Lab Stealing Joules

Lab “Stealing Joules” Introduction: The transfer of heat is an important process in our understanding of chemistry. In order for chemists to determine the change in thermodynamic quantities, we use a calorimeter, which isolates transfer of mass and heat. In our lab setting, we used a coffee cup calorimeter made of two styrofoam cups placed in a beaker, a thermometer, and a lid, which would prevent mass and heat from transferring outside the system. We found the heat of neutralization reaction between 2M NaOH(aq) and 2M HCl(aq). Then we found the specific heat capacity of an unknown metal “E.” Lastly, we found the molar heat of ammonium chloride. This helped us better understand 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 HCl: Used as acid for neutralization reaction. 6) 2M NaOH: Used as base for neutralization reaction. 7) Scale: Used to measure the mass of substances. 8) Hot plate: Used to heat water until boiling point. 9) Ammonium chloride: Used as salt to determine its molar heat. 10) Timer: Helped us keep track of time. Observations and Experimental: Part 1- Heat of a Neutralization reaction Mass of cup and lid: 9.55g 50 mL of 2M HCl added to calorimeter / Temperature: 23.0°C / Mass: 62.09g 50 mL of 2M NaOH added to calorimeter / Temperature: 35.4 / Mass: 107.73g This is the graph of temperature after adding 50 mL of 2M NaOH to 50 mL of 2M HCl. Part 2- Specific heat capacity of a metal Unknown metal type E: 21.61g Temperature of cup before metal added: 18.9°C Temperature of cup after adding metal: 25.1°C Mass of calorimeter + water: 60.02g, water=50.47 This is the graph of temperature of water after adding the metal that was placed in boiling water. Part 3- Molar heat of solution of a salt Weight of calorimeter: 170.07g Weight of calorimeter + 50 mL of water: 223.31g Initial temperature: 23°C Mass of NH4Cl: 2g This is the graph of temperature after adding 2 grams of NH4Cl to water. Discussion and Conclusion: In the lab, we used a calorimeter to measure the heat produced by the reaction of HCl and NaOH for part 1, the heat capacity of an unknown metal for part 2, and the heat of a solution of a salt for part 3. This lab was essential in our understanding of the transfer of heat between a system and its surroundings. It also showed us how substances with lower heat capacities are able to more easily change in temperature, than those with higher heat capacities such as water. We may have had errors throughout the experiment, in that we may not have used the correct amount of each substance, or that we may not have let the water heat up fully to 100°C. However, our main error occurred in part 2, because we did not take the temperature of the unknown metal, and as a result we were unable to predict the correct identity of the solution. Overall, this lab was able to expand my understanding of the process of calorimetry. Focus Questions: 1) The heat of neutralization for a strong acid/base reaction is: J ∗10.9 °C=4567.86 J ≈ 4.6 kJ q=mc Δ T , q=107.73 g∗3.89 g °C q= qrxn kJ 4.6 kJ ≈ 23 2 M∗0.1 L=0.2 n , nrxn mol 0.2n 2) The identity of our unknown metal is Aluminum. 25.207°C - 18.9°C=6.307°C. mc ΔT =mc ΔT /50.47∗4.184∗6.307 =21.61∗c∗(95.3−25.2) c=0.879 , this value is close to the c value of 1331.83J= c∗1514.86 g °C aluminum, which is 0.897. 3) The molar heat of the ammonium chloride salt solution is: 23°C-20.638°C=2.362°C Mass of solution=53.24g 2g =0.0374 mol N H 4 Cl 35.45 g qrxn (2.362° C)(4.184 J / g ° C)( 53.24 g ) =14068.18 J ≈ 14.1 kJ Δ H solution= 0.0374 mol moles of salt References: 1) Parker, V. B., Thermal Properties of Uni-Univalent Electrolytes, Natl. Stand. Ref. Data Series— Natl. Bur. Stand.(U.S.), No .2, 1965. 2) Nave, Richard. “Law of Dulong and Petit.” Law of Dulong and Petit, hyperphysics.phyastr.gsu.edu/hbase/thermo/Dulong.html. Post Lab Questions: 1) From our measurements, we obtained 14.1 kJ/mol for the molar heat of NH4Cl. Using online references, I was able to find that the molar heat of NH4Cl should be 14.78 kJ/mol. 14.78 −14.1 ∗100 %=4.6 % We can calculate my percent error to be 14.78 Possible reasons for this percent error can be that we exposed the calorimeter to the outside by exposing the lid for too long. Next, we may have plotted our graph incorrectly, and this could have caused us to use an incorrect change in temperature. Lastly, we did not stir the solution throughout the trial, which may have been a reason that it did not fully dissolve, and cause the temperature of the water to change. 2) The scientific heat capacity of Mg is roughly twice as large as that of Ti because Mg is almost half the atomic size of Ti, therefore the specific heat capacity must be doubled. Similarly, we can see that the specific heat capacities of lead and gold are nearly identical because they have roughly the same atomic mass. −¿(aq) → H 2 O ( l) +¿(aq)+O H ¿ 3) H¿ The calculated heat of neutralization would not change, because if we are doubling the concentrations, we would double the value of q, and the moles in the solution, which would cancel out. The temperature would change, because we are creating more products, and releasing more energy. 4) M = mass of water + mass of solute = (84.00g)+(2.80g)= 86.80g ΔT = T f −T i = 25˚C-23.07˚C= 1.93˚C q= (86.8g)(4.184 J/mol℃)(1.93˚C)=700.920416 J N H 4 N O3 =(2.8 g)(1 mol/80.04 g) = 0.03498 mol 700.920416 J ❑ ΔH = 20.038 kJ/mol, so the enthalpy of dissolution is solution = .03498 mol ≈ 20 kJ /mol Moles of