Experiment 9 Entropy and Spontaneity

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University of Nebraska, Lincoln *

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Chemistry

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Feb 20, 2024

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Experiment 9 Entropy and Spontaneity Total Points: 46 Student’s Name: Adam Miller Lab Section: 804 NOTE: You must show your work for all calculations; no work, no credit. Data and Observations (10pts) Calcium Temperature Grams Water q = 4.18 x ΔT x m (m = mass of Water) ΔH = -q/moles Grams Mole T i T f ΔT Trial 1 1.01 .0252 293.9 309.5 15.6 100 6520.8 -258761 Trial 2 1.00 .0250 294.0 312.6 18.6 100 7774.8 -310992 Average ΔH -284877 1. (2pts) Balance the chemical equation of calcium with water to produce calcium hydroxide and hydrogen gas. Ca (s) + 2H 2 0 (l) -> Ca(OH) 2(aq) + H 2(g) 2. (12pts) List the reference values for ΔG° f and S° of each reaction component and calculate ΔG° rxn and ΔS theoretical for the reaction. Using ΔG° rxn and the average of the experimentally determined value for ΔH in this experiment, report semi-experimental value for ΔS (assume standard temperature). Compare this value to the theoretical value (ΔS theoretical ) and calculate a percent error. What would cause any variance? (You must show your calculations.) Reaction Component ΔG° f (kJ/mol) S° (J/mol*K) Ca(s) 0 41.6 H 2 O(l) -237.13 69.9 Ca(OH) 2 (s) -897.5 83.4 H 2 (g) 0 131.0 Causes of Variance a) Heat is lost to surroundings because it is an open system b) Impure Calcium c) Temperature might be measured incorrectly
Experiment 9 Entropy and Spontaneity Total Points: 46 Comparison ∆G rxn = ∆G products - ∆G reactants (-897.5 + 0) – (2(-237.13) + 0) = -423.24 = ∆G rxn ∆S theoretical = ∆S products - ∆S reactants (83.4 + 131.0) – (41.6 + 2(69.9)) = 33 J/(mol*K) = ∆S theoretical ∆G rxn = ∆H – 298.15∆S rxn -423240 = -284877 – (298.15 * ∆S) ∆S rxn = 464.07 J/mol (464.07 – 33) / 33 * 100 = 1306.27% 3. (2pts) Does the sign on the experimental entropy value agree with the type of change? Yes, gas was produced which makes the entropy value increase. Titration Part (2pts) Molarity HCl Initial Buret Reading Final Buret Reading Volume of HCl Trial 1 6 0 8.81 8.81 Trial 2 6 10 17.60 7.60 4. (2pts) Calculate the number of moles of OH - formed. (Must show calculation for each trial.) Moles HCl = moles OH - Mol HCl * ((1mol Ca(OH) 2 )/(2mol HCl)) * ((2mol OH)/(1mol Ca(OH) 2 )) Trial 1 - 6M * .00881 L = .05286 mol OH - Trial 2 - 6M * .00760 L = .0456 mol OH - 5. (2pts) Determine the ratio of moles of OH - / moles of Ca 2+ . (Must show your calculation for each trial.) Trial 1 - .05286 / .0252 = 2.1 Trial 2 - .0456 / .0252 = 1.8
Experiment 9 Entropy and Spontaneity Total Points: 46 Post-lab Questions NOTE: You must show your work for all calculations; no work, no credit. 1. (2pts) State the first and second laws of thermodynamics The first law says that the energy is constant in the universe. If energy leaves a system, it enters the surrounding. The energy can neither be destroyed nor created. The second law says the processes occurs spontaneously if the entropy change is greater than one. If the entropy is positive, then the spontaneous process can occur. 2. (4pts) When ΔG = 0, a system is at equilibrium. Use values for the ΔH° and ΔS° to determine the temperature at which the following system would reach equilibrium. (See appendix at the end of your textbook for relevant thermodynamic values.) N 2(g) + O 2(g) 2NO (g) ∆H = 2 * 90.37 – 0 ∆H = 180.74 kJ 180740 J/mol ∆S = 2*210.6 – (191.5 + 205) ∆S = 24.7 J/mol ∆G = ∆H – T∆S 0 = 180740 – 24.7T T = 7317.41 K 3. (8pts) Complete the following table: Enthalpy Change ΔH Entropy Change ΔS Predicted Sign for ΔG Spontaneity + - Low Temp. + Never Spontaneous High Temp. + Never Spontaneous + + Low Temp. + Non-Spontaneous High Temp. - Spontaneous - - Low Temp. - Spontaneous High Temp. + Non-Spontaneous - + Low Temp. - Always Spontaneous High Temp. - Always Spontaneous
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