First initial of last name Chem. 1B Midterm 1 Version B February 1, 2019 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 14 possible) Problem 2 (of 20 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 Given the following equilibrium constants at 427 C Na 2O(s) 2Na(l) + ½O 2(g) K 1 = 2 10 25 NaO(g) Na(l) + ½O 2(g) K 2 = 2 10 5 Na 2O 2(s) 2Na(l) + O 2(g) K 3 = 5 10 29 NaO 2(s) Na(l) + O 2(g) K 4 = 3 10 14 Determine the value for the equilibrium constant for 2NaO(g) Na 2O 2(s) 2(NaO(g) Na(l) + ½ O 2(g)) 2 K 2 2Na(l) + O 2(g) Na 2O 2(s) 2NaO(g) Na 2O 2(s) 1 K3 2 K 2 = (2 10 5 2 ) K 3 5 10 29 = 8 1018 2) 6 pts Steam reforming of methane (CH 4) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 100. L and with 46. mol of methane gas and 5.5 mol of water vapor at 35.0 C. He then raises the temperature, and when the mixture has come to equilibrium, measures the amount of hydrogen gas to be 12 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to 2 significant digits. H 2O(g) + CH 4(g) CO(g) + 3H 2(g) H2O CH4 CO H2 I(mol) 5.5 46 0 0 C(mol) x x +x +3x E(mol) 1.5 42 4 12 From the H2: 0 3x = 12 therefore, x =4 E(M) 0.015 0.42 0.040 0.12 K = [CO][H 2] 3 [H 2 O][CH 4 ] = (0.040)(0.12)3 (0.015)(0.42) = 0.011 3) 6 pts A solution has a ph of 4.90, what is the poh, [H + ], and [OH ]? poh = 14.00 ph = 14.00 4.90 = 9.10 [H + ] = 10 ph = 10 4.90 = 1.3 10 5 M [OH 1.0 10 14 ] = 1.3 10 5 = 7.7 10 10 M 2
4) 6 pts A chemist must prepare 350.0 ml of nitric acid solution with a ph of at 1.60 at 25 C. He will do this in three steps: Fill a 350.0 ml volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (6.0 M) stock nitric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated nitric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits. Determine the final concentration of H + concentration HNO 3 (a strong acid) [H + ] = [HNO 3 ] = 10 ph = 10 1.60 = 0.025 M Determine the amount of 6.0 M stock solution needed to make 350.0 ml of 0.025 M HNO 3 M 1 V 1 = M 2 V 2 (6.0 M)(V 1 ) = (0.025 M)(350.0 ml) V 1 = 1.5 ml 5) 6 pts What is the ph of the following: 1 10 12 M NaOH Because the concentration [OH ] is much less than 1 10 7 M the majority of the OH comes from water and the ph = 7.0 0.20 M magnesium hydroxide Mg(OH) 2 is a strong base [OH ] = 2[Mg(OH) 2 ] [OH ] = 2(0.20 M) = 0.40 M poh = log[oh ] log(0.40 M) = 0.40 ph = 14.00 poh = 14.00 0.40 = 13.60 6) 6 pts When the equilibrium is disturbed, in which direction will the reaction proceed? Circle the correct answer. H 2(g) + Br 2(g) 2HBr(g) exothermic Remove H 2 Reactants Products No Change Reduce the volume Reactants Products No Change Decrease the temperature Reactants Products No Change Add Ar 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. 1) 14 pts There are 7 samples, each containing 1.0 M solutions of the following: KOH, NaHC 6H 6O 6, C 2H 5NH 3CN, HNO 3,CH 3NH 3ClO 2, LiClO 4, HF Put the solutions in order from smallest to largest ph. You must show work that can be followed to receive credit for this problem. KOH Strong Base HNO 3 Strong Acid HF Weak Acid K a=7.2 10 4 LiClO 4 Salt from strong acid and strong base neutral LiClO 4(s) Li + (aq) +ClO 4 (aq) Li + + H 2O LiOH + H + (LiOH strong base) Li + is neutral OCl 4 + H 2O LiClO 4 is neutral HOCl 4 + OH (HOCl 4 strong acid) OCl 4 is neutral NaHC 6H 6O 6(s) Na+(aq) + HC 6H 6O 6 (aq) Na + + H 2O NaOH + H + (NaOH is a strong base) Na + is neutral HC 6H 6O 6 + H 2O HC 6H 6O 6 + H 2O H 2C 6H 6O 6 + OH Suggest basic C 6H 6O 6 2 + H 3O + Suggest acidic HC 6H 6O 6 (aq) H + (aq) + C 6H 6O 6 2 (aq) K a=1.6 10 12 HC 6H 6O 6 (aq) + H 2O(l) H 2C 6H 6O 6(aq) + OH (aq) K b = K w = 1.0 10 14 K a 7.9 10 NaHC 6H 6O 6 is basic because K b>k a 5 = 1.3 10 10 C 2H 5NH 3CN(s) C 2H 5NH 3+ (aq) + CN (aq) C 2H 5NH + 3 + H 2O C 2H 5NH 2 + H 3O + C 2H 5NH + 3 is acidic K a = K w 1.0 10 14 = = 1.8 10 11 K b 5.6 10 4 CN + H 2O HCN + OH CN is basic K b = K w 1.0 10 14 = = 1.6 10 5 K a 6.2 10 10 C 2H 5NH 3CN is basic because K b>k a CH 3NH 3ClO 2(s) CH 3NH 3+ (aq) + ClO 2 (aq) CH 3NH + 3 + H 2O CH 3NH 2 + H 3O + CH 3NH + 3 is acidic K a = K w 1.0 10 14 = = 2.3 10 11 K b 4.38 10 4 ClO 2 + H 2O HClO 2 + OH OCl 2 is basic K b = K w = 1.0 10 14 K a 1.2 10 CH 3NH 3ClO 2 is acidic because K b<k a 2 = 8.3 10 13 Bases: KOH (K b>1), C 2H 5NH 3CN (K b=1.6 10 5 ), NaHC 6H 6O 6 (K b=1.3 10 10 ) The larger the K b the larger the ph Neutral: LiClO 4 Acids: HNO 3 (K a>1), CH 3NH 3ClO 2 (K a=2.3 10 11 ), HF (K a=7.2 10 4 ) The smaller the K a the larger the ph HNO 3 < HF < CH 3NH 3ClO 2 < LiClO 4 < NaHC 6H 6O 6 < C 2H 5NH 3CN < KOH 4
2a) 9 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( 1 atm ) = 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 ) V 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 0.2231 atm = 0.0258 atm P PCl3 = P Cl2 = x = 0.2231 atm K P = P PCl 3 P Cl2 = (0.2231)(0.2231) = 1.93 P PCl5 0.0258 2b) 11 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 0.2231 atm 0.2231 atm Added (n) 0 mol 0 mol 0.250 mol Added (P = nrt ) V 0 atm 0 atm 5.36 atm Initial 0.0258 atm 0.2231 atm 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 ± ( 7.73)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 = 0.2231 atm x = 0.2231 atm 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 multiplechoice question is worth 5 points. 1. How are K and KP related for the following reaction? CO(g) + 1/2O2(g) A) KP = K CO2(s) B) KP = RTK C) KP = K D) KP = K E) None of the above 2. Consider the equation A(aq) + 2B(aq) 3C(aq) + 2D(aq). 42.0 ml of 0.046 M A is mixed with 24.0 ml 0.105 M B. At equilibrium, the concentration of C is 0.0416 M. Calculate K. A) 4.8 B) 0.0029 C) 0.074 D) 2.9 10 4 E) 0.033 3. There are two beakers one beaker has a ph of 2.0 (more H + ions in solution) and the other beaker has a ph of 4.0 (less H + ions in solution). You know that the beakers contain either HCl (strong acid) or HF (weak acid). Match the ph to the correct beaker. A) The beaker with a ph=4.0 is HCl B) The beaker with a ph=2.0 is HCl C) Cannot tell from the information given 4. The salt BX, when dissolved in water, produces an acidic solution. Which of the following could be true? A) HX is a strong acid. B) Both HX and the cation B + are weak acids. C) The cation B + is a weak acid. D) HX is a weak acid. E) All of these could be true. 6
5. The conjugate acid of HPO4 2 is: A) H2PO4 B) H3PO4 C) HPO4 2 D) H3O + E) PO4 3 6. For a particular system at a particular temperature, there are equilibrium constant(s) and equilibrium position(s). A) an infinite number of, an infinite number of B) one, one C) an infinite number of, one D) one, an infinite number of E) None of the above E,E,C,E,A,D 7