Exam+1+-+Key+300+section

Checkpoint Exam 1 – Answer Key and Explanations Question 2 For the reaction: P 4 (s) + 6 HF (g) → 2 PF 3 (g) + 2 PH 3 (g) 2.1 Use the enthalpies of formation to determine the enthalpy of reaction, ∆ H° rxn , in kJ. Enthalpies of Formation ∆ H° f,P ₄ (s) = 0 kJ/mol ∆ H° f,HF(g) = –254 kJ/mol ∆ H° f,PF ₃ (g) = –915 kJ/mol ∆ H° f,PH ₃ (g) = +9 kJ/mol The enthalpy of reaction is calculated the following way: ∆࠵?° !"# = % ࠵? $ ∆࠵?° %,$ ’!()*+,- − % ࠵? $ ∆࠵?° %,$ !./+,/#,- For the reaction listed above: ∆࠵?° !"# = )࠵? 01 ! (3) ∆࠵?° %,01 ! (3) + ࠵? 05 ! (3) ∆࠵?° %,05 ! (3) + – )࠵? 0 " (-) ∆࠵?° %,0 " (-) + ࠵? 51(3) ∆࠵?° %,51(3) + ∆࠵?° !"# = -2 ࠵?࠵?࠵? 2−915 67 8(9 6 + 2 ࠵?࠵?࠵? 2+9 67 8(9 67 – -1 ࠵?࠵?࠵? 20 67 8(9 6 + 6 ࠵?࠵?࠵? 2−254 67 8(9 67 ∆࠵?° ࠵?࠵?࠵? = −࠵?࠵?࠵? ࠵?࠵? 2.2 Predict the sign of the ∆ S° rxn from the reaction equation given above. Give your answer (positive, negative, or zero) and use one sentence to explain the reasoning for your answer using concepts from the course. Because the number of molecules decreases from reactants to products and, specifically, the number of gas molecules decreases, the system is becoming more ordered (less disordered) so the entropy change will be negative . Question 3 For the reaction of ethylene gas and water to give ethanol: C 2 H 4 (g) + H 2 O (l) → CH 3 CH 2 OH (l) ∆ H° rxn = -52 kJ ∆ S° rxn = -144 J/K 3.1 Which of the following describes the reaction? The enthalpy of the reaction is favorable. Exothermic reactions and those that have increasing disorder both contribute to a more negative free energy of reaction, and are therefore favorable properties.

3.2 This reaction is only spontaneous at sufficiently low temperatures. Below what temperature in Kelvin will this reaction be spontaneous? Reaction will be spontaneous when free energy is negative (∆G° < 0). ∆࠵?° = ∆࠵?° − ࠵?∆࠵?° 0 > −52000 ࠵? − ࠵? 2−144 7 = 6 ࠵? < −52000 ࠵? −144 7 = = ࠵?࠵?࠵? ࠵? 3.3 If the same reaction involved water vapor (gaseous H 2 O) instead of liquid water: C 2 H 4 (g) + H 2 O ( g ) → CH 3 CH 2 OH (l) How would the enthalpy of reaction ( ∆ H° rxn ) and entropy of reaction ( ∆ S° rxn ) change? The enthalpy would decrease. Gases have a higher enthalpy than liquids (you need to add heat to go from a liquid to a gas). The entropy would decrease. The reactants are more disordered (higher entropy) to start with as a gas instead of a liquid. Question 4 4.1 How many intermediates does this reaction have? Two (2) intermediates One at point C and one at point E – intermediates are energy minima that are not reactants or products 4.2 Which reaction listed below represents the slow step (rate determining step)? A → C A step in a reaction goes from one minimum energy state to the next minimum energy state. 4.3 According to the Hammond Postulate, which of the following will the last transition state most closely resemble? Point E The transition state most closely resembles the reactant of product of the reaction (or step of reaction) that is closest in energy to the transition state.

6.2 How many moles of H 2 will be produced in 15.0 minutes of this reaction? 7 moles of H 2 are produced for every 2 moles of CO 2 produced. ࠵?࠵?࠵? ࠵? H = 5.68 8(9 @A # 8B# × 7 ࠵?࠵?࠵? ࠵? H 2 ࠵?࠵?࠵? ࠵?࠵? H × 15.0 ࠵?࠵?࠵? = ࠵?࠵?࠵? ࠵?࠵?࠵? ࠵? ࠵? Question 7 Molecule X and Y can react with each other. Which of the following would be predicted by Frontier Molecular Orbital theory? Because the HOMO of X is closer in energy to the LUMO of Y (∆E = 0.510 Hartrees vs ∆E = 0.634 Hartrees for the HOMO of Y and LUMO of X), we know: Molecule X will act as a nucleophile Molecule X will react using its HOMO Molecule Y will react using its LUMO Question 8 Write one sentence explaining why Frontier Molecular Orbital theory does not allow two molecules to react, both using their HOMO orbitals. Can be viewed two ways: 1. In FMO theory, when two full orbitals overlap, both the resulting bonding and antibonding orbitals that are created are filled , therefore there is no energy benefit (the bond order would be zero for the new bond between reactants). Must be clear that the new orbitals are filled. A high-energy antibonding orbital is created for any interaction including a HOMO-LUMO interaction, but the orbital is not filled. 2. For FMO theory, there is sharing of an electron pair from a donor (nucleophile) to an acceptor (electrophile), so one molecule needs an empty orbital and HOMOs are, by definition, filled. Must indicate that an electron pair is being shared in the overlap, not just that one orbital needs to be empty and the other filled.

Question 9 Frontier molecular orbital theory analysis determines that in the reaction of carbonyl sulfide (SCO) and hydrogen fluoride (HF), it is the HOMO of carbonyl sulfide and the LUMO of hydrogen fluoride that are involved in the reaction. The greatest overlap of these two orbitals will occur between the H atom of HF and the S atom of SCO because that’s where the electron density of the two orbitals is greatest. An S–H bond forms .