Lab Report 5

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University of Toronto *

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Chemistry

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Apr 3, 2024

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Experiment 5: Qualitative and Quantitative Aspects of Equilibrium Report Sheet You are welcome to seek guidance on the content of this lab report from your TA during your lab session and from the lab coordinator during lab office hours. You are also welcome to work constructively with your peers on the general content and understanding of the material. However, the work submitted here in this report sheet must be your own. For more details on academic integrity and potential penalties see the Code of Behaviour on Academic Matters . I certify that this submitted laboratory report represents entirely my own efforts. I have read and understand the University of Toronto policies regarding, and sanctions for, academic honesty. I regret that I violated the Code of Behaviour on this assessment and would like to admit that now so that I can take responsibility for my mistake. A. Heat capacity of calorimeter 1. Insert your two plots of temperature vs. time for the calibration of the calorimeter and complete the table. Plot #1
Plot #2 Run # Initial Temperature of HCl ( o C) Initial Temperature of NaOH ( o C) Extrapolated Final Temperature ( o C) 1 18.6 20.4 26.6 2 19.5 20.2 26.3
2. Was your solution acidic or basic after the run was complete? What does this tell you about the reaction? After the run was complete the solution was highly basic at pH 12. This means the reaction was a strong acid-base reaction and is exothermic which can be seen from the rise in the temperature at the beginning. 3. Show your calculation of the heat capacity of the calorimeter (for run #1). C = n/V n = CV n HCl = (1.0 M) (0.75) n NaOH = (1.1 M) (0.75) = 0.075 mol = 0.0825 mol (HCl = limiting reagent) qReaction = nΔH 0 H3O + OH = (0.075 mol) (-58985 J/mol) = 4423.88 ( -58985 J/mol is found using the table in lab manual page 5-7 for the initial temperature 18.6 o C ) HEAT CAPACITY OF CALORIMETER - nΔH 0 H3O + OH = - (C CAL ΔT CAL + C WATER V ACID ΔT ACID + C WATER V BASE ΔT BASE - (4.13 J/ o C -1 mL -1 * 28.1 o C + 4.13 J/ o C -1 mL -1 * 75 mL* 28.1 o C + 4.183 J/ o C -1 mL -1 * 75 mL * 28.1 o C = - 5431.91 nΔH = - C CAL ΔT - 2C ACID VΔT - 4423.88 = - C CAL (6.9) - 2(4.18)(75 mL)(6.9) - 4423.88 = - C CAL (6.9) - 4326.3 C CAL (6.9) / (6.9) = - 4326.3 + 4423.88 / 6.9 C CAL = 14.14 J/ o C
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3. Complete the table: Run # Heat Capacity of Calorimeter 1 14.14 J/ o C 2 33.1 J/ o C Average 23.62 J/ o C B. Enthalpy of formation of magnesium oxide 4. Insert your plot of temperature vs time for the Mg trial. 5. Complete the table: Reactant Mass (g) Initial Temperature of HCl ( o C) Extrapolated Final Temperature ( o C) ΔH∞ for the reaction with HCl Mg 0.28 20.5 26.9 - 2707.42 J mol -1
6. Show your calculation of ΔH∞ for the reaction Mg + 2HCl MgCl 2 + H 2 Limiting reagent - Mg, nMg = 0.0012 mol 0.012*ΔH = -((28.1*4.13) + (4.183*150*4.13)) ΔH = - 2707.42 J mol -1 7. Insert your plot of temperature vs time for the MgO trial. 8. Complete the table: Reactant Mass (g) Initial Temperature of HCl ( o C) Extrapolated Final Temperature ( o C) ΔH∞ for the reaction with HCl MgO 0.28 20.5 26..9 - 43487.94 J mol -1
9. Show your calculation of ΔH∞ for the reaction, MgO + 2HCl MgCl 2 + H 2 O Limiting reagent - MgO, nMgO = 0.0018mol 0.018*ΔH = -((28.1*6.9) + (4.183*150*6.9)) ΔH = - 43487.94 J mol -1 10. Use your experimental data to calculate the enthalpy of formation, ΔH f ∞, of MgO. To help with the calculation, first, summarize the relevant ΔH∞ values in the table below. Note that some are experimental (meaning you determined them using your data) and some are from the literature. Reaction Δ H (kJ/mol) Mg(s) + 2HCl(aq) → MgCl 2 (aq) + H 2 (g) - 2707.42 J mol -1 MgO(s) + 2HCl(aq) → MgCl 2 (aq) + H 2 O(l) - 43487.94 J mol -1 H 2 (g) + ½ O 2 (g) H 2 O (l) - 285826 J mol -1 11. Then write the chemical equation for the enthalpy of formation of MgO: Mg (s) + ½ O 2 (g) = MgO (s) 12. Show your calculation of the enthalpy of the formation of MgO: We need Mg and ½ O 2 on the left side and MgO on the right - It is possible to deduce the enthalpy of formation by adding the sum of the enthalpy reaction between Mg and HCl and the negative enthalpy of the reaction between MgO and HCl, as we need MgO towards the right and the enthalpy of formation of water. ΔH f = (- 2707.42) + (-285826) - (-43487.94) = -245045.48 J mol -1 ≈ -245045 J mol -1 ≈ 254.045 KJ mol -1 13. How does your experimental value for the enthalpy of formation of MgO compare with the literature value? How can you account for any discrepancies?
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The enthalpy of formation for MgO in the literature is -601 kJ mol, which is rather similar to the value we discovered. This mismatch may be the result of random error brought on by erroneous temperature readings, the heat lost to the environment when adding ingredients or setting up the next step of the experiment, or both. 14. List any references used. “Thermodynamics Enthalpy of Formation Magnesium Oxide - Lab Report Experiment Thermodynamics:” StuDocu , https://www.studocu.com/row/document/quaid-i-azam-university/chemistry/thermodynamics-enthalpy-of- formation-magnesium-oxide/17338623.