The Battle for Chemical Domination On the Open Seas Group Slaughter The Battle for Chemical Domination On the Open Seas

Set-Up: Number 1-10 on your piece of paper. Each team member must print their name next to ONE number only. Don’t let any other team see what your numbers are! These are your battleships.

The Rules You gain the ability to take shots at other teams’ battleships by getting questions correct. Questions are worth 1, 2, or 3 shots at the team of your choice. When one of your battleships is sunk, the person who’s battleship was shot will be rescued by their teammates. THE TEAM ONLY DIES WHEN EVERY MEMBER HAS BEEN SHOT. Once a team is dead, if they get the next question right, they will come back to life as angry ghosts! Ghosts get double shots for every question they get right.

The Rules: Part II Winners will be determined by which team has the most LIVING battleships at the end of the class. All students MUST participate. You show this by making sure everyone’s color appears in the answer. Sharing pens results in immediate disqualification for that question. Answers are only accepted if written on the group’s white board! NO FRIENDLY FIRE ;)

May the odds be ever in your favor… Last Questions? Let the games begin! May the odds be ever in your favor…

Question 1: for 2 shots A student constructs a galvanic cell based on the half-reactions above. Write a balanced equation for the reaction that occurs in the cell. Calculate Eo for the cell.

Question 1: for 2 shots Write a balanced equation for the reaction that occurs in the cell. Calculate Eo for the cell.

Question 2: for 1 shot Given the following reaction: 2 NO2 + F2  2 NO2F The experimentally determined rate law is rate = k[NO2][F2] A suggested mechanism for the reaction is NO2 + F2  NO2F + F Slow F + NO2  NO2F Fast Is this an acceptable mechanism? (That is, does it satisfy all requirements?) Justify your answer.

Question 2: for 1 shot Yes, this is an acceptable mechanism! It meets both requirements: 1. Both steps added together gives the overall rxn: NO2 + F2  NO2F + F F + NO2  NO2F 2 NO2 + F2  2 NO2F 2. The experimentally determined rate law matches the rate law determined from the slow step of the mechanism: first order with respect to both reactants, because they both have a coefficient of 1 in the slow step; rate = k[NO2][F2]

Question 3: for 2 shots H2O(l) → H2O(s) Identify the sign of ∆H, ∆S, and ∆G at 25oC for the following reaction, and justify your answers: H2O(l) → H2O(s)

Question 3: for 2 shots H2O(l) → H2O(s) ∆H: −, because the system is going from a higher energy state of matter to a lower energy state, so the system must have released energy ∆S: −, because the system is going from a liquid (more entropy) to a solid (less entropy) ∆G: +, because water does NOT spontaneously become ice at room temp

Question 4: for 1 shot Copper may be used for electroplating, with a half-reaction of Cu2+ + 2 e−  Cu(s) How many moles of electrons must be transferred in this reaction to produce 5.16 g of copper metal?

Question 4: for 1 shot How many moles of electrons must be transferred in this reaction to produce 5.16 g of copper metal? Cu2+ + 2 e−  Cu(s)

Question 5: for 1 shot 2 H2O2  2 H2O + O2 The decomposition of hydrogen peroxide is shown above and has the following proposed mechanism. H2O2 + I−  H2O + IO− H2O2 + IO−  H2O + O2 + I− What substance serves as the catalyst in this reaction, and how do you know?

Question 5: for 1 shot H2O2 + I−  H2O + IO− H2O2 + IO−  H2O + O2 + I− I− serves as the catalyst in this reaction, because it occurs as a reactant in an earlier step and is reproduced as a product in a later step

Question 6: for 1 shot Use the information provided and the balance equation to determine the standard heat of formation, ΔHof, of CCl4(g).

Question 6: for 1 shot

Question 7: for 2 shots Copper may be used for electroplating, with a half-reaction of Cu2+ + 2 e−  Cu(s) If a current of 10.0 amp is applied to a Cu2+ solution for 60.0 minutes, what mass of copper will be plated out?

Question 7: for 2 shots If a current of 10.0 amp is applied to a Cu2+ solution for 60.0 minutes, what mass of copper will be plated out?

Question 8: for 1 shot For the following reaction: 2 NO(g) + Br2(g)  2 NOBr2(g) This data was collected. Experiment [NO]i (M) [Br2]i (M) Initial Rate (M-1min-1) 1 0.020 0.10 5.20 x 10−3 2 0.020 0.20 1.04 x 10−2 3 0.040 0.10 2.08 x 10−2 What is the rate law for this reaction? Justify.

Question 8: for 1 shot Rate = k[NO]2[Br2] Experiments 1 & 2: The reaction is second order with respect to NO because 2 x [NO] while [Br2] constant quadruples the rate of reaction. Experiments 1 & 3: The reaction is first order with respect to Br2 because 2 x [Br2] while [NO] constant doubles the rate of reaction.

Question 9: for 1 shot Which species was oxidized and which was reduced in the balanced reaction below? Justify with oxidation states. ;)

Question 9: for 1 shot Oxidized: I− (oxidation state −1 → 0) Reduced: Cr2O72− (Cr oxidation state +6 → +3)

Question 10: for 1 shot Given the following reaction at equilibrium: When the temperature of the reaction mixture is increased, what is the effect on [CO2]? Explain.

Question 10: for 1 shot Increasing the temperature in an endothermic reaction will cause the reaction to shift right, since heat is the equivalent of a reactant. This will cause [CO2] to increase.

Question 11: for 1 shot Answer AND justify. X  2Y + Z The concentration of X over time is used to create the three graphs below. What is the rate law for this reaction?

Question 11: for 1 shot Answer AND justify. rate = k[X] or rate = k[X]1 The reaction is first order with respect to X because the graph of ln(x) vs time is linear with a negative slope.