lab 17 (1).docx

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InterAmerican Recinto Metropolitano *

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Chemistry

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Feb 20, 2024

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Lab 17 introduction Transfer of heat is one of the important processes in chemistry. to determine the change in thermodynamic quantities, one of the methods used by chemist is a calorimeter. In this lab setting, we are using coffee cup calorimeter made of two Styrofoam cups with a lid, a thermometer, to experiment the principles behind the transfer of energy in the form of heat. Materials: 1) Lab manual: Used to learn lab procedures, and guidelines. 2 Excel: Used to create charts for the data 3) Styrofoam cups: Used as the calorimeter. 4 Thermometer: Used to measure the temperature of the solutions. 5) 2M HCI: Used as acid for neutralization reaction. 6) 2M NaOH: Used as base for neutralization reaction. 7) Ammonium chloride: Used as salt to determine its molar heat 8) Hot plate: Used to heat water until boiling point. 9) Scale: Used to measure the mass of substances. 10) Timer: Helped us keep track of time Observations and Experimental: Part 1: - Heat of a Neutralization reaction Mass of cup and lid: 8.8g 50 mL of 2M HCI added to calorimeter / Temperature: 25.2°C / Mass: 42.8g 50 mL of 2M NaOH added to calorimeter / Temperature: 25.4 / Mass: about 90.8g 30 second after, 25.5°C

60 second after, 25.6°C 90 second after, 25.7°C 120 second after, 25.7°C 150 second after, 25.7°C Part 2: - Specific heat capacity of a metal Unknown metal type A: 60.1g The mass of the calorimeter plus water 55.1g Temperature of cup before metal added: 20.1°C Temperature of cup after adding metal: 29.3°C 30 seconds after, 29.2°C 60 seconds after 29.1°C 90 seconds after 29.0°C 120 seconds after 28.9°C 150 seconds after 28.8°C 180 seconds after 28.7°C

210 seconds after 28.7°C Part 3 - Molar heat of solution of a salt Mass of calorimeter plus of water: 52.30g Temperature of water 24.3 Initial temperature: 21.8°C 30 seconds after 21.9°C 60 seconds after 22.0°C 90 seconds after 22.0°C 120 seconds after 22.0°C

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Discussion and conclusion Through the experiment of this lab, we were able to used calorimeter to measure the heat produced by the reaction of HCl and NaOH, additionally we were able to determine the heat capacity of the unknowns metal, and the final operation of procedure that we carried out was the molar heat of solution of a salt, the lab was essential in that it showed us how substances with lower heat capacities are able to more easily change in temperature than those with higher heat capacities like water. Furthermore, this lab enhances our understanding of the transfer of heat between a system and its surroundings. Although we were able to complete every part of this lab with success, there might have been some issues with our overall experiment, for example we might had use more of less of the substance in the procedure, also while measuring the temperature we noticed that the number in our thermometer kept changing slightly. Focus Questions: 1)The heat of neutralization for a strong acid/base reaction is: Q=MC AT-> q=107.73 g * 3.89J/g°C * 10.9°C=4567.86 J 4.6 kJ ∆ ≈ Q=q/n->2M*0.1L=0.2n . →4. 6/0. 2≈23??/?𝑜? 2) The identity of our unknown metal is lead. 29.3°C – 28.7°C=6.307°C

mc AT =mc AT/50g *4.184 * 0.6 = -125.52J -125.52J= 60.1g(C)(28.7-100) C = 0.12959=12959 this value is close to the C value of Lead, which is 0.130. ≈0. 130 3) The molar heat of the ammonium chloride salt solution is: 22°C-21.8°C=0.2°C Mass of solution = 52.30g 2g/35.45g = 0.0374mol NH4Cl H= (0.2°C*4.184*52.30)/0.0374mol = 1170.177J/mol ∆ Reference: Smeureanu, G. & Geggier, S. (2022) General Chemistry Laboratory. New York, NY Nave, Richard. "Law of Dulong and Petit." Law of Dulong and Petit, hyperphysics.phy- astr.gsu.edu/hbase/thermo/Dulong.html Post Lab Questions 1. Compare your results for the molar heat of your salt (kJ/mol) to the literature value. What is your percent error? try to give three reasons why your value (kJ/mol) may be different from the tabulated values. -The molar heat of our salt was 1170J/mol and the literature value was 1478 kJ/mol, The calculated percent error is :(1478-1170) *100%/1478 = 20.8% Among the reason why our value is different than literature value: a-it might had been a flaw in building our calorimeter b-we did not the take the temperature right away c-our Styrofoam cups wasn’t properly isolated. 2. In 1819, French physicists Dulong and Petit found experimentally that for many solids at room temperature, c ~ 3R (R = gas constant in SI units). Show that Dulong- Petit's rule is in agreement with the specific heat capacities of metals in the table.

Explain why the specific heat capacity of magnesium is roughly twice as large as the one of titanium, and why the specific heat capacities of lead and gold are nearly identical. The scientific heat capacity of Mg is about twice as large as the one of Ti, because it is almost half the atomic size of Ti, and as a result the specific capacity must be double, just as the specific heat capacity of lead and gold, they are almost identical in value, because their molar masses are approximately close to each other. 3. Write the net ionic equation for the reaction between hydrochloric acid and aqueous sodium hydroxide reaction. How would the temperature change and the calculated heat of neutralization vary if the concentration of the acid and the base used in this experiment will double? Explain. HCl (aq) + NaOH (aq) -> H2O (l) + NaCI (aq) Net ionic equation: H+ (aq) + OH (aq) -> H2O (l) If the concentration of the acid and the base is increased or doubled, the temperature would increase as well because an increase in concentration leads to an increase in heat that means a higher temperature and no change in the heat of neutralization. 4. In a coffee-cup calorimeter, 3.20g of NH4NO: is mixed with 86.00 g of water at an initial temperature of 23.53°C. After dissolution of the salt, the final temperature of the calorimeter contents is 21.32°C. Assuming the solution has a heat capacity of 4.184 J/g* °C and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of NH4NO3, in units of kJ/mol. Answer: Molar mass of NH4NO3 = 84.00g 2.80g * 1mol/84.00g = 0.03 mol NH4NO3 2.80g + 84.00g = 86.8g Q = MCAT Q= 90.1 g *4.184 J/g* °C * (21.32 °C - 23.53 °C) Q= 86.8 g *4.184 J/g* °C * (23.07 °C - 25 °C) Q = 833.12J H = 833.12/0.04 = 20828j/mol ∆ ≈ 20. 82??/?𝑜?

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