Lab Report 5

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University of Toronto *

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135

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Chemistry

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Apr 3, 2024

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pdf

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7

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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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