Calorimetry Lab (1)

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Calorimetry Lab

Introduction Calorimetry is a technique used in chemistry to measure the heat involved in a chemical reaction or physical process. The heat in a reaction or process is related to the change in the system's internal energy, which can be measured using a calorimeter. Several scenarios were observed using the calorimeter, such as a strong acid-strong base reaction, hot water mixed with room temperature water, ammonium chloride added to water, and inserting a hot metallic object into the water. This experiment aimed to determine the reaction's enthalpy change (ΔH). The enthalpy change of a chemical reaction is a thermodynamic property that describes the amount of heat released or absorbed during the reaction at constant pressure. The enthalpy change can be positive (endothermic) or negative (exothermic) depending on whether the reaction absorbs or releases heat. A simple coffee cup calorimeter was used in this experiment to measure the heat released during the reactions. The heat released from the reactions was used to heat a known amount of water in the calorimeter, which allowed the calculation of the enthalpy change for the reaction using the equation: ΔH = q / n Where ΔH is the enthalpy change, q is the heat released by the reaction, and n is the number of moles of the limiting reagent. Another useful equation that can be used is: q = mcΔT q (J) represents the amount of heat energy transferred between the system and its surroundings, m(g) represents the mass, c (J/g·°C) is the specific heat capacity of the substance, and ΔT(°C) represents the temperature change. Methods and Materials Since this lab has four parts, each will be sectioned off with its letter. For all sections, when the calorimeter cup is mentioned, it refers to two styrofoam cups stacked on each other and a styrofoam lid with two holes for the thermometer and stirring rod. A The materials used in this lab section include the calorimeter cup, a graduated cylinder, a 125mL Erlenmeyer flask, a hot plate, a top-loading balance, approximately 200mL of distilled water, a thermometer, and a stirring rod. The procedure below was performed two times. Ensure that the calorimeter cups are dried in between uses.

50 mL of water, measured using a graduated cylinder, was added to a 125 mL Erlenmeyer flask. The flask was then heated on a hot plate until the water temperature had risen to about 20°C above room temperature. Then, the flask was removed from the hot plate. We weighed the dry calorimeter to 0.1 g accuracy and recorded the value. Next, 50 mL of distilled water at room temperature was poured into the calorimeter, which was then re-weighed and recorded. The temperature of the water in the calorimeter was also measured and recorded. The hot water temperature was recorded and immediately poured into the calorimeter. We replaced the calorimeter cover and began stirring. Temperature readings were taken every 5 seconds for 2 minutes or until the temperature change became constant. Finally, the calorimeter and contents were weighed to determine the mass of the hot water added. B The materials used in this lab section include the calorimeter cup, two graduated cylinders, 50 mL of 5.00 M HCl, 50 mL of 5.00M NaOH, a thermometer, and a stirring rod. The procedure below was performed two times. Ensure that the calorimeter cups are washed and dried in between uses. We measured 50 mL of 5.00M HCl in one labelled graduated cylinder. The same was done for 5.00M NaOH in another labelled graduated cylinder. The NaOH solution had its temperature taken, but not the HCl solution. Similarly to section A, we mixed and stirred the solutions, and all temperature changes were recorded over 5-second intervals for 2 minutes. The final weight of the calorimeter cup was recorded. C The materials used in this lab section include the calorimeter cup, a 400mL beaker filled halfway with water, a hot plate, a top-loading balance, string, a retort stand, a metal sample, distilled water, a graduated cylinder, a thermometer, and a stirring rod. The procedure below was performed two times. Ensure that the calorimeter cups are dried in between uses. We placed the beaker with water on top of the hot plate set on high. The retort stand was set up so that the clamp hung over the water, and the metallic object (weighed) could hang and not touch the sides of the beaker. Enough time was given so the water could boil and the object would be at the same temperature. We set up the calorimeter with 100 mL of distilled water and placed the hot metal into the water. The same procedure of recording measurements from the previous sections was repeated. While resetting the calorimeter cup, the metallic object was placed again into the boiling water.

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D The materials used in this lab section include the calorimeter cup, distilled water, 12.0g of ammonium chloride, a graduated cylinder, a thermometer, and a stirring rod. The procedure below was performed two times. Ensure that the calorimeter cups are washed and dried in between uses. We poured 80 mL of water into the calorimeter using a graduated cylinder. We then read and recorded the temperature of the water. Next, we weighed 12.0g of ammonium chloride and added it to the water in the calorimeter. We replaced the cover on the calorimeter and began stirring rapidly. The same procedure of recording measurements from the previous sections was repeated. After the recordings were done, the calorimeter was weighed and recorded. Results The first experiment conducted was the mixing of hot and cold water. Two trials were done, and below are the major points of information: A Weight of cold water Weight of hot water Temp. increase in calorimeter Temp. decrease in hot water Trial 1 49.2 g 48.5 g 8.0°C 15.3°C Trial 2 48.3 g 47.7 g 9.6°C 7.9°C Table 1: Section A reading values. For the calculations, there will be an example for each step following trial 1. To begin, the heat lost from the hot water will be determined using q = mcΔT, where the heat capacity of water is used (4.18 J/°C g). 𝑄 = ??Δ𝑇 𝑄 = 48. 5𝑔(4. 18 𝐽/𝑔°𝐶)15. 3°𝐶 𝑄 = 3101. 77𝐽 A similar process is done with the cold water readings to reach 1938.18 J. To determine how much heat was lost to the calorimeter during the experiment, we subtracted the cold water readings from the hot water with a result of 1250.87 J. That value was then divided by the temperature increase in the calorimeter to produce a heat capacity of 138.99 (J/°C). The same process was repeated with trial 2. The final heat capacity of the calorimeter resulted in 50.59 ±125.02(J/°C). This value will be used to calculate items in further sections. The neutralisation reaction of HCl and NaOH was measured for the next experiment in the lab. Two trials were done, and below are the major information points:

B Temp. change in calorimeter Weight of final solution Trial 1 37.2°C 110.0 g Trial 2 37.3°C 108.0 g Table 2: Section B reading values. The heat absorbed by the residual 2.5M NaCl solution for trial 1 was found to be 14,690.28 J through using q = mcΔT. To determine how much heat the calorimeter absorbed, we use the value found from part A: 𝐻??? ???𝑜???? = 37. 2°𝐶 / 50. 59(𝐽/°𝐶) = 1, 881. 95 𝐽 With these two values, the total heat created in the reaction is the sum of how much heat was absorbed through the calorimeter and the solution. The result for trial one is 16, 572.23 J was created in the reaction. We know that 0.25 moles of H 2 O were produced during the reaction. This is how we determined the heat of neutralisation per mole of H 2 O: 𝐻𝑜?(𝐻2?) = 16, 572.23𝐽 0.25 ?𝑜? 𝐻2? = 66, 288. 90 𝐽/?𝑜?? The same process was repeated with trial two, resulting in an average molar heat of neutralisation of -65.8 ± 0.59 kJ/mole. Next was the experiment, where a metal object was heated and placed into the calorimeter. Two trials were done, and below are the major information points: C Mass of metal Temp. change in calorimeter Trial 1 99.2 g 6.7°C Trial 2 99.2 g 6.7°C Table 3: Section C reading values. Since both values are identical, the following calculations apply for both trials. The volume of water used in both trials was 100mL or 100g. To calculate the heat gained by the water, we use q = mcΔT again to get a result of 2800.6J. Assuming that the metal was heated to 100°C in the boiling water before being placed in the calorimeter, the temperature change in the metal would result in 100 - 6.7 = 93.3°C. The heat lost by the metal is what was gained from the water, so it would be -2800.6J. With

this information, the heat capacity of the metal can be calculated by manipulating q = mcΔT where m is the mass of the metal, q is already known, and the temperature is 93.3°C. The result is a heat capacity of 0.282 (J/°C). Since both values are the same, there is no error value. Finally, the reaction of ammonium chloride with water. Two trials were done, and below are the major information points: D Weight of final solution Temp. change in calorimeter Trial 1 89.2 g 9.7°C Trial 2 90.3 g 9.7°C Figure 4: Section D reading values. To determine the heat lost by the ammonium chloride solution, we again use q = mcΔT using the values from the table, and c = 3.59 (J/°C g). Trial 1 has a result of 3106.21J. A similar process to before is used to determine the heat lost from the calorimeter. Multiplying the temperature change by the calorimeter heat capacity from part A. = 9.7°C(50.95 J/°C) = 490.72J. The total heat absorbed is the sum of the calorimeter and ammonium chloride solution: = 490.72J + 3106.21J = 3596.93J To find the heat of solution per mole of NH 4 Cl, we divide this value by the number of moles used in the experiment (0.220 moles). The final result for trial 1 is a heat of solution per mole of NH 4 Cl of 16,365.57J. Repeating the same process for trial 2, the average result is 16.4 ± 0.11 kJ/mole. Discussion The lab was mildly successful overall, attributed to many factors throughout the experiments. The fundamental problem strung throughout each trial is the calorimeter cup itself. Two holes exist for the temperature probe and stirring rod to fit through. These holes are not snug fits, thus leaving a lot of room for air to enter. This is not a closed system; there is plenty of room for heat to escape, especially through the larger hole for the stirring rod. Knowing the basics of thermodynamics, it is clear that having an open system while trying to measure temperature changes will lead to inaccurate results. For future experiments, it is advised that students use a magnetic stirring bar and a thermometer that can be used without breaking the system.

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The next source of error comes from the measurements of fluids. Throughout the experiment, an attempt was made to have precise measurements using graduated cylinders and pipettes; however, the mass difference shows how often fewer millilitres of water were added to the calorimeter than expected. Although the error was minor, incorrect ratios, such as those in the neutralisation reaction, may severely affect the results when looking for specific values. It is best to weigh all substances added to the experiment as accurate as possible to account for poor measurement practices. The final source of error stems from human error. Yes, it is not entirely acceptable to source human error. However, it is easy to make mistakes when taking measurements at precise intervals or multitasking while writing values down. An ideal setting would involve a computer performing all the measurements so that no timing errors or accidental misinterpretations occur. Moving onto further discussion questions, what is a joule since it is the primary unit of energy? A joule is the S.I. unit of energy and is defined as the amount of work done when a force of one newton is applied over a distance of one meter. A joule is an energy required to move an object one meter with a force of one newton. To convert 6.68 calories to joules, we can use the conversion factor 1 calorie = 4.184 joules. Therefore, 6.68 calories x (4.184 joules/1 calorie) = 27.98 joules. So, 6.68 calories are equivalent to 27.95 joules. Below are a few individual questions answered in their own sections: Will a weak acid/strong base neutralisation have the same molar heat of neutralisation as the strong acid/strong base neutralisation? In a strong acid/base neutralisation, the acid and base completely dissociate in water to form ions. The reaction between them is highly exothermic, meaning much heat is released. The molar heat of neutralisation for strong acid/strong base neutralisation is a constant value, typically around -55.9 kJ/mol(Doubtnut). On the other hand, in a weak acid/strong base neutralisation, the acid does not completely dissociate in water. Only a small fraction of the acid molecules react with the base. As a result, less heat is released during the reaction, leading to a lower molar heat of neutralisation compared to strong acid/strong base neutralisation. The values obtained from part B showed a molar heat of neutralisation of -65.8 ± 0.59 kJ/mole. The percent error is 17.7%.

If you have used 300 mL of water and 45.0 g of NH4Cl in your experiment to determine ΔH of solution, would you expect to get a larger, smaller, or identical temperature change? A smaller temperature change is expected if 300 mL of water is used instead of 100 mL. This is because a larger quantity of water would require more heat energy to raise its temperature by a given amount than a smaller quantity. Why do you use the heat capacity of the 2.5 M NaCl solution rather than the heat capacities of the original acid and base solutions? Using the heat capacity of the solution rather than the heat capacities of the original acid and base solutions is necessary because the neutralisation reaction changes the properties of the original solutions, such as their concentrations, densities, and volumes. The solution's heat capacity considers these changes and provides a more accurate measurement of the heat released or absorbed during the reaction. Why is the dissolving of ammonium chloride endothermic? Would you get similar results for dissolving all salts? The dissolving of ammonium chloride is endothermic because the process of dissolving requires energy input to break the ionic bonds between the ammonium cations and chloride anions as a solid. The heat of solution of a salt depends on the nature of the ions and the interactions between the ions and the solvent molecules. Some salts may have a positive heat of solution(endothermic), like the dissolving of ammonium chloride. In contrast, others may have a negative (exothermic) process where energy is released during dissolving. One example of a negative heat of solution is NaCl. It is exothermic because the interactions between the ions and water molecules are stronger than between the sodium and chloride ions. Suppose the calorimeter used for this experiment had been made of a heat-conducting material (such as a metal) rather than Styrofoam. Would the measured calorimeter constant be larger or smaller? A heat-conducting material would allow for greater heat transfer between the contents of the calorimeter and the surroundings, leading to less heat being absorbed by the calorimeter itself. This would result in a smaller calorimeter constant than if a less heat-conductive material like Styrofoam had been used to make the calorimeter.

Conclusion Although the lab was conducted as carefully as possible, the results show highly inaccurate results from what is expected. The few sources of error are most likely the cause for such discrepancies. Although no results were what was expected (negative values when not normal, errors larger than the value itself, actual percent error being large), the lab allowed a proper practice of theory learned. More than just practising calculations, the discussion provided an opportunity to think deeper about what is being taught and the connection between certain topics.

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Works Cited Doubtnut. (2021, May 1). Enthalpy of neutralisation of acetic acid by NaOH . doubtnut. Retrieved March 15, 2023, from https://www.doubtnut.com/question-answer-chemistry/enthalpy-of-neutralisation- of-acetic-acid-by-naoh-is-506-kj-mol-1-calculate-deltah-for-ionisation-of-6441192 82

Calorimetry Lab (1)

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