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Page 1: Calorimetry Presentation

CALORIMETRYDenielle B. BaybayCharmaine Joy A. Cabaña

Page 2: Calorimetry Presentation

Theoretical Framework

Calorimetry• the science of measuring heat changes

in physical and chemical processes• essential since all reactions involve

changes in energy

What you need to know!

Heat released = Heat a

bsorbed!

Heat transfers fro

m one object

to another until

they achieve

equilibriu

m.

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How to measure change in Enthalpy (∆H)?• By constructing a “surrounding”

which retains heat (so temperature may be observable)

CALORIMETERis the answer!!!

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Calorimeter as “the surrounding”!

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Definition of Terms• Specific heat capacity (s)

• intensive property• Amount of heat required to raise the temperature of

one gram of a substance by one degree Celcius.• Heat capacity (C)

• extensive property• amount of heat required to raise the temperature of a

given quantity of the substance by one degree Celcius.

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Experimental• Part A: Calorimeter Constant

This is the final /equilibrium temperature.

This is the initial temperature of calorimeter and water.

10mL of tap water

Calorimeter with thermometer

Record temperature until

3 identical readings

50mL of tap water

Heat to 50oC (use another

thermometer)

10mL of hot water

Calorimeter with tap water

Record temperature every 3s until temp

becomes constant for 4 successive

readings.

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Results• Part A:

Time (sec) 0s 3s 6s 9s 12s 15sTemperatur

e(°C)

39 39 40 40 40 40

Mass of tap water 4.08g

Mass of hot water 10.57g

Initial temperature of tap water

29°C

Initial temperature of hot water

50°C

Equilibrium Temperature 40°C

Calorimeter Constant 5.53 cal°C

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Plot of Temperature vs. Time

0 3 6 9 12 1538.4

38.6

38.8

39

39.2

39.4

39.6

39.8

40

40.2

Te

mp

era

ture

(C

)

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Part A: Calorimeter Constant• Used to calculate amount of heat

required to raise contents of calorimeter by change in temperature. (Ccal = qcal/∆T)

• qtotal = 0

• qhot water + qtap water + qcalorimeter = 0

• *q = m s ∆ T; qcal = -qhot - qtap

• qcal = -mtaps∆Ttap – mhots∆ThotRemember this?Heat transfers from one object to another until they achieve EQUILIBRIUM.

And in an isolated system, heat gained by one thing is equal to the lost of another.

Cham
Page 10: Calorimetry Presentation

Part A: Calculations

• qcal = -mtaps∆Ttap – mhots∆Thot

= -[(4.08g) (1.00 cal/goC) (40-29)oC] – [(10.57g)((1.00 cal/goC) (40-50) oC]

= 60.82 cal

• Ccal = qcal/∆T= 60.82cal / (40-29)oC = 5.53 cal/oC

where s: 1 cal/goC;

(specific heat of

water)

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Experimental• Part B: Heat of Fusion

This is the final /equilibrium temperature.

Determine weight of beaker with tissue and weight of ice.

This is the initial temperature of calorimeter and water.

10mL of tap water

Calorimeter with

thermometer

Record temperature

until 3 identical readings

Weigh 10g of ice in beaker with tissue

paper at bottom

Calorimeter with tap water

Record temperature every 3s until temp

becomes constant for 4 successive readings.

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Results• Part B:

Time (s)

Temp() Time (s) Temp(oC)

0 17 24 53 15 27 46 13 30 49 11 33 3.3512 11 36 3.915 11 39 3.918 5 42 3.921 5 45 3.9

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Mass of tap water 4.08gMass of ice + beaker + tissue

38.27g

Mass of beaker + tissue 28.65g

Mass of ice 9.62g

Initial temperature of tap water

27

Initial temperature of ice 0

Equilibrium temperature 3.9

Calorimeter constant 5.53 cal/°CHeat of fusion of ice 138.90 J/g

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0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 450

2

4

6

8

10

12

14

16

18

Te

mp

era

ture

(C

)

Time (s)

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Part B: Heat of Fusion of Ice• (a.k.a. enthalpy of fusion or specific melting heat)

energy required to change one mole of a substance from ice(solid) to water(liquid) at its melting point (0 °C)

The following relationships will be used in part B of the experiment:

A. Heat lost by water in the calorimeter = original mass of water in the calorimeter x specific heat of the water x change in temperature of the water

q = m s ∆T

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B. heat given off by the water = heat absorbed by the ice

C. heat absorbed by the ice = heat of fusion of ice x mass of melted ice

q = ∆Hfusion x Mice

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Part B: Calculations

• ∆Hfusion = qfusion ÷ mice

• qcal + qtap + qice +qfusion= 0

• ∆Hfusion = (-qice – qtap – qcal) ÷ mice

= {[-9.62g(4.184J/goC)(3.9-0)oC] – [4.08g(4.184J/goC)(3.9-29)oC] – [(10.14cal/oC)(3.9-29)oC(4.184J/goC)]} ÷ 9.62g

= 138.90 J/g or 33.20 cal/g

i

q = m s

∆Twhere s: 4.184J/goC;

(specific heat of

water)

Page 18: Calorimetry Presentation

Heat of Reaction

• heat liberated or absorbed when a chemical reaction takes place

• Heat (or enthalpy) of neutralization (∆H)• heat evolved when an acid and a base react to form a

salt plus water• Heat measurements -> performed by carrying out the

reaction in a calorimeter

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• Heat (Q) given off by the neutralization reaction • absorbed by the reaction solution and the

calorimeter. • Both the solution and calorimeter increase in

temperature due to the absorbed heat • increase - can be measured with a

thermometer.• ∆H is negative – heat is evolved

positive if heat is absorbed.

Page 20: Calorimetry Presentation

Part C. Heat of Reaction

Set I

5.5 ml 6M HCl + 4.5 ml 6M NaOH

Set II

5.5 ml 6M HOAc+ 4.5 ml 6M NaOH

Set III

5.5 ml 6M HCl + 4.5 ml 6M NH4OH

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Record temperature of each reagent

If temperatures are not equal, place in

water bath

Record as initial temperature

Measure volume of all

solutions using a syringe

Mix all reagents in the

calorimeter

Take temperature

readings every 3 seconds until

a constant temperature is

obtained.

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Results for Set I (Part C)

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Time (s) Temperature (oC)

0 40

3 42

6 43

9 43

12 43

15 43.5

18 43.5

21 43.5

24 43.8

27 43.8

30 43.8

33 43.8

• Mols of HCl: 0.033 mols• Mols of NaOH: 0.027 mols• Mols of limiting reagent: 0.027 mols• ∆H of neutralization: - 1.980 x 103 cal/mol

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Calculations:

• 5.5 ml 6M HCl • 4.5 ml 6M NaOH

• HCl(aq) + NaOH H2O(l) + NaCl(s)

To find the concentration of the acid in the solution,

M1V1 = M2V2 3.3M

M1 = M2V2 0.01L

V1 = 0.033 mol/L

M1 = (6M)(5.5ML)

10ML

M1 = 3.3M

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To find the concentration of the base in the solution,

M1V1 = M2V2 3.3M

M1 = M2V2 0.01L

V1 = 0.027 mol/L

M1 = (6M)(4.5ML)

10ML

M1 = 2.7M

*by merely inspecting the concentration of the acid and the base in the solution, it can be deduced that the BASE is the LIMITING REAGENT.

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To find the heat of neutralization,

• ∆Hneut = -Qcal – Qsol’n÷ mollimiting

• ∆Hneut = (-Ccal∆t - msol’ncsp∆t) ÷ mollimiting

• ∆Hneut = [-msol’ncsp(tf-ti) – Ccal(tf-ti)] ÷ mollimiting

Substitung the values,

∆Hneut = [(-10g)(1.00cal/goC)(43.8-40oC) - (4.07cal/oC)(43.8-40)oC] ÷ 0.027 mol

∆Hneut = - 1.980 x 103 cal/mol

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Results for Set II (Part C)

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Time (s) Temperature (oC)

0 60

3 82

6 82

9 82

12 82

• Mols of HOAc: 0.033 mols• Mols of NaOH: 0.027 mols• Mols of limiting reagent: 0.027 mols• ∆H of neutralization: -1.146 x 104 cal/mol

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Calculations:

• 5.5 ml 6M HOAc• 4.5 ml 6M NaOH

• HOAc + NaOH NaOAc + H2O

To find the concentration of the acid in the solution,

M1V1 = M2V2 3.3M

M1 = M2V2 0.01L

V1 = 0.033 mol/L

M1 = (6M)(5.5ML)

10ML

M1 = 3.3M

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To find the concentration of the base in the solution,

M1V1 = M2V2 3.3M

M1 = M2V2 0.01L

V1 = 0.027 mol/L

M1 = (6M)(4.5ML)

10ML

M1 = 2.7M

*by merely inspecting the concentration of the acid and the base in the solution, it can be deduced that the BASE is the LIMITING REAGENT.

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To find the heat of neutralization,

• ∆Hneut = -Qcal – Qsol’n÷ mollimiting

• ∆Hneut = (-Ccal∆t - msol’ncsp∆t) ÷ mollimiting

• ∆Hneut = [-msol’ncsp(tf-ti) – Ccal(tf-ti)] ÷ mollimiting

Substitung the values,

∆Hneut = [(-10g)(1.00cal/goC)(82-60oC) - (4.07cal/oC)(82-60)oC] ÷ 0.027 mol

∆Hneut = -1.146 x 104 cal/mol

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Results for Set III (Part C)

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Time (s) Temperature (oC)

0 30

3 30

6 37

9 36

12 36

15 36

18 36

• Mols of HOAc: 0.033 mols• Mols of NaOH: 0.027 mols• Mols of limiting reagent: 0.027 mols• ∆H of neutralization: -3.13 x 103 cal/mol

Page 34: Calorimetry Presentation

Calculations:

• 5.5 ml 6M HCl• 4.5 ml 6M NH4OH

• NH4OH + HCl NH4Cl + H2O

To find the concentration of the acid in the solution,

M1V1 = M2V2 3.3M

M1 = M2V2 0.01L

V1 = 0.033 mol/L

M1 = (6M)(5.5ML)

10ML

M1 = 3.3M

Page 35: Calorimetry Presentation

To find the heat of neutralization,

• ∆Hneut = -Qcal – Qsol’n÷ mollimiting

• ∆Hneut = (-Ccal∆t - msol’ncsp∆t) ÷ mollimiting

• ∆Hneut = [-msol’ncsp(tf-ti) – Ccal(tf-ti)] ÷ mollimiting

Substitung the values,

∆Hneut = [(-10g)(1.00cal/goC)(36-30oC) - (4.07cal/oC)(36-30)oC] ÷ 0.027 mol

∆Hneut = -3.13 x 103 cal/mol

Page 36: Calorimetry Presentation

Comparison of values

• ∆Hfusion = 138.90 J/g

• ∆Hneut(SET I) = - 1.980 x 103 cal/mol

• ∆Hneut(SET II) = - -1.146 x 104 cal/mol

• ∆Hneut(SET III) = -3.13 x 103

cal/mol

• 333.55 J/g• -13.48 x 103 cal/mol

• -13.42 x 103 cal/mol

• -11.92 x 103 cal/mol

Experimental Values Theoretical Values

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Conclusions and Recommendations

• Calorimetry is helpful in determining heats of reaction in chemistry experiments. Heat is related with temperature and can moreover be defined in many ways

• Calorimeter constant – specific heat of the calorimeter. Specific heats of different calorimeters vary from each other.

• Heat of Fusion – amount of thermal energy which must be absorbed or evolved for 1 mole of a substance to change state

- may be positive or negative with respect to the substance

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Conclusions and Recommendations

• Heat of neutralization – amount of heat involved (either given off or absorbed) in the neutralization reaction of acids and bases to form salt and water

• To minimize error, implementation of proper use of the apparatus for the calorimetry setups are recommended

• Other unavoidable sources of error served to cause a large percent error for each part of the experiment. However, the techniques associated with calorimetry were practiced, providing valuable experience.