Answers_to_assn_2

Answers_to_assn_2 - Spring 06 Chem 123 Assignment 2 Friday...

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Spring 06 Chem 123 Assignment 2 Friday July 7 th at the beginning of class 1. Order the following acids from strongest to weakest. A) HI B) CH 3 OH C) CH 3 CH 2 COOH D) C 6 H 5 OH E) CH 3 CHClCOOH A) HI > E) CH 3 CHClCOOH > C) CH 3 CH 2 COOH > D) C 6 H 5 OH 2. Order the following bases from strongest to weakest. A) (CH 3 ) 3 CO¯ B) HO¯ C) CH 3 NH 2 D) CH 3 NH¯ E) CH 3 CH 2 ¯ F) C 6 H 5 NH 2 E) CH 3 CH 2 ¯ > D) CH 3 NH¯ > (CH 3 ) 3 CO¯ > B) HO¯ > C) CH 3 NH 2 > F) C 6 H 5 NH 2 3. In the following cases, you are given equilibrium constants for the two indicated reactions at 25°C, using atm for partial pressures of the gases. Use the data to calculate K p for the third reaction. N 2 (g) + 1 / 2 O 2 (g) N 2 O(g) K p, 1 = 7.1 × 10 19 N 2 (g) + O 2 (g) 2 NO(g) K p, 2 = 4.23 × 10 31 2 N 2 O(g) + O 2 (g) 4 NO(g) K p , 3 = ? Reactions 1 and 2 can be combined as follows to give reaction 3: 2N 2 O(g) 2N 2 (g) + O 2 (g) K 1,new = (1 / K p ,1 ) 2 2N 2 (g) + 2O 2 (g) 4 NO(g) K 2,new = ( K p ,2 ) 2 2 N 2 O(g) + O 2 (g) 4 NO(g) K 3 = K 1,new × K 2,new = 3.55 × 10 25 4. For the reaction CO(g) + 2H 2 (g) Æ CH 3 OH(g), Δ H = 90.2 kJ (per mole of CO) and K p = 2.2 × 10 4 at 298 K (when the pressures are expressed in atmospheres). What is K p at 500 K? () 31 500 41 1 90.2 10 J mol 11 ln 500 K 298 K 2.2 10 8.3145 J K mol K −− −× ⎛⎞ =− ⎜⎟ × ⎝⎠ K 500 = 9.0 × 10 3 Note that, for an exothermic reaction, the equilibrium constant decreases as the temperature increases.
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CHEM 123: Practice Problems page 2 of 18 2 5. Trimethylamine, (CH 3 ) 3 N, is a weak base that ionizes in aqueous solution: ( C H 3 ) 3 N(aq) + H 2 O(l) (CH 3 ) 3 NH + (aq) + OH (aq) A 0.120 M solution of (CH 3 ) 3 N(aq) is 2.29% ionized at 25 O C. (a) Calculate [OH ], [(CH 3 ) 3 NH + ], [H 3 O + ] and the pH for a 0.120 M (CH 3 ) 3 N(aq) solution at 25 o C. (b) Calculate K b for (CH 3 ) 3 N at 25 o C. (c) Calculate the degree of ionization, α , of a 0.096 M solution of trimethylamine. Does the degree of ionization increase, decrease, or remain unchanged as the concentration of (CH 3 ) 3 N decreases? Give reasons for your answer. (a) (CH 3 ) 3 N( aq ) + H 2 O( l ) (CH 3 ) 3 NH + ( aq ) + OH ( aq ) initial: 0.120 M 0 0 change: x + x + x equil: 0.120 x x x () 31 2.29 % ionization = 100% 0.120 2.75 10 mol L 0.120 100 x x −− ⎛⎞ ×⇒ = = × ⎜⎟ ⎝⎠ [OH ] = [(CH 3 ) 3 NH + ] = 2.75 × 10 3 mol L 1 pOH = 2.561 [(CH 3 ) 3 N] = (0.120 2.75 × 10 3 ) mol L 1 = 0.117 mol L 1 pH = 14 pOH = 11.439 (b) 33 5 [(CH ) NH ][OH ] (0.00275)(0.00275) 6.44 10 [(CH ) N] (0.117) b K +− == = × (c) For a 0.096 M solution, we have: 2 23 3 0.096 0 2.45 10 mol L 0.096 2.45 10 % ionization 100 2.56% 0.096 bb b x Kx K x K x x =⇒ + = = × × = 1 Notice that the degree of ionization is higher in the more dilute solution. This is consistent with Le Chatelier’s principle: to make a more dilute solution, we add water. As we add water, the equilibrium shifts to the right and thus the degree of ionization increases.
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CHEM 123: Practice Problems page 3 of 18 3 6. Use the K a values of oxalic acid, H 2 C 2 O 4 , to decide whether an aqueous solution of NaHC 2 O 4 will be acidic, basic or neutral. Look up the K a values in Appendix D of the text. Note: It is not necessary to do any pH calculations.
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This note was uploaded on 12/01/2010 for the course CHEM CHEM 123 taught by Professor Oakley during the Winter '08 term at Waterloo.

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Answers_to_assn_2 - Spring 06 Chem 123 Assignment 2 Friday...

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