3Fe +2 (aq) + 2PO 4. NaHCO 3 + H 2 O. Chem 121 Quiz 3 Practice Fall 2017

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1 Chem 121 Quiz 3 Practice Fall 2017 The following quiz contains 22 questions (and bonuses!) valued at 1 point/question Name KEY G = H T S PV = nrt P 1 V 1 = P 2 V 2 P 1 /T 1 = P 2 /T 2 V 1 /T 1 = V 2 /T 2 K = o C R =.0821 L atm/mole K 1 atm = 760 mm STP: 1 mol = 22.4 L C g = kp g % w/v = (g solute/ml solution) Write the net ionic equation when solutions of NaNO 3 and Na 3 PO 4 are mixed. If there is no reaction, write N.R. NR 2. Write the net ionic equation when solutions of FeSO 4 and Na 3 PO 4 are mixed. If there is no reaction, write N.R. 3Fe +2 (aq) + 2PO 4-3 (aq) Fe 3 (PO 4 ) 2 (s) Working backwards from product to reactant really helps show the numbers of ions that must come out of solution to generate a given formula. Fe = +2 since SO 4 = -2 always 3. Please give the products of the following acid-base reaction as shown (do not change lead coefficients). If there is no reaction, write N.R. H 2 CO 3 + NaOH NaHCO 3 + H 2 O NaOH is the limiting reagent not enough to generate Na 2 CO 3

2 4. Please give the products of the following already balanced acid-base reaction. If there is no reaction, write N.R. H 2 CO 3 + HCl NR both acids 5. Please balance the following acid-base reaction. 1 H 3 PO NaOH 1 Na 3 PO H 2 O 6. What is the oxidation state of chlorine in sodium hypochlorite (household bleach) NaOCl? a. +2 b. +1 c. 0 d. -1 e. -2 Na = +1, O = -2, Cl = +1 sum of assigned oxidation numbers = 0 since compound is neutral 7. Not only is nitric acid a strong acid, it is also an oxidizing agent (recall we used nitric acid in the copper to copper lab to oxidize Cu 0 to Cu +2 ). What is the oxidation state of nitrogen in nitric acid? Hint: If you need help with the formula for nitric acid, consult the table on page 1. HNO 3 H = +1, O = -2 x 3, N = +5 Again, the sum of assigned oxidation numbers = 0 since compound is neutral Oxidation State: In the reaction between nitrogen and hydrogen to produce ammonia, how much hygrogen is needed to generate 34 g of ammonia, assuming a quantitative (100%) reaction? 3H 2 + N 2 2NH 3 a. 2.0 g b. 2.7 g c. 4.0 g d. 6.0 g e. 34 g Mass to moles moles to moles moles to mass baby! (34 g NH 3 )(mole NH 3 /17 g NH 3 )(3 mole H 2 /2 mol NH 3 )(2.0 g H 2 /mole H 2 ) = 6.0 g H 2

3 9. Referring to the question immediately above, if 6.0 g of H 2 were reacted in excess N 2 and 17 g of NH 3 were obtained, what would be the % yield? a. 17 % b. 22 % c. 50 % d. 100 % e. 200 % H 2 limits the reaction, and according to the above question, should yield 34 g NH3 in theory. Since only 17 g are obtained, the % yield is (actual/theoretical)100 = (17 g/34 g)100 = 50 % 10. Reactions that proceed only with the input of free energy are termed a. Exothermic b. Endothermic c. Exergonic d. Endergonic 11. Is the following reaction spontaneous at 300K? Please show your work H 2 (g) + Br 2 (l) 2HBr (g) H = kcal/mol, S = 27.2 x 10-3 kcal/mol K G = H T S = kcal/mol (300K)( 27.2 x 10-3 kcal/mol K) G = kcal/mol Spontaneous? Yes, G < 0 Bonus (1 EC): Is the reaction between H 2 and Br 2 spontaneous at all temperatures? If so, briefly explain. If not, at what temperature does the reaction change from spontaneous to non-spontaneous? Since the reaction is exothermic with an increase in entropy, it will be spontaneous at any temperature Alternatively, setting G = 0 and solving for transition [from spontaneous to non-spontaneous] temperature, a negative value for the absolute temperature would result ridiculous!

4 12. Which of the following is not a means by which enzymes increase the rate of reaction? a. Increase temperature b. Improve orientation of collision c. Lower the energy barrier for reaction, E a (energy of activation) d. Increase fraction of molecules with E > E a e. N/A; all are means by which enzymes increase reaction rate Tough question since enzymes are proteins they tend to unfold and stop functioning at higher temperatures. By making bonds while bonds are being broken, they are lowering E a, and consequently, more molecules would have E > E a. Generally, catalysts avoid having to crank up the temperature to speed up the reaction since they lower the energy barrier between reactants and products 13. Write the equilibrium expression for the following reaction HF (aq) + H 2 O (l) F - (aq) + H 3 O + (aq) K = [ F ][ H O 3 [ HF][ H 2 ] O] 14. Consider reaction 1 leading to the synthesis of glycogen from glucose Reaction 1: Glucose-1-PO 4 + UTP Glucose-1-UDP + PPi Reaction 2: Glucose-1-UDP + Glucose n Glucose n+1 + UDP PPi (pyrophosphate, 2 phosphates stuck together) is removed by hydrolysis to 2 phosphates after formation in reaction 1. What is the effect of removing PPi on reaction 1? a. Shifts equilibrium towards reactants b. Shifts equilibrium towards products c. Raises E a d. Lowers E a e. All of the above Simply put, removal of products shifts reaction towards products (reestablishing the constant equilibrium product to reactant ratio)

5 15. What is the pressure in a constant temperature system initially at 760 mm Hg and L that is compressed to L? a x 10 6 mm Hg b x 10 5 mm Hg c x 10 4 mm Hg d x 10-4 mm Hg e x 10-5 mm Hg f x 10-6 mm Hg Boyle s law problem volume down by a factor of 1000, pressure up by a factor of x 10 3 = 7.60 x A sample of gas at constant pressure occupies 10.0 L at 333 K. What is the volume occupied if the temperature is changed to 1.00 x 10 3 K? a L b L c L d. 60.0L e x 10 3 L Charle s law problem ABSOLUTE temperature up by a factor of 3, volume up by a factor of A closed steel drum exerting a pressure of 2.00 atm at o C is cooled to 10.0 o C. What is the final pressure? a atm b atm c atm d atm e atm Law of Guy-Lussac problem, but temperature must be converted first. P 1 /T 1 = P 2 /T 2 or P 2 = P 1 (T 2 /T 1 ) = 2.00 atm(283 K/373 K) = 1.52 atm. T decreases, P decreases

6 18. The closed steel drum in the preceding question is heated to 3.73 x 10 2 K. What is the final pressure? a atm b atm c atm d atm e atm Returning the closed system to its initial temperature returns the system to its initial pressure Bonus (1 EC): At 30 o C, the Henry s law constant for O 2 dissolved in water drops to 1.0 x 10-3 mol/l atm. What mass of oxygen is dissolved in a 1.00 x 10 3 L fish tank at 30 o C? Assume sea level conditions: P T = 1.00 atm, 21 % O 2 (1.0 x 10-3 mol/l atm)(.21 atm)(1.00 x 10 3 L)(32 g/mol) Answer: 6.7 g O 2 In class, we solve the same problem at 20 o C and obtain a value of 8.7 g O 2, as the Henry s law constant for oxygen in 20 o C is 1.3 x 10-3 mol/l atm. As temperature goes up, the ability to dissolve gases goes down poor fishes! 19. Which of the following would you expect to have the highest boiling point? a. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 b. CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 c. CH 3 CH 2 CH 2 CH 2 CH 3 d. CH 3 CH 2 CH 2 CH 3 e. CH 3 CH 2 CH 3 Larger size allows more contact between molecules and thus greater Van der Waals forces to develop 20. Which of the following would you expect to exert the highest vapor pressure? a. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 b. CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 c. CH 3 CH 2 CH 2 CH 2 CH 3 d. CH 3 CH 2 CH 2 CH 3 e. CH 3 CH 2 CH 3

7 The lower the boiling point the higher the vapor pressure a given compound exerts. Alternatively stated, the normal boiling point occurs when the vapor pressure of a compound exceeds standard atmospheric pressure, so compounds exerting high vapor pressures at low temperatures will have low boiling points 21. Which of the following would you expect to have the greatest solubility in water? Hint: drawing Lewis structures may help a. HOCH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 OH b. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 NH 2 c. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 NHCl d. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 NCl 2 e. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 NH 3 Cl Close examination of the right hand portion of option (e) shows that it is ionized, since N may only have 4 bonds (Cl - is hanging out next to it for charge neutrality). Ion-dipole interactions are energetically very favorable Temperature ( o C) Vapor Pressure (mm Hg) What is the relative humidity on a 30 o C day if the gas pressure due to water is 9.2 mm Hg? R.H. = (actual H 2 O vapor pressure equilibrium H 2 O vapor pressure)100 R.H. = (9.2 mm Hg/31.8 mm Hg)100 Answer: 29 % Bonus (1 EC): What is the dew point on a 30 o C day if the gas pressure due to water is 9.2 mm Hg? Answer: 10 o C If the temperature falls to this temperature, the equilibrium (100 % R.H.) condition would be met. Any lower and dew would have to form to reestablish equilibrium