Chem. 1A Midterm 2 ersion B November 9, 2016 First initial of last name Name: Print Neatly. You will lose 1 point if I cannot read your name or perm number. Perm Number: All work must be shown on the exam for partial credit. Points will be taken off for incorrect or missing units. Calculators are allowed. Cell phones may not be used as calculators. On fundamental and challenge problems you must show your work in order to receive credit for the problem. If your cell phone goes off during the exam you will have your exam removed from you. Fundamentals (of 36 possible) Problem 1 (of 16 possible) Problem 2 (of 18 possible) Multiple Choice (of 30 possible) Midterm Total (of 100 possible) 1
Fundamental Questions Each of these fundamental chemistry questions is worth 6 points. You must show work to get credit. Little to no partial credit will be awarded. Make sure to include the correct units on your answers. 1) 6 pts Balance the following equation in acidic solution Cu(s) + NO 3 (aq) Cu 2+ (aq) + NO(g) Cu Cu 2+ Cu Cu 2+ + 2e NO 3 NO NO 3 NO +2H 2O NO 3 + 4H + NO +2H 2O NO 3 + 4H + + 3e NO +2H 2O 3(Cu Cu 2+ + 2e )= 3Cu 3Cu 2+ + 6e 2(NO 3 + 4H + + 3e NO +2H 2O )= 2NO 3 + 8H + + 6e 2NO +4H 2O 3Cu(s) + 2NO 3 (aq) + 8H + (aq) 3Cu 2+ (aq) + 2NO(g) +4H 2O(l) 2) 6 pts What are the (a) molecular, (b) complete ionic, and (c) net ionic reactions when the following are mixed? If no reaction occurs write no reaction. Do not forget to include states. Fe(NO 3) 3(aq) and (NH 4) 2SO 4 No reaction CaCl 2(aq) and K 2SO 4(aq) Molecular: CaCl 2(aq) + K 2SO 4(aq) 2KCl(aq) + CaSO 4(s) Complete Ionic Equation: Ca 2+ (aq)+2cl (aq)+2k + 2 (aq)+so 4 (aq) 2K + (aq)+2cl (aq)+caso 4(s) Net Ionic Equation: Ca 2+ 2 (aq) + SO 4 (aq) CaSO 4(s) 3) 6 pts Consider the following samples of gases at the same temperature. Arrange each of these samples in order from lowest to highest. a. Pressure ii = vi < i = iv = v = viii < iii = vii b. Average kinetic energy All sample have same kinetic energy c. Density ii < i = iv =vi < iii = v = viii < vii d. Root means square velocity v = vi = vii = viii < i = ii = iii = iv 2
4) 6 pts Each sketch below shows a flask with some gas and a pool of mercury in it. The gas is at a pressure of 0.5 atm. A Jshaped tube is connected to the bottom of the flask, and the mercury can freely flow in or out of this tube. (You can assume that there is so much more mercury in the pool than can fit into the tube that even if the Jtube is completely filled, the level of mercury in the pool won't change.) Notice also that in the left sketch the Jtube is open at its other end, so that air can freely flow in or out of the tube. On the other hand, in the right sketch the Jtube is closed at its other end, and you should assume there is no gas between the mercury and the closed end of the tube. To answer this question, you must decide what the mercury level will be when the mercury finally stops flowing in or out of the tube and draw the level in on the picture. You must show calculation to get full credit. Open Tube Close Tube Open Tube: Gravity: 1.5 m P Bulb : 0.5 atm ( P Ex : ( ) = 380 mmhg = 0.380 mhg ) = = 0.760 mhg Total Rise: 1.5 m + 0.380 m + 0.760 m = 1.12 m Close Tube: Gravity: 1.5 m P Bulb : 0.5 atm( ) = 380 mmhg = 0.380 mhg Total Rise: 1.5 m + 0.380 m = 1.88 m 5) 6 pts During a titration a 0.02500 L sample of H 2SO 4 required 24.16 ml of 0.106 M NaOH to reach the equivalence point. What is the initial concentration of the H 2SO 4? 2NaOH(aq) + H 2SO 4(aq) Na 2SO 4(aq) + 2H 2O(l) 24.16 ml NaOH ( 1 L mol NaOH ) (0.106 ) = 0.00256 mol NaOH 1000 ml 1 L NaOH 0.00256 mol NaOH ( 1 mol H 2SO 4 2 mol NaOH 2SO 4 M = n 0.000128 mol = = 0.0512 M 0.02500 L 6) 6 pts When the equilibrium is disturbed, in which direction will the reaction proceed? Circle the correct answer. 2CO(g) + O 2(g) 2CO 2(g) exothermic Remove CO 2 Reactants Products No Change Reduce the volume Reactants Products No Change Increase the temperature Reactants Products No Change Add Ne Reactants Products No Change Add O 2 Reactants Products No Change 3
Challenge Problems Each of the following short answer questions are worth the noted points. Partial credit will be given. You must show your work to get credit. Make sure to include proper units on your answer. 1a) 6 pts A mixture of 1.00 g H2 and 8.60 g O2 is introduced into a 1.500 L flask at 25 C. What is the total gas pressure in the flask? P tot = n totrt Determine n Determine P tot = H 2 = (n H 2 + n O2 )RT n H2 = 1.00 g ( 1 mol H 2 2.0158 g H 2 ) = 0.496 mol H 2 n O 2 n O2 = 8.60 g ( 1 mol O 2 ) = 0.269 mol O 31.998 g O 2 2 L atm (0.496 mol + 0.269 mol)(0.08206 mol K )(298K) = 12.5 atm 1.500 L 1b) 10 pts When the mixture is ignited, an explosive reaction occurs in which water is the only product. What is the total gas pressure when the flask is returned to 25 C? (The vapor pressure of water at 25 C is 23.8 mmhg.) Reaction of Interest 2H2(g) + O2(g) 2H2O(g & l) This is a limiting reagent problem H2 (L.R.) O2 H2O I (mol) 0.496 0.269 0 C (mol) 2x = 0.496 x = 0.248 +2x = +0.496 F (mol) 0 0.4962x=0 0.021 0.496 x=0.248 P tot = P H2 + P O2 + P H2 O Calculate P H2 P H2 = 0 atm Calculate P O2 P O2 = n L atm O 2 RT (0.021 mol)(0.08206 mol K = )(298K) = 0.342 atm 1.500 L Calculate the P H2 O At 25 C the majority of the water will be in the liquid state. The pressure of H2O comes from the vapor pressure of water. P H2 O = 23.8 mmhg ( ) = 0.0313 atm Calculate Ptot P tot = P H2 + P O2 + P H2 O = 0 atm + 0.342 atm + 0.0313 atm = 0.373 atm 4
2a) 8 pts A 2.4156 g sample of PCl 5 was placed in an empty 2.000L flask and allowed to decompose to PCl 3 and Cl 2 at 250.0 C: PCl 5(g) PCl 3(g) + Cl 2(g) At equilibrium the total pressure inside the flask was observed to be 358.7 torr. Calculate the partial pressure of each gas at equilibrium and the value of K P at 250.0 C. P tot = P PCl5 + P PCl3 + P Cl2 = 358.7 torr( ) = 0.4720 atm (at equilibrium) 760 torr PCl 5 PCl 3 Cl 2 Initial (m) 2.4156 g 0 g 0 g Initial (n = m M x ) 0.011601 mol 0 mol 0 mol Initial (P = nrt ) 0.24894atm 0 atm 0 atm Change x +x +x Equilibrium 0.24894x x x Therefore P tot = P PCl5 + P PCl3 + P Cl2 = 0.24894 x + x + x = 0.4720 x = 0.2231 P PCl5 = 0.2489 atm x = 0.2489 atm 0.223 = 0.0258 atm P PCl3 = P Cl2 = x = 0.223 K P = P PCl 3 P Cl2 = (0.2231)(0.2231) = 1.93 P PCl5 0.0258 2b) 10 pts What are the new equilibrium pressures if 0.250 mole of Cl 2 gas is added to the flask? PCl 5 PCl 3 Cl 2 Equilibrium 0.0258 atm 0.223 0.223 Added (n) 0 mol 0 mol 0.250 mol Added(P = nrt ) 0 atm 0 atm 5.36 atm Initial 0.0258 atm 0.223 5.58 atm Change +x x x Equilibrium 0.0258+x 0.2231x 5.58x K P = P PCl 3 P Cl2 (0.2231 x)(5.58 x) = = 1.93 P PCl5 0.0258 + x x 2 5.80x + 1.24 = 1.93 0.0258 + x x 2 5.80x + 1.24 = 0.0498 + 1.93x x 2 7.73x + 1.19 = 0 x = b ± b2 4ac = 7.73 ± ( 7.63)2 4(1)(1.19) = 7.57 and 0.156 2a 2(1) Since the pressure cannot be negative x=0.156 Using guess and check x = 0.1578 with significant figures x = 0.158 Partial pressure at equilibrium P PCl5 = 0.0258 atm + x = 0.0258 atm + 0.156 atm = 0.182 atm P PCl3 = 0.223 x = 0.223 0.156 atm = 0.067 atm P Cl2 = 5.58 atm x = 5.58 atm 0.156 atm = 5.42 atm 5
Multiple Choice Questions On the ParScore form you need to fill in your answers, perm number, test version, and name. Failure to do any of these things will result in the loss of 1 point. Your perm number is placed and bubbled in under the ID number. Do not skip boxes or put in a hyphen; unused boxes should be left blank. Bubble in your test version (B) under the test form. Note: Your ParScore form will not be returned to you, therefore, for your records, you may want to mark your answers on this sheet. Each multiple choice question is worth 5 points. 1. How is the observed pressure of a gas related to the ideal pressure? A) The relationship depends on the gas. B) The observed pressure is less than the ideal pressure. C) The observed pressure is greater than the ideal pressure. D) They are equal. E) None of the above 2. Consider the equation 2A(g) 2B(g) + C(g). At a particular temperature, K = 1.6 x 10 4. If you mixed 5.0 mol B, 0.10 mol C, and 0.0010 mol A in a 1L container, in which direction would the reaction initially proceed? A) To the left. B) To the right. C) The above mixture is the equilibrium mixture. D) We cannot tell from the information given. 3. Given the following two reactions what is the equilibrium constant of 2A(aq) D(aq) + E(aq)? A(aq) + B(aq) D(aq) + C(aq) K1 = 5.0 2C(aq) + D(aq) 2B(aq) + E(aq) K2 = 2.0 A) 10. B) 2.5 C) 0.40 D) 20. E) None of the above 4. The following experiment was carried out using a newly synthesized chlorofluorocarbon. Exactly 50 ml of the gas effused through a porous barrier in 157 s. The same volume of argon effused in 76 s under the same conditions. Which compound is the chlorofluorocarbon? A) C2Cl5F B) C2Cl4F2 C) C2Cl2F4 D) C2ClF5 E) C2Cl3F3 6
5. The equilibrium constant K for the following reaction at 900. C is 0.0028. What is KP at this temperature? CS2(g) +4H2(g) CH4(g) + 2H2S(g) A) 0.28 B) 2.7 10 1 C) 2.9 10 3 D) 3.3 10 6 E) None of the above 6. Which of the following reactions does not involve oxidationreduction? A) Mg + 2HI MgI2 + H2 B) MnO2 + 4HI I2 + 2H2O + MnI2 C) 2Na + 2H2O 2NaOH + H2 D) LiOH + HCl H2O + LiCl E) All of the above are oxidationreduction reactions Solutions: B, A, E, C, E, D 7