IB Chemistry Lab Report Measurements the Amounts of Heat Released Using a Coffee Cup Calorimeter in These Three Related Exothermic Reactions

Energy changes occur in all chemical reactions; energy is either absorbed or released. If energy is released in the form of heat, the reaction is called exothermic. If energy is absorbed, the reaction is called endothermic. In this experiment, we will measure the amounts of heat released using a coffee cup calorimeter in these three related exothermic reactions:

(1) NaOH(s) Na+(aq) + 0H-(aq) + x1 kcal

(2) NaOH(s) + H+(aq) + C1-(aq) H2O+ Na+(aq) + C1-(aq) + x2 kcal

(3) Na+(aq) + 0H-(aq) + H+(aq) + Cl-(aq) H20 + Na+(aq) + Cl-(aq) + x3 kcal

Using our results we will verify Hess' law. The law states that energy changes are state functions. The amount of energy depends only on the states of the reactants and the state of the products, but not on the intermediate steps. Energy (enthalpy) changes in chemical reactions are the same, regardless whether the reactions occur in one or several steps. The total energy change in a chemical reaction is the sum of the energy changes in its many steps leading to the overall reaction.

Objective

Measure the heat released in three different exothermic reactions.

Demonstrate that heats of reaction are additive, hence verifying Hess's law.

Equipment used

Safety goggles

1 plastic-foam cup

1 100-mL graduated cylinder

1 400-mL beaker

1 50-mL beaker

1 thermometer

8 centigram balances/class

1 spatula

Material required

Sodium hydroxide pellets

1.0M sodium hydroxide

0.5M hydrochloric acid

1.0M hydrochloric acid

Distilled water

Procedure

Procedure note: After each reaction, dispose of the solution and rinse the cup and thermometer with water.

Reaction I

1. Measure 100 mL of distilled water into a plastic-foam cup. Place the cup inside a 400-mL beaker for support. This assembly, together with a thermometer, will serve as your calorimeter.

2. Measure and record the mass of a 50-mLbeaker to the nearest 0.01g. Using a spatula, add as close to 2.00 g as possible of sodium hydroxide pellets to the beaker. Measure and record the combined mass of the beaker and sodium hydroxide to the nearest 0.01g. (Do this operation as quickly possible to avoid error due to absorption of water by the NaOH.)

3. Measure and record the temperature of the water in the foam cup to the nearest 0.5°C. Add the weighed NaOH pellets to the water in the calorimeter. Stir the mixture gently with the thermometer until all the solid has dissolved. CAUTION: Hold the thermometer with your hand at all times. Record the highest temperature reached during the reaction.

Reaction 2

4. Measure 100 mL of 0.5M HCI into the plastic-foam cup, and place the cup inside a400-mL beaker.

5. Using a spatula, measure out 2.00 g of solid NaOH pellets.

6. Measure and record the temperature of the HC1 solution in the foam cup. Add the NaOH pellets to the acid solution and stir gently until the solid is dissolved. Measure and record the highest temperature reached by the solution during the reaction

Reaction 3

7. Place the plastic-foam cup inside a 400-mL beaker. Measure 50 mL of 1.0M HC1 into the cup. Rinse the graduated cylinder and fill with 50 mL of 1.0M NaOH.

8. Measure and record the temperature of the HC1 solution (in the cup) and the NaOH solution (in the cylinder) to the nearest 0.5°C. Rinse the thermometer between measurements

9. Pour the NaOH solution into the foam cup. Stir the mixture gently, measure and record the highest temperature reached

Data collection

Quantitative data

Reaction 1

Table 1: Data recorded from reaction 1

Nature of data recordedReaction 1symbol

Mass of beaker28.4±0.01gm3

Mass of beaker+NaOH30.38±0.01gm2

Volume of water100.2±0.1mLVw

Initial temperature 296.5±0.5KTi

Final temperature 302.5±0.5KTf

Reaction 2

1st trial

Table 2: Data recorded from reaction 2

Nature of data recordedReaction 1symbol

Mass of beaker28.4±0.01gm5

Mass of NaOH pellets2.02±0.01gm4

Volume of 0.5M HCl100.2±0.1mLV2

Initial temperature297.1±0.5KTi2

Final temperature 309.5±0.5KTf2

2nd trial

Table 3: Data recorded from 2nd trial of reaction 2

Nature of data recordedReaction 1symbol

Mass of beaker28.4±0.01gm5'

Mass of NaOH pellets2.00±0.01gm4'

Volume of 0.5M HCl100.2±0.1mLV2'

Initial temperature297.1±0.5KTi2'

Final temperature 308.2±0.5KTf2'

Reaction3

Volume of 1.0M HCl: VHCL= 50.2±0.1mL

Volume of 1.0M NaOH: VHCL= 50.2±0.1mL

Temperature of NaOH solution: TB=297.0±0.5ºC

Temperature of HCl solution: TA=297.0±0.5ºC

Highest temperature reached after mixing the acid and the base: Tm2= 30.0±0.05ºC

Qualitative data

While removing the NaOH pellets to weigh them, they became moist only seconds after being exposed to the air.

Data processing and presentation

For all reactions

qSYSTEM=0

qCALORIMETER+qSOLUTION+qRXN=0

qRXN= -(qCALORIMETER+qSOLUTION)

qCALORIMETER= 0

qRXN=-qSOLUTION

= -mSOLUTION∙SSOLUTION∙∆T

= -dSOLUTION∙VSOLUTION∙SSOLUTION∙∆T

Specific heat capacity of solution: S1= 4.186 j/g∙°C

Density of solution: d1≈1.00g/dm3

Part 1

NaOH(s) Na+(aq) + 0H-(aq

Calculation of ∆T

∆T=Tf-Ti

= (302.5±0.5K)- (296.5±0.5K)

= 5.5±1K

Calculation of ∆H1 of reaction

Volume of solution: Vw=100.2±0.1mL=100.2∙10-3±0.1dm3

qDISS= -d1∙ Vw∙ S1

=-(1.00g/dm3)∙(100.2∙10-3±∙10-4dm3)∙[4.186 j/(g∙°K)] ∙(5.5±1ºK)

=-2.3069±4∙10-4

qDISS=∆H1

∆H1=-2.3069±4∙10-4j

Determining mass of NaOH pellets

Mass of NaOH pellets and beaker: m2=30.38±0.01g

Mass of beaker: m3=28.4±0.01g

Mass of NaOH:

m1=m2-m3

m1= (30.38±0.01g)-(28.4±0.01g)

m1= 1.98±0.02g

Determining ∆H molar

n(NaOH)=m1(NaOH)/M1(NaOH)

M1=40g∙mol-1

n(NaOH) =(1.98±0.02g)/(40g∙mol-1)

=0.0495±5∙10-4mol

∆H1=-2.3069±4∙10-4

∆Hmolar2=∆H1/ n(NaOH) = -46.5±0.5j∙mol-1

Part 2

NaOH(s) + HCl(aq) H2O(l)+ Na+(aq) + C1-(aq)

Calculation of ∆H2 of reaction

qRXN=-d1∙ V2∙SSOLUTION∙∆T

Volume of solution: V2=0.1dm3

Average ∆T of both trials: ∆Ť = ŤF2-Ť I2

=12±1K

qRXN= -5.0282±4∙10-4j

qDISS=∆H2

∆H2=-5.0282±4∙10-4j

Finding limiting reagent of reaction

Average mass of sodium hydroxide pellets: m(NaOH)= (m4+ m4')/2

= 2.01±0.01g

n(NaOH)= (2.01±0.01g)/ 40g∙mol-1

n(NaOH)=0.0503±0.0003mol

C(HCl)=0.5mol/dm3

Average volume of HCl: V(HCL)= 100.1±0.1mL=100.1∙10-3±1∙10-4dm3

n(HCl)= C∙V

=0.05005±5∙10-5 mol

n(HCl)< n(NaOH), hence HCl is the limiting reagent

Determining ∆H molar2

∆Hmolar=∆H2/ n(HCl)

∆Hmolar2=-110.4±0.1j∙mol-1

Part 3

NaOH(aq) + HCl(aq) H2O(l)+ Na+(aq) + C1-(aq)

Calculation of ∆H3 of reaction

qRXN=-d1∙ V3∙SSOLUTION∙∆T

Volume of HCl solution: V3=0.05dm3±1∙10-4dm3

Volume of NaOH solution: V4=0.05dm3±1∙10-4dm3

Volume of both solutions in calorimeter: V5= 0.1±2∙10-4dm3

of ∆T2

Initial temperature of NaOH and HCl solutions is 297.0±0.5K

∆T2=Tf-Ti

= (303.0±0.5K)- (297.0±0.5K)

= 6.0±1K

qRXN= -2.5116±4∙10-4j

qDISS=∆H3

∆H3= -2.5116±4∙10-4j

Finding limiting reagent of reaction

C(HCl)=1M

C(NaOH)=1M

Determining ∆H3 molar

∆H3molar=∆H3/ n(HCl)

∆H3molar=-110.4±0.1j∙mol-1

Resources: http://www.library.tedankara.k12.tr

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html

http://world.casio.com/edu/resources/program_lib/ea200/pdf/12_p36_37.pdf