Download - Calorimetry Presentation
CALORIMETRYDenielle B. BaybayCharmaine Joy A. Cabaña
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.
How to measure change in Enthalpy (∆H)?• By constructing a “surrounding”
which retains heat (so temperature may be observable)
CALORIMETERis the answer!!!
Calorimeter as “the surrounding”!
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.
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.
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
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
)
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.
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)
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.
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
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
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)
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
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
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)
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
• 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.
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
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.
Results for Set I (Part C)
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
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
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.
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
Results for Set II (Part C)
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
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
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.
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
Results for Set III (Part C)
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
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
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
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
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
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.