Lab Report 1 (che165) v4368

UNIVERSITI TEKNOLOGI MARA FAKULTI KEJURUTERAAN KIMIA CHEMISTRY LABORATORY (CHE 165) NAME / STUDENT NO. : MOHD IZWAN BIN ABD WAHID / 2008543089 MOHD ZULFADHLI BIN FAIZOL AZMI / 2008504423 NUR SYAFIQA BINTI AMAN / 2008511857 PUTERI NURULBAITY BINTI MEGAT YUNUS / 2008548465 EXPERIMENT

: LAB 1 (HEAT OF NEUTRALIZATION)

DATE PERFORMED

: 21st JULY 2009

SEMESTER

: JULY 2009 - NOVEMBER 2009

PROGRAMME / CODE : DIPLOMA IN CHEMICAL ENGINEERING / EH110 No 1 2 3 4 5 6 7 8 9 10 11 12 Remarks:

Title Abstract Introduction Objective Theory Apparatus Methodology Results Discussion Calculation Conclusion Recommendation Reference Total

Allocated marks (%) 5 5 5 5 5 10 10 20 10 5 5 5 90

Marks (%)

Checked by: by: (MISS HAMIZURA BINTI HASSAN)

Rechecked

ABSTRACT The objective of this experiment was to determine the energy changes accompanying neutralization reactions by using a calorimeter. The energy changes of neutralization by using a calorimeter was used by thermometer at three times to get the average of the temperature. From the graft given, the experiment was successful.

INTRODUCTION Every chemical change is accompanied by a change in energy, usually in the form of heat. The energy change of a reaction that occurs at constant pressure is termed the heat of reaction of the enthalpy change. The symbol ∆H (the symbol ∆ means “change in”) is used to denote the enthalpy change. If heat is evolved, the reaction is exothermic (∆H < 0); and if heat is absorbed, the reaction is endothermic (∆H > 0). In this experiment, the heat of neutralization (or the enthalpy of neutralization) will be measured when an acid and a base react to form water. This quantity of heat is measure experimentally by allowing the reaction to take place in a thermally insulated vessel called a calorimeter. The heat liberated in the neutralization will cause an increase in the temperature of the solution and of the calorimeter. If the calorimeter were perfect, no heat would be radiated to the laboratory. The concern in this experiment is with the heat of the reaction and because some heat is absorbed by the calorimeter itself, the amount of heat absorbed by the calorimeter must be known. This requires the determination of the heat capacity of the calorimeter. “Heat capacity of the calorimeter” means the amount of the heat (that is the number of joules) required to raise its temperature 1 Kelvin, which is the same as 1˚C. In this experiment, the temperature of the calorimeter and its contents is measured before and after the reaction. The change in the enthalpy, ∆H, is equal to the negative product of the temperature change, ∆T, times the heat capacity of the calorimeter and its contents:

∆H = -∆T (heat capacity of calorimeter + heat capacity of contents) [1] Note that the numerical difference on the Celsius scale is the same as the numerical difference on the Kelvin scale where ∆T is the difference between the final and initial temperatures: ∆T = T f – Ti. Because ∆H is negative for an exothermic reaction whereas ∆T is positive, a negative sign is required is Equation [1].

OBJECTIVES To measure, using a calorimeter, the energy changes accompanying neutralization reactions.

THEORY The heat capacity of the calorimeter is determined by measuring the temperature change that occurs when a known amount of hot water is added to a known amount of cold water in the calorimeter. The heat lost by the warm water is equal to the heat gained by the cold water and the calorimeter (assume no heat is lost to the laboratory). For example, if T1 equals the temperature of a calorimeter and 50 mL of cooler water, if T2 equals the temperature of 50 mL of warmer water added to it, and if Tf equals the temperature after mixing, then the heat lost by the warmer water is heat lost by warmer water = (T2 – Tf) X 50 g X 4.18 J/K-g [2] The specific heat of water is 4.184 J/K-g, and the density of water is 1.00 g/mL. The heat gained by the cooler water is heat gained by cooler water = (Tf – T1) X 50 g X 4.18 J/K-g [3]

The heat lost to the calorimeter is the difference between heat lost by the warmer water and that gained by the cooler water: (heat lost by warmer water) – (heat gained by cooler water) = heat gained by the calorimeter Substituting Equations [2] and [3]: [(T2 – Tf) X 50 g X 4.18 J/K-g] – [(Tf – T1) X 50 g X 4.18 J/K-g] = (Tf – T1) X heat capacity of calorimeter [4] Note that the heat lost to the calorimeter equals its temperature change times its heat capacity. Thus by measuring T1, T2, and Tf, the heat capacity of the calorimeter can be calculated from Equation [4]. Once the heat capacity of the calorimeter is determined, Equation [1] can be used to determine the ∆H for the neutralization reaction.

APPARATUS & CHEMICALS Thermometers (2)

Split one hole stopper

Styrofoam cups (2)

250 mL beaker (2)

50 mL graduated cylinder (2)

400 mL beaker

Square cardboard with hole in center

Hot plate

1 M HCl 1 M NaOH 1 M acetic acid (HC2H 3O2)

METHODOLOGY Part A. Heat Capacity of Calorimeter 1.

A calorimeter was constructed similar to the one shown if Figure 1 by nesting two

Styrofoam cups together. A cork borer was used to make a hole in the lid just big enough to it the thermometer and slip the thermometer into a split one-hole rubber stopper to prevent the thermometer from entering too deeply into the calorimeter. The thermometer should not touch the bottom of the cup. The entire apparatus was rested in a 400 mL beaker to provide stability. 2.

50.0 mL of tap water was placed exactly in the calorimeter cup, and the cover and

thermometer was replaced. 5 to 10 minutes was allowed for the system to reach thermal equilibrium; then the temperature (to the nearest 0.1˚C) was recorded. Temperature after thermal equilibrium is reached = TEquilibrium. 3.

50.0 mL of water was placed exactly in a 250 mL beaker and the water was heated with

a low flame until the temperature is approximately water will be lost, leading to an erroneous result. 4.

The hot water was allowed to stand for a minute or two, quickly record its temperature

i.e. t0 = 0 s (to the nearest 0.1˚C) and pour it as completely as possible into the calorimeter. The lid was replaced with the thermometer and the water was carefully stirred with the thermometer. The temperature was recorded as a function of time i.e. the first 3 temperature readings are taken every 5 seconds and for the next two minutes the temperature readings was taken every 15 seconds. A temperature-versus-time curve was constructed from the data in Table 1.

Part B. Heat of Neutralization of HCL-NaOH 1.

The calorimeter and the thermometer were dried with a towel. 50.0 mL of 1.0 M NaOH

was carefully measured and added it to the calorimeter. The lid was placed on the calorimeter but leave the thermometer out. 50.0 mL of 1.0 M HCl was measured out exactly into a dry beaker. Allowed it to stand near the calorimeter for 3 to 4 minutes. The temperature of the acid (THCl) was measured and recorded. 2.

The thermometer with tap water was rinsed, and wipe dried. The thermometer was

insert into the calorimeter and the temperature of the NaOH solution was measured. The temperatures of the NaOH and the HCl should not differ by more than 0.5˚C. If the difference is greater than 0.5˚C, adjust the temperature of the HCl by either warming it by holding the beaker in your hands or cooling the outside of the beaker with tap water until the temperature of HCl is within 0.5˚C of that of the NaOH. The temperature of the NaOH solution i.e. t 0 = 0 s was recorded. 3.

The lid was lift and carefully added the 1.0 M HCl all the once. Be careful not to splash

any on the upper sides of the cup. The solution was stirred gently with the thermometer and the first 3 temperature readings were recorded as the function of time every 5 seconds. Then, for the next 3 minutes the temperature readings were taken every 15 seconds. The NaCl solution may be assumed. To have the same density and specific heat as water. A temperature-versustime curve was construct from the data in the Table 2. Part C. Heat of Neutralization of HC2H3O2–NaOH 1.

The same procedure as in Part B was followed, but substitute 1.0 M HC 2H3O2 in place

of 1.0 M HCl. Waste Disposal Instructions The stock solutions were handled carefully. A wet sponge or paper towel may be used to clean up any spills. The reaction mixtures produced in the Styrofoam cups contain harmless salts. They should be disposed of in the designated receptacles.

RESULTS & CALCULATIONS Results for Part A Table 1: Data for experiment of Heat Capacity of Calorimeter Notation

Time, Temperature, Temperature, Temperature, s ˚C (1) ˚C (2) ˚C (3) 26.0 25.0 27.0 0 40.0 40.0 40.0 5 38.5 37.0 37.5 10 36.0 35.0 34.0 15 36.0 35.0 34.0 30 36.0 35.0 34.0 45 36.0 35.0 34.0 60 36.0 35.0 34.0 75 36.0 35.0 34.0 90 36.0 35.0 34.0 105 36.0 35.0 34.0 120 36.0 35.0 34.0 135 36.0 35.0 34.0

TEquilibrium t0 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10 t11

Temperature, ˚C (Average) 26.0 40.0 37.5 35.0 35.0 35.0 35.0 35.0 35.0 35.0 35.0 35.0 35.0

Results for Part B Table 2: Data for experiment of Heat of Neutralization of HCl-NaOH

Notation THCl t0 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10 t11

Time, Temperature, Temperature, Temperature, Temperature, s ˚C (1) ˚C (2) ˚C (3) ˚C (Average) 27.0 26.0 28.0 27.0 0 26.5 26.0 27.0 26.5 5 28.0 27.0 29.0 28.0 10 28.0 28.0 28.0 28.0 15 28.0 28.0 28.0 28.0 30 28.0 28.0 28.0 28.0 45 28.0 28.0 28.0 28.0 60 28.0 28.0 28.0 28.0 75 28.0 28.0 28.0 28.0 90 28.0 28.0 28.0 28.0 105 28.0 28.0 28.0 28.0 120 28.0 28.0 28.0 28.0 135 28.0 28.0 28.0 28.0

Results for Part C Table 3: Data for experiment of Heat of Neutralization of HC2H3O2–NaOH Notation THC2H3O2 t0 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10 t11

Time, Temperature, Temperature, Temperature, s ˚C (1) ˚C (2) ˚C (3) 27.0 26.0 28.0 0 26.5 26.5 26.5 5 30.0 29.0 31.0 10 33.0 33.0 33.0 15 33.0 33.0 33.0 30 33.0 33.0 33.0 45 33.0 33.0 33.0 60 33.0 33.0 33.0 75 33.0 33.0 33.0 90 33.0 33.0 33.0 105 33.0 33.0 33.0 120 33.0 33.0 33.0 135 33.0 33.0 33.0

Temperature, ˚C (Average) 27.0 26.5 30.0 33.0 33.0 33.0 33.0 33.0 33.0 33.0 33.0 33.0 33.0

DISCUSSION The need to repeat the experiments or the readings of the temperature differently is required. This is because, we can get the accurate readings by the average of the temperature readings. Then, the delinquency, for instant, if there are any mistakes during the first experiment can be seen and will be corrected. Other than that, any unexpected behavior; such as entered the laboratory not at the time or not punctual must not be happened. Then, students should not play and must be serious during the experiment. Moreover, do not play with the chemicals especially the things which is easy burning. And always practice the safety first that is required during the experiment and while we entered the laboratory.

CONCLUSIONS Part A. Heat Capacity of Calorimeter Determine ∆T from the curve and calculate: 1.

Temperature of calorimeter and water before mixing

26 ˚C

2.

Temperature of warm water

40 ˚C

3.

Maximum temperature determined from the curve

39 ˚C

4.

Heat lost by warm water

522.5 J

= ( T2 – Tf ) x 50g x 4.14 J/K-g = ( 40 ˚C - 37.5 ˚C) x 50g x 4.18 J/K-g = 522.5 J 5.

Heat gained by cooler water

2403.5 J

= ( Tf – T1 ) x 50g x 4.18 J/ K-g = ( 37.5 ˚C – 26 ˚C ) x 50g x 4.18 J/K-g = 2403.5 J 6.

Heat gained by the calorimeter = ( heat lost by warmer water ) – ( heat gained by cooler water )

-1941 J

= 522.5 J – 2463.5 J = -1941 J 7.

Heat capacity of calorimeter

-168.8 J/K

= heat gained by the calorimeter ( Tf – T1 ) = -1941___ ( 37.5 – 26 ) = -168.8 J/K

Part B. Heat of Neutralization of HCl-NaOH Determine ∆T from the curve and calculate: 1.

Temperature of calorimeter and NaOH

26.5 ˚C

2.

∆T determined from curve after adding HCl to the NaOH

28 ˚C

3.

Heat gained by solution

418 J

the heat gained by the solution , where ∆T = 28 ˚C 1K x 100g x 4.18 J/K-g = 418 J 4.

Heat gained by calorimeter

-168.8 J

the heat gained by the calorimeter , where ∆T = 28 ˚C – 27 ˚C, is 1K x ( -168.8 J/K) = -168.8 J 5.

Total joules released by reaction ([3] + [4])

2234.7 J

2403.5 J + (-168.8 J) = 2234 J 6.

Complete this reaction: HCl + NaOH  NaCl + H2O

7.

The number of moles of HCl in 50 mL of 1.0 M HO (show calculations) 0.05mol HCL No of moles = MV 1000

= ( 1.0 m ) ( 50 ml ) 1000 = 0.05 mol HCL 8.

The number of moles of H2O produced in reaction of: 50 mL 1.0 M HCl and 50 mL 1.0 M NaOH (show calculations):

0.05mol H2O

0.05 mol HCL x 1 mol H2O 1 mol HCL 9. Joules released per mole of water formed

45294 J/mol

= 2234.7 J 0.05mol = 45294 J/mol

Part C. Heat of Neutralization of HC2H3O2 –NaOH 1.

Temperature of calorimeter and NaOH

26.5 ˚C

2.

∆T determined from curve after adding HC2H3O2 to the NaOH

30 ˚C

3.

Heat gained by solution

1254 J

The heat gained by the solution , where ∆T = 30 ˚C – 27 ˚C 3K x 100g x 4.18 J/K-g = 1254 J 4.

Heat gained by calorimeter

-506.4 J

The heat gained by the solution , where ∆T = 30 ˚C – 27 ˚C 3K x ( -168.8 J/K ) = -506.4 5.

Total joules released by reaction ([3] + [4]) 2403.5 J + (-168.8 J) = 2234 J

6.

Complete this reaction: HC2H3O2 + NaOH NaC2H3O2 + H2O

7.

The number of moles of H2O produced in reaction of: 50 mL 1.0 M HC2H3O2 and 50 mL 1.0 M NaOH (show calculations): No of moles = MV 1000 = (1.0M)(100mL) 1000

0.1 mol

= 0.1 mol 8.

Joules released per mole of water formed

22347 J/mol

= 2234.7 J 0.1 mol = 22347 J/mol

RECOMMENDATION The recommendations of this experiment are the technique and apparatus should be improved. Moreover, we must make sure that the observation that cause errors during the experiment; such as parallax erratum, did not happened.

REFERENCES Eng Ngnang Heng / Focus Super Chemistry / 2007 Raymond Chang / Chemistry Ninth Edition / 2008 Douglas C Giancoli / Sixth Edition Physics / 2005