FORMULA SHEET (tear off)

Transcription

1 FORMULA SHEET (tear off) N A = x C = ( 5 / 9 ) ( F - 32) F = ( 9 / 5 )( C) amu = x kg C = K K = C atm = 760 torr = 760 mm Hg 1 atm = bar pv = nrt R = L atm/mol K 1 L atm = J R = J/mol K 1 J= 1 kg m 2 /s 2 ln(p) = - H vap + C ln(p 2 /p 1 ) = - ( H vap /R) { (1/T 2 ) - (1/T 1 ) } T p A = X A p A [B] = k p B p A = X B p A T b = K b m B T f = K f m B = [B]RT H = U + pv G = H - TS G rxn = G rxn + RT ln Q ln K = - G rxn /RT K p = K C (RT) n If ax 2 + bx + c = 0, then x = ( - b [b 2-4ac] 1/2 ) 2a K a K b = K w = 1.0 x (at T = 25 C)

2 GENERAL CHEMISTRY 2 SECOND EXAM November 3, 2017 Name KEY Version 2 Panthersoft ID Signature Part 1 (20 points) Part 2 (32 points) Part 3 (48 points) TOTAL (100 points) Do all of the following problems. Show your work.. 2

3 Part 1. Multiple choice. Circle the letter corresponding to the correct answer. There is one and only one correct answer per problem. [4 points each] 1) For the chemical reaction Cl 2 (g) + 2 NO 2 (g) 2 NO 2 Cl(g) K C = 1.8 at a temperature T. For a particular set of starting conditions and the same temperature Q C = As the system approaches equilibrium, which of the following will occur? a) The moles of NO 2 Cl(g) will increase b) The moles of Cl 2 (g) will increase A c) The moles of NO 2 (g) will increase d) Both b and c e) None of the above 2) Consider the following three reactions I II III H 2 (g) + Cl 2 (g) 2 HCl(g) 3 Fe(s) + 4 H 2 O(g) Fe 3 O 4 (s) + 4 H 2 (g) C(s) + H 2 O(g) CO(g) + H 2 (g) For which of the above reactions will the numerical value for K C and K p be equal? a) Reaction I only. b) Reaction II only. D c) Reaction III only. d) Both reaction I and reaction II e) Both reaction I and reaction III 3) Which of the following hydroxide compounds is a strong soluble base? a) KOH (potassium hydroxide) b) AgOH (silver hydroxide) A c) Ni(OH) 2 (nickel II hydroxide) d) Both a and b e) Both a and b and c 4) From the substances HI, H 2 S, and H 2 Se a) HI is the strongest acid and H 2 Se is the weakest acid b) HI is the strongest acid and H 2 S is the weakest acid B c) H 2 S is the strongest acid and HI is the weakest acid d) H 2 S is the strongest acid and H 2 Se is the weakest acid e) H 2 Se is the strongest acid and H 2 S is the weakest acid 5) In the reaction Al(H 2 O) 6 3+ (aq) + H 2 O( ) Al(H 2 O) 5 OH 2+ (aq) + H 3 O + (aq) 3+ the Al(H 2 O) 6 ion functions as a) the conjugate base of H 2 O b) the conjugate acid of H 2 O D c) a Bronsted base d) a Bronsted acid e) none of the above Version 1: C, E, C, E, B Version 3: A, E, B, D, A Version 4: C, D, B, A, C 3

4 Part 2. Short answer. 1) Consider the following chemical reaction CO(g) + H 2 (g) H 2 CO(g) H rxn = kj/mol A system at constant volume containing CO, H 2, and H 2 CO is initially at equilibrium. For each of the following changes to the system indicate whether the number of moles of H 2 CO in the system will increase, remain the same, or decrease as the system returns to equilibrium. Circle your answer. [4 points each] a) moles of CO(g) is added to the system moles of H2CO moles of H 2 CO moles of H 2 CO increases remains the same decreases b) The temperature of the system is increased by 20.0 C moles of H2CO moles of H 2 CO moles of H 2 CO increases remains the same decreases 2) Values for K b for several weak bases are given below, at T = 25. C. NH 3 K b = 1.8 x 10-5 (C 2 H 5 ) 2 NH K b = 5.4 x 10-4 C 5 H 5 N K b = 1.7 x 10-9 C 6 H 5 NH 2 K b = 3.8 x From the lists below circle the correct answer. There is one and only one correct answer per problem. [4 points each] a) The strongest weak base NH 3 (C2H5)2NH C 5 H 5 N C 6 H 5 NH 2 b) The strongest conjugate acid NH 4 + (C 2 H 5 ) 2 NH 2 + C 5 H 5 NH + C6H5NH3 + 3) An aqueous solution of a weak acid has ph = 5.18 at T = 25. C. What are the values for poh and [H 3 O + ] for the solution? [4 points each] poh = 8.82 [H 3 O + ] = 6.6 x 10-6 M Version 1: poh = 8.12, [H 3 O + ] = 1.3 x 10-6 M Version 3: poh = 9.75, [H 3 O + ] = 5.6 x 10-5 M Version 4: poh = 9.28, [H 3 O + ] = 1.9 x 10-5 M 4

5 4) Rubidium hydroxide (RbOH) is a strong soluble base. Find the ph of a M aqueous solution of rubidium hydroxide at T = 25. C. [8 points] [OH - ] = mol RbOH 1 mol OH - = M L 1 mol RbOH poh = - log 10 [OH-] = - log 10 (0.014) = 1.85 ph = poh = = Version 1: ph = Version 3: ph = Version 4: ph = Part 3. Problems. 1) Using the information given below find the numerical values for G rxn and K for the following reaction. You may assume that T = 25. C. [16 points] H 2 (g) + I 2 (g) 2 HI(g) Substance H f (kj/mol) G f (kj/mol) S (J/mol K) H 2 (g) HI(g) I 2 (g) G rxn = [2 G f (HI(g)) ] - [ G f (H 2 (g)) + G f (I 2 (g)) ] = [ 2 (1.70) ] - [ ] = kj/mol ln K = - G rxn = - ( kj/mol) (1000 J/1 kj) = 6.43 RT (8.314 J/mol K)(298. K) K = e 6.43 =

6 2) Hypochlorous acid (HOCl) is a weak acid, with K a = 3.5 x 10-8 at T = 25. C. a) Give the conjugate base of HOCl (correct formula and charge) [4 points] OCl - Version 1: OBr - Version 3: OCl - Version 4: OBr- b) What is the ph and the percent dissociation for a M aqueous solution of HOCl at T = 25. C? [12 points] Reaction is HOCl(aq) + H 2 O( ) H 3 O + (aq) + OCl - (aq) K a = [H 3 O + ] [OCl - ] = 3.5 x 10-8 [HOCl] Initial Change Equilibrium H 3 O + 0 x x OCl - 0 x x HOCl x x So (x) (x) = 3.5 x 10-8 ( x) If we assume x << , then we get x 2 = 3.5 x 10-8 x 2 = (3.5 x 10-8 ) ( ) = 3.15 x ( ) x = (3.15 x ) 1/2 = 1.78 x 10-5 ph = - log 10 (1.78 x 10-5 ) = 4.75 % ionization = [OCl-] eq x 100% = (1.78 x 10-5 ) x 100 % = 0.20 % [HOCl] initial ( ) Version 1: ph = 5.32, % dissociation = 0.05 % Version 3: ph = 5.08, % dissociation = 0.42 % Version 4: ph = 5.65, % dissociation = 0.11 % 6

7 3) Because the F-F single bond in F 2 is a weak bond, diatomic fluorine will dissociate at high temperatures. The reaction may be written as F 2 (g) 2 F(g) K C = at T = K A closed system at T = K initially has [F 2 ] = mol/l. No fluorine atoms are initially present in the system. What will be the value for [F], the concentration of fluorine atoms in the system, when equilibrium is reached? [16 points] K C = [F] 2 = [F 2 ] Initial Change Equilibrium F 0 2x 2x F x x So (2x) 2 = ( x) If we assume x << , then 4x 2 = x 2 = (0.064) (.1400) = 2.24 x 10-3 x = (2.24 x 10-3 ) 1/2 = But is not small compared to (it is not at least 10 times smaller), so we will have to use another method to find x (2x) 2 = x 2 = (0.064) ( x) = x ( x) 4x x = 0 So x = [ (0.064) 2-4 (4) ( ) ] 1/2 = , (4) The underlined root gives positive concentrations for F 2 and F. Therefore [F] = 2x = 2 (0.040) = M Version 1: [F] = M Version 3: [F] = M Version 4: [F] = M 7