2. Failure to submit this paper in its entirety at the end of the examination may result in disqualification.

Transcription

1 Memorial University of Newfoundland St. John s, Newfoundland and Labrador Chemistry 101 Intersession 007 Midterm Exam May 8 th, 007 Time: 0 Minutes Name: MUN #: Dr. Peter Warburton READ THE FOLLOWING CAREFULLY! 1. This exam has 7 pages with setions. SECTION A is omprised of longer answer questions and SECTION B is omprised of short answer questions. Ensure that this examination paper is omplete, i.e. that all pages are present.. Failure to submit this paper in its entirety at the end of the examination may result in disqualifiation. 3. A Periodi Table and physial onstants are provided. These are on the last page of the examination and may be detahed for use during the examination. 4. Answer eah question in the spae provided. Should you require more spae, use the bak of the page and indiate learly where this has been done.. When answering questions show all relevant formulae, all alulations, and justify all simplifying assumptions. 6. Corret units should be maintained in all steps of a alulation. 7. Numerial answers should be reported with the orret signifiant digits. Do not write in the enlosed area below. Question Value Mark A1 A 7 A3 9 Setion B 1 Total 4 Page 1 of 7

2 Chemistry 101 Midterm Exam Setion A Longer Answer Questions (1 Marks) The value of eah part of a question is given in the square brakets in the left margin beside eah question. [] A1. What is the a value at C for propioni aid (C H COOH) if the perent ionization of a 0040 M lution of propioni aid is.6%? C H COOH (aq) + H O (l) C H COO - (aq) + H 3 O + (aq) Initial 0040 M 0 M 0 M Change -x +x +x Eqm 0040 M - x +x +x By the definition of perent ionization, we know that % ion [H 3 O + ] eqm / [C H COOH] initial Sine [H 3 O + ] eqm x this means ( [C H COOH] initial ) ( % ion ) (0040 M)(.6%) M + [ C H COO ][ H O ] ( 000 ) -8 3 x 4. 0 x 10 a. -3 [ CHCOOH] x x x [7] A. An equilibrium mixture at a partiular temperature ontains 310 M CO (g), 140 M Cl (g) and 780 M COCl (g) in a.00 L ontainer. Calulate the number of moles of Cl (g) whih must be added to the equilibrium mixture in the.00 L flask to redue the onentration of CO (g) to 0 M when equilibrium is re-established at the same temperature for the hemial system shown below: For our original system at equilibrium [ COCl] Cl CO Cl (g) + CO (g) COCl (g) ( 780) ( 140)( 310) [ ][ ] 18.0 (all in mol L -1 ) Cl (g) + CO (g) COCl (g) Initial [ ] x Change in [ ] -y -y +y Equilibrium [ ] x - y 310 y y We an see that y 060 M [ COCl] [ Cl ][ CO] 0 ( y) ( ) ( x - y)( 0) ( x - 060)( 0) ( 840) 18.0 ( 0 + x)( 0) ( 840) M ( 18.0)( 0) Whih means sine we have.00 L ontainer we needed to add 14 mol of Cl Out of interest, this question represents an appliation of Le Chatalier s priniple. If we add Cl to the equilibrium mixture, the system will respond by trying to get rid of the added Cl. The only way it an do this is by reating me of the new Cl with CO, dereasing the amount of CO and inreasing the COCl in the new equilibrium mixture. Page of 7

3 Chemistry 101 Midterm Exam [9] A3. The reation SO Cl (g) SO (g) + Cl (g) has the equilibrium onstant 011 at a ertain temperature. A.0 L flask starts with 30 mol of SO, 16 mol of Cl, and 0 mol of SO Cl. Calulate the equilibrium onentrations of the gases. If we start with a reation quotient alulation in terms of onentration, we need the initial onentrations of the gases. Sine onentration is number of moles divided by volume, the initial onentrations of the gases will be 1 mol L -1 of SO, 0 mol L -1 of Cl, and mol L -1 of SO Cl [ SO ][ Cl ] [ SO Cl ] ( 1)( 0) ( ) Q. > Sine Q >, we expet to lose produts and gain reatants, the ICE table looks like (all in mol L -1 ) SO Cl (g) SO (g) + Cl (g) Initial 1 0 Change +x -x -x Equilibrium + x 1 - x 0 - x Rearranging, we get [ SO][ Cl] [ SO Cl ] ( 1 - x)( 0 - x) ( + x) ( + x) ( 1 - x)( 0 - x) x + x x 4 x This is a quadrati equation of the form ax + bx + 0, whih has lutions given by 0 0 b ± b 4a a (-41) ± (-4 ) 4(1)( ) 1 (1) (41) + (41) + (41) (-41) 4(1)( + 009) (41) or 096 or 037 (41) or (41) 01 or mol L or (-41) 4(1)( ) 037 (41) mol L 0 (41) 14 or If we hoose 174 mol L -1 as the answer, then [SO ] 1 mol L -1 - (193 mol L -1 ) -043 mol L -1, whih isn t physially possible. Therefore 048 mol L -1. This means our equilibrium onentrations are [SO Cl ] mol L -1 + (048 mol L -1 ) 9 8 mol L -1 [SO ] 1 mol L -1 - (048 mol L -1 ) 10 mol L -1 [Cl ] 0 mol L -1 - (048 mol L -1 ) 03 mol L -1. We should hek our answer: [ SO ][ Cl ] ( 10)( 03) 011 [ SOCl ] ( 98) This is the equilibrium onstant we were given, within rounding errors, we should have the right answer. Page 3 of 7

4 Chemistry 101 Midterm Exam Setion B Short Answer (1 Marks) The value of eah part of a question is given in the square brakets in the left margin beside eah question. B1. What is the hemial formula for slaked lime? Ca(OH) [3] B. Phosgene is formed from arbon monoxide and hlorine: CO (g) + Cl (g) COCl (g) H -10 kj Briefly explain whether the amount of phosgene inreases, dereases or remains the same when an equilibrium system for the above reation is disturbed by: a) dereasing the volume. Dereasing the volume inreases the total pressure of the system. The system responds by trying to derease the pressure by shifting towards the side with less moles of gas. For this reation this is the produts, the amount of phosgene inreases. b) inreasing the temperature. Sine the reation is exothermi, inreasing the temperature favours the reverse reation (it s like inreasing the onentration of the produt heat ) and the amount of phosgene dereases. ) adding sulphur dioxide (SO ) to the ontainer. Sulphur dioxide reats with hlorine to form SO Cl. Sine sulphur dioxide reats with hlorine, it dereases the amount of hlorine available to be used in the first reation. Dereasing the amount of a reatant shifts the reation from right to left and the amount of phosgene dereases. B3. Xenon forms ompounds with highly eletronegative fluorine and oxygen. In these ompounds the xenon often is found in one of four ommon oxidation states. Give two of these four oxidation states. Only the first two will be marked The ommon oxidation states of xenon are +, +4, +6 and +8. B4. What is the struture of the oxoaid H SeO 3 [selenous aid]? Page 4 of 7

5 Chemistry 101 Midterm Exam [] B. What are the definitions of a Brønstead-Lowry base and an Arrhenius base. Give an example of a Brønstead-Lowry base that IS NOT al an Arrhenius base. A BL base is a proton aeptor. An Arrhenius base disiates in water to give OH -. NH 3 is an example of BL base while not being an Arrhenius base. [] B6. Suppose in a sealed ontainer that two gas phase equilibria our at the same time: GeO (g) + W O 6 (g) GeWO 4 (g) 7.0 x 10 3 Reation GeO (g) + W O 6 (g) GeW O 7 (g) 3.8 x 10 4 Reation What is the equilibrium onstant for the following reation? GeO (g) + GeW O 7 (g) GeWO 4 (g) Reation When we add reations and we get reation. Adding together equilibria means we multiply equilibrium onstants, x / 7.0 x 10 3 / 3.8 x [3] B7. What is the ph of a 0 mol L -1 lution of CaO at C? Sine CaO is an alkaline earth metal oxide, it is a strong base and will reat with water ompletely suh that [CaO] i [OH - ] eq [OH - ] eq (0 mol L -1 ) 16 mol L -1 and ph poh (-log [OH - ] eq ) (- log 16) ph [] B8. Hydrogen gas an reat with dium to form dium hydride. Hydrogen gas an al be used in the formation of methanol by the redution of arbon monoxide. What are the balaned equations for these two reations? Na (s) + H (g) NaH (s) CO (g) + H (g) CH 3 OH (g) [3] B9. The reation PCl (g) PCl 3 (g) + Cl (g) has the equilibrium onstant p 11. at 300 C. Calulate the equilibrium onstant for this reation at 300 C. p (RT) n where n (moles of produt gases moles of reatant gasses) ( 1) p / (RT) p / (RT) 1 p / (RT) (11.) / [(06)(73)] note T is in elvin and the speifi value of R used! 44 Page of 7

6 Chemistry 101 Midterm Exam B1 What is the harateristi olour of dium in a Bunsen burner flame? Yellow B11. What is the balaned equation for the prodution of liquid barium from the reation of liquid barium oxide by liquid aluminum? 3 BaO (l) + Al (l) Al O 3 (s) + 3 Ba (l) B1. Whih two noble gases are used in lasers? r and Xe The End Page 6 of 7