University:
Glendale Community College
Course:
CHM152 | General Chemistry II
Academic year:

2019
Views:

219

Pages:

1
Author:

unidosenluz0y

Thermodynamics Key 1. Predict the sign of ΔSrxn (positive or negative) for the following reactions: a. 2 Mg (s) + O2 (g) → 2 MgO (s) ΔS__-__ ΔS__+__ b. C6H12O6(s) J 2C2H5OH(l) + 2CO2(g) c. 2NH3(g) + CO2(g) J H2O(l) + NH2CONH2(aq) ΔS___-__ o o ΔS__-_ d. Cu(s) (100 C) J Cu(s) (25 C) 2. Which two variables can indicate if a reaction is spontaneous or not? ____ΔG and ΔStot__ 3. The second law of thermodynamics tells us that: a. Energy is conserved. b. The entropy of a pure solid crystal at 0 Kelvin is zero. c. The total enthalpy of the universe is increasing. d. The total entropy of a spontaneous process is increasing. e. The free energy of a spontaneous process is increasing. 4. Which combination of ΔH and ΔS for a reaction is not spontaneous at low temperatures and becomes spontaneous at higher temperatures? Answer: ΔH _positive_ and ΔS _positive 5. The process of a liquid boiling has a: a. Positive enthalpy change and a positive entropy change. b. Positive enthalpy change and a negative entropy change. c. Negative enthalpy change and a positive entropy change. d. Negative enthalpy change and a negative entropy change. e. Not enough information is given. 6. Calculate the temperature at which this reaction changes from being spontaneous to non spontaneous: Mg(s) + O2(g) ' MgO(s). Is the reaction spontaneous above or below this temperature? (Hint – balance the rxn and use data from Appendix B) ΔHorxn = [ 2(-601.7)] – [0] = -1203 kJ ΔSorxn = [ 2(26.9)] – [ 1(32.7) + 1(205.0) ] = -216.6 J/K 0 = (-1203 kJ) – T (-0.2166 kJ/K) T = -1203 kJ / -0.2166 kJ/K T = 5550 Kelvin or 5280oC spont below this temp 7. Calculate ΔGorxn for this reaction: 2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l) Given: ΔHfo CH3OH (l) = -238.6 kJ/mol ΔHfo H2O (l) = -285.83 kJ/mol o S CH3OH (l) = 126.8 J/molK ΔHfo CO2 (g) = -393.5 kJ/mol So O2 (g) = 205.0 J/molK So CO2 (g) = 213.6 J/molK o S H2O (l) = 69.91 J/molK (only use this given data for your solution) ΔHorxn = [ 2(-393.5) + 4(-285.83)] - [ 2(-238.6) + 0] = -1453.12 kJ ΔSorxn = [ 2(213.6) + 4(69.91)] - [ 2(126.8) + 3(205)] = -161.76 J/K = -0.16176 kJ/K ΔGorxn = -1453.12 kJ/mol - (298 K)-0.16176 kJ/molK = -1405 kJ