CHEM 123L Exp 2 - experiment 2 for the Chem 123 lab PDF

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experiment 2 for the Chem 123 lab...

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Experiment 2 The purpose of this lab was to determine the molar enthalpy of the NaOH solution, measure the heat capacity of the calorimeter, and measure the enthalpy of neutralization reactions. In this experiment, calorimetry was the technique used to measure the amount of heat transferred, absorbed, or affected by the system (Calorimetry, 2017). To measure the change in temperature, an apparatus called a calorimeter was used. With this device, a chemical or physical process occurs in isolation, meaning that no heat is lost or gained (Calorimeter, 2017). This allows the processes heat flow to be measured by determining the change in temperature (Calorimeter, 2017). The calorimeter apparatus used in this experiment was not a sealed system, meaning that the internal pressure was equal to the atmospheric pressure. Therefore, the heat of the reaction (q) and the enthalpy change (ΔH) are equal. This can be expressed through the following equation: ΔH (J) = q = mass (g) x specific heat (J deg-1 g-1) x ΔT (deg). To find the heat required to raise the temperature of a sample, the following equation is used: ΔH = q = ms ΔT or ΔH = q = C ΔT. For accurate measurements, the calorimeter constant, which is the amount of heat needed to raise the temperature of the calorimeter by 1 degree, should be found. To find this value, the following equation is used: CcalΔT = - (msΔT)heated H2O - (msΔT)cooled H2O. The final temperature is the result of both hot and cold water together, and the mass of both hot and cold water are found by using the known volumes and assuming that 1 mL is equal to 1 gram, as well as by using the specific heat of water, being 4.184 J ℃-1 g-1. The specific heat of NaOH and HCl can be determined experimentally by mixing it in the calorimeter with a known amount of water and recording the changes in temperature (Calorimetry, 2017). Eventually, the water and the metal will reach the

2 same temperature, which means that the heat gained by the water is equal to the heat lost by the metal (Calorimetry, 2017). When measuring the enthalpy of neutralization reactions, for all reactions involving strong electrolytes, the heat of neutralization (qneut) will be constant – producing the same amount of heat; q = -55.90kJ of heat per mol H2O (Heat of Neutralization, 2012). When the system is under constant pressure, which in this case it is, the heat of the reaction is equal to the enthalpy change, thus ΔHneut is also equal to -55.90kJ of heat per mol H2O (Miller et al, 1947). This value represents that heat is being released from the system, indicating that the reaction is exothermic (Miller et al, 1947). For reactions involving weak electrolytes, the heat of reaction is the same as that for a strong electrolyte reaction, meaning that it will also produce -55.90kJ of heat per mol H2O (Heat of Neutralization, 2012). In contrast to strong electrolytes, the heat of reaction in the dissociation step of the reaction is dependent upon the structure of the weak electrolytes and can be endothermic or exothermic, meaning that the heat of reaction for the process will be greater or less than -55.90kJ per mol H2O (Miller et al, 1947). The following equation is used to differentiate between strong and weak electrolytes: ΔH = qneut = (msΔT). In determining the heat of solution of a salt, after the reaction has finished, the system is composed of the calorimeter and an aqueous solution, having heat correlating with each. With no heat being lost or gained from the environment, the following can be used: qcal = CcalΔT and qsoln = CsolnΔT = msoln ssoln (Tf-Ti). The specific heat capacity of the aqueous solution can either be determined experimentally, or by assuming that it is equal to 4.184 J ℃-1 g-1. The molar enthalpy of the solution is often expressed and determined through the equation: ΔHsoln = qrxn / n. The molar enthalpy of the solution is correspondent to the heat of the solution per mole of solute (Miller et al, 1947).

3 Procedure: The experimental procedure used for this experiment was outlined in the CHEM 123L lab manual, Experiment #2. All steps were followed without deviation. Experimental Observations: Part A: Mass of empty calorimeter  52.45g Mass of NaOH  10.37g Initial temperature of water  23.5℃ Table 1: Temperatures after mixing NaOH at 10 second intervals Time (s) 10 20 30 40 50 60

Temperature (℃) 24.8 26.0 31.0 37.7 38.1 40.8

Time (minutes) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Temperature (℃) 33.3 33.6 33.5 33.4 33.4 33.3 33.2 33.2 33.2 33.1 33.1 33.0 33.0 32.9 32.9 32.8 32.8 32.7 32.7 32.6 32.6 32.5 32.5

Table 2: Temperatures after mixing NaOH at 1 minute intervals

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Part B: Mass of calorimeter with 125mL heated DI water  122.64g Mass of calorimeter with 250mL DI water (after reaction is complete)  246.35g Initial temperature of cool water  22.6℃ Temperature of cooled and heated H2O 34.1℃ Table 3: Temperatures of heated water prior to mixing Time (minutes) 1 2 3 4 5

Temperature (℃) 57.4 56.0 55.1 54.3 53.7

Table 4: Temperatures of mixture of heated and cool water at 1 minute intervals Time (minutes) Temperature (℃) 1 34.3 2 34.3 3 34.3 4 34.3 5 34.2 6 34.2 7 34.2 8 34.1 9 34.0 10 34.0 (℃) Time (s) Temperature 5 10 15 20 25 30 35

27.4 27.4 27.4 27.4 27.4 27.4 27.4

Part C: Concentration of HCl  0.9713 mol/L Temperature of HCl before mixing  22.4℃ Temperature of NaOH  22.6℃ Table 5: Temperature after mixing HCl and NaOH at 5 second intervals

5 40 45 50 55 60

27.4 27.4 27.4 27.4 27.4

Table 6: Temperature after mixing HCl and NaOH at 30 second intervals Time (minutes) .30 1.00 1.30 2.00 2.30 3.00 3.30 4.00 4.30 5.00 5.30 6.00 6.30 7.00 7.30 8.00 8.30 9.00

Temperature (℃) 27.4 27.3 27.3 27.3 27.3 27.3 27.3 27.3 27.3 27.3 27.2 27.2 27.2 27.2 27.2 27.2 27.2 27.2

Results and Calculations: Part A: a) q = - (mH2O + mNaOH) (SH2O) (ΔT) = - (250g + 10.37g) (4.184 J ℃-1 g-1) (40.8℃ - 24.2℃) = - 18083.842 J ΔH = q / [( nH2O + nNaOH ) / M] = (- 18083.842 J) / [(250g +10.37g / 58.01239 g mol-1)] = (- 18083.842 J) / (4.4882 mol) = - 4036.96 J mol-1

6 b) Ccal = 0.306 kJ ℃-1 (determined in Part B) qrxn = - (Ccal + msolnSsoln) (ΔT) = - (303.862 J ℃-1 + (260.37g) (4.184 J ℃-1 g-1)) (40.8℃ - 24.2℃) = - 23127.951 J Part B: Ti of cool H2O  22.6℃ Ti of heated H2O from extrapolation  52.7℃ Tf of both cooled and heated H2O from extrapolation  35℃ mhotH2O = mcalorimeter 125mL heated H2O - mcalorimeter = (122.64g) – (52.45g) = 70.19g mcoldH2O = mcalorimeter 250mL H2O after rxn - mcalorimeter = (246.35g) – (52.45g) = 193.90g qcalorimeter = - (msΔT)heated H2O - (msΔT)cooled H2O = - ((70.19g) (4.184 J ℃-1 g-1) (35℃ - 52.7℃)) – ((193.90) (4.184 J ℃-1 g-1) (35℃ jdbsuguishf22.6℃)) = - (- 5198.047 J) – (10059.842 J) = - 4861.795 J CcalΔT = - (msΔT)heated H2O - (msΔT)cooled H2O = - ((70.19g) (4.184 J ℃-1 g-1) (35℃ - 52.7℃)) – ((193.90) (4.184 J ℃-1 g-1) (35℃ jdbsuguishf22.6℃)) = - (- 5198.047 J) – (10059.842 J)

7 = 4861.795 J / ΔT = 4861.795 J / 16℃ = 303.862 J ℃-1 = 0.304 kJ ℃-1 Part C: CNaOH = 10.00gmol / 39.997g / 0.25L = 1.00 mol L-1 ΔHH2O = -55.9kJ mol-1 H2O HCl + NaOH  HOH + NaCl Note: Assume that NaOH is the limiting reagent in the equation above qneut = (msΔT)NaOH = (250g) (4.184 J ℃-1 g-1) (40.8℃ – 24.8℃) = 16736 J = 16.736 kJ qneut / ΔHH2O = 16.736 kJ / 55.9kJ mol-1 H2O = 0.2994 mol H2O = 0.2994 mol NaOH CNaOH = mol/L = 0.2994 mol / 0.25 L = 1.198 mol L-1 qneut = (16.736 kJ) – (0.304kJ ℃-1) (-17.57) = 16.73kJ – 5.38kJ = 11.35kJ / ΔHH2O = 11.35kJ / 55.90kJ

8 = 0.203 mol / 0.25 L = 0.812 mol L-1

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12 Summary: Part A q (without Ccal): -18083.842 J

Part B Ti of cool H2O: 22.6℃

Part C ΔHH2O: -55.9kJ mol-1 H2O

ΔH: -4036.96 J mol-1

Ti of heated H2O from

CNaOH: 1.00 mol L-1

qrxn (with Ccal): -23127.951 J

extrapolation: 52.7℃

qneut: 16.736 kJ

Tf of both cooled and heated

CNaOH2: 1.198 mol L-1

H2O from extrapolation:

qneut: 0.812 mol L-1

35℃ mhotH2O: 70.19 g mcoldH2O: 193.90 g qcalorimeter: -4861.795 J Ccal: 0.304 kJ ℃-1

Discussion: In this experiment, when performing calculations, it was assumed that for when measuring the enthalpy of neutralization reactions involving strong and weak electrolytes, the heat of reaction is equal to -55.90kJ per mole of H2O (Miller et al, 1947). When determining the heat of solution of a salt, it is assumed that the specific heat capacity of the aqueous solution is equal to 4.184 J ℃-1 g-1. By making this assumption, we conclude that each solution used has the same specific heat capacity as water, which would increase the chances of getting altered results. When finding the mass of both hot and cold water, it is assumed that with using the known volume, 1 mL is equal to 1 gram. The apparatus used was a calorimeter made out of a styrofoam cup, also known as an expanded polystyrene calorimeter. Since this calorimeter is not a sealed system, the internal and atmospheric pressure are equal, which may allow errors to occur, such as heat leaving or entering the system. This error has a great effect on both the temperatures and values determined. Some other possible errors that may have occurred when performing this lab include having the

13 magnetic stirrer strike the thermometer when recording temperature changes, having dissimilar NaOH and HCl temperatures when mixing, as well as having too much heat lost to the surroundings due to the apparatus and methods. Firstly, if the magnetic stirrer was striking the thermometer when recording temperature changes, the temperatures would be inaccurate due to the stirrer creating a ‘vortex’ where it spins – decreasing the temperatures recorded (Miller et al, 1947). Secondly, if the NaOH and HCl solutions had dissimilar temperatures (being more than 0.5℃ apart), then the solution after mixing would be either too cold or too hot, which would affect the recordings and calculations. Lastly, it is possible that more heat than expected was lost to the surroundings due to the apparatus used, as well as the cover having a hole in the middle. This would allow the temperatures to be a lesser value than anticipated. As a result, it is possible that systematic, instrumental, or method errors may have occurred during this lab which would have affected the results and calculations made in this experiment. Conclusion: In conclusion, the molar enthalpy of the NaOH solution was measured, the heat capacity of the calorimeter, and the enthalpy of neutralization reactions were determined through various calculations. The molar enthalpy of the solution was determined to be -4036.96 J mol-1, therefore meaning that heat is released during the reaction, as well as that there is constant pressure acting on the system. Assuming no heat was lost to the calorimeter, the q value was found to be -18083.842 J. Including the heat lost to the calorimeter, the q value was found to be -23127.951 J, which is a larger value due to the loss of heat. The second value was determined using the Ccal constant which was determined in part B of the calculations. This constant was calculated to be 0.304 kJ ℃-1, which was done by using the ΔT values found by extrapolating graph 2. In part B, the qcalorimeter value was determined to be -4861.795 J, which seems reasonable since all qcalorimeter values must be a negative value when qrxn = -qcalorimeter. Due to the apparatus used in this lab not being a sealed system, the internal pressure is equal to the atmospheric pressure, therefore allowing for more possible errors to occur during the experiment.

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References Calorimeter. (2017, June 16). Retrieved from http://www.oxfordreference.com/view/10.1093/oi/authority.20110803095543561 Calorimetry. (2017, June 16). Retrieved from http://www.oxfordreference.com/view/10.1093/oi/authority.20110803095543563 Department of Chemistry. CHEM 123L Laboratory Manual; University of Waterloo: Waterloo, Canada, 2019; p 33-35. Heat of Neutralization: HCl(aq) NaOH(aq). (2012). Retrieved from https://chemdemos.uoregon.edu/demos/Heat-of-Neutralization-HClaq-NaOHaq Miller, J. G., Lowell, A. I., & Lucasse, W. W. (1947, 03). Calorimetric studies of neutralization reactions. Journal of Chemical Education, 24(3), 121. doi:10.1021/ed024p121...