kJ/mol 2

1 Mg( s) + O 2 ( g ) ⟶ MgO(s) 2

∆ H = 176.4668+(-102.6742)+(-285.820)

= -212.0274 kJ/mol

The theoretical value of enthalpy formation of MgO is -601.7 kJ/mol Percentage error =

−212.0274 −601.7

× 100

= 35.07 %

The factor that might affected the result of this experiment is the equipment might be broken or I did not used it correctly that giving me a huge percentage error. Besides, human error might be one of the reasons such as parallax error that might occur during the measurement of the temperature or the weight. The heat capacity of coffee cup that obtained by my friend is 42.13 J/˚C while I got 35.07 J/˚C. The differences between the values might due to the different calculating when plotting the graph as we need to find perfect straight line across the curve. Standard enthalpy of formation of magnesium oxide calculating by my friend is -348.18kJ/mol while the value that I got is -212.0274 kJ/mol. The difference between the value is small but the difference might occur due to the temperature changes. When conducting the experiment, we might do some error when measuring the temperature or the weight that cause the difference in the values.

CONCLUSION From the experiment that was conducted, We can compare the heat capacity of coffee cup calorimeter and the heat capacity of copper cup calorimeter as the result of both calorimeters turn out to be different between each other. Both calorimeters have different capability to conduct heat as copper calorimeter conducted heat better than coffee cup calorimeter. We can also determine the standard enthalpy of formation of magnesium oxide by calculating using

the Hess`s law which is -212.027 kJ/mol. During the reaction, there is a chance that heat will be lost to the surroundings. The negative sign in the result shows that this is exothermic reaction thus the heat was released to the surrounding during the experiment. This is because all the results that was obtained in this experiment show that the temperature increase from initial to final temperature. QUESTIONS 1) a) Which of the two calorimeters: coffee cup calorimeter or copper calorimeter has a higher specific heat capacity? Copper calorimeter b) What conclusion can you make regarding the relationship of heat capacity of calorimeter with temperature change of the reactions? The heat capacity of calorimeter increases of the temperature change of the reactions decreases. 2) For the following problems, assume that the volume of the final solution is 200 cm 3, the density of the solution is 1.00 g mL-1 and the heat capacity of the solution is the same as water (4.184 J g-1 C). a) When 0.800 g of Ca metal is added to 200 cm3 of 0.500 M HCI (aq) according to the method described in Procedure B, a temperature increase of 13.0 C  is observed. What is ∆Hrxn at room temperature for the reaction of Ca(s) + 2H+ (aq)? Ca + HCl → CaCl2 + H2 No mole of Ca =

mass molar mass

=

0.800 g 40.078 gmol−1

= 0.020 mol Ca (Limiting

reagent) No mole of HCl = MV = (0.500 M) (0.2L) = 0.100 mol HCl

1 mol of Ca

= 1 mol of CaCl2

0.020 mol Ca = 0.020 mol of CaCl2

Q=mc∆T = (200g) (4.184 J g-1 C)  (13 C)  = 10.868 kJ

∆Hrxn =

10.868 kJ 0.020 mol

= -543.40 kJ/mol

b) When 1.12 g of CaO is added to 200 cm 3 of 0.500 M HCI (aq) according to the method described in Procedure B, a temperature increase of 4.62 C  is observed. What is ∆Hrxn at room temperature for the reaction of CaO(s) + 2H+?

CaO + 2HCl → CaCl2 + H2O No mole of CaO =

mass molar mass

=

1.12 g 56.08 gmol−1

= 0.0294 mol CaO

(Limiting reagent) No mole of HCl = MV = (0.500 M) (0.2L) = 0.100 mol HCl

1 mol of Ca0

= 1 mol of CaCl2

0.0294 mol Ca0 = 0.0294 mol of CaCl2

Q=mc∆T = (200g) (4.184 J g-1 C)  (46.2 C) = 3.862 kJ

∆Hrxn =

3.862 kJ 0.0294 mol

= -131.36 kJ/mol

3) A student carelessly inserts the thermometer while assembling the coffee cup calorimeter and a large hole is torn in the plastic lid. How will this affect his experimental results? The heat of water will be released to the surroundings through the large hole that had been formed. As the result, the temperature of hot water will be reduced faster than usual. Thus, there will be an increase in the temperature change.

REFERENCES 1. Copper Alliance (n.d), Copper: Properties and Application. Retrieved October 23,2021

from https://copperalliance.org.uk/knowledge-base/education/education-resources/copperpropertiesapplications/#:~:text=Copper%20is%20a%20good%20conductor,silver%2C %20copper%20is%20the%20best. 2. Khan Academy (n.d), Endothermic vs. Exothermic Reactions. Retrieved October 23,2021

from https://www.khanacademy.org/testprep/mcat/chemicalprocesses/thermochemistry/a/endoth ermic-vs-exothermic-reactions 3. LibreTexts (February 4,2021), Heat Capacity, Enthalpy and Calorimetry. Retrieved

October 23,2021 from https://chem.libretexts.org/Bookshelves/General_Chemistry/Map %3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/UNIT_4%3A_EQUILIBRIUM _IN_CHEMICAL_REACTIONS/12%3A_Thermodynamic_Processes_and_Thermochemi stry/12.3%3A_Heat_Capacity%2C_Enthalpy%2C_and_Calorimetry 4. Lumen Physics. Temperature change and heat capacity.Retrieved October 24,2021 from https://courses.lumenlearning.com/physics/chapter/14-2-temperature change-and-heatcapacity/...


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