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Unformatted text preview: Section 6/ Chemical Equilibria / Suggested solutions / SKH Bishop Baker Secondary School ALevel Chemistry Section 6 Chemical Equilibria 81 IB Q.5 (b) (i) K p = P HI 2 ; K c = [HI] 2 P H2 x P I2 [H 2 ][I 2 ] [1] + [1] (ii) K c = (3.6) 2 = 29.5 [1] + [1] (4  1.8)(2  1.8) (1 mark for the correct substitution; 1 mark for the correct answer.) 82 IB Q.9 (a) (i) K p = P H2S P NH3 P x = partial pressure [1] P H2S = P NH3 = 0.54/2 [ ] + [ ] K p = (0.54/2) 2 = 0.0729 atm 2 ( mark for numerical answer; mark for unit.) [1] (ii) either P H2S = y + 0.335 [ ] P NH3 = y [ ] or P H2S x P NH3 = K p y(y + 0.335) = 0.0729 [1] y = 0.15 atm ( mark for numerical answer; mark for unit.) [1] 83 IA Q.1 (d) To BiOCl(s) add small amounts of conc. HCl with shaking to dissolve the solid. [1] Add large excess of water. [1] 83 IB Q.9 (b) (i) K p = P H2O P H2 P O2 [ ] (iii) At equilibrium, P H2O = 1.18 x 10 40 P H2 P O2 [ ] Since P H2O P H2 P O2 the reaction goes almost to completion at equilibrium. [1] 84 IB Q.8 (a) (i) By decreasing the temperature, or increasing the pressure. (Also accept the increase of concentration of the reactants or the removal of product as soon as it is formed. Any one [1] Additional wrong answer : 1 mark) (ii) High yield of Z is favoured by a low temperature. [1] However, the time for reaching the equilibrium at a low temperature will be too long , [1] (Rate of reaction is too slow.) so a compromise in temperature (723 K) is chosen [ ] so that the percentage of product is not too low, while the time to reach equilibrium is not too long. High yield of Z is favoured by high pressure . [1] However, high pressure is economically expensive , [1] so a compromise is again chosen. [ ] (iii) 1 Section 6/ Chemical Equilibria / Suggested solutions / Effect : 1. Time to reach equilibrium is longer . [1] Reaction rate will be much slower. 2. Percentage of product is unaffected . [1] (iv) xX(g) + yY(g) == zZ(g) At equilibrium, k 1 [X] x [Y] y = k1 [Z] z [1] K = [Z] z = k 1 [X] x [Y] y k 1 [1] (No working  1 mark) 85 IA Q.1 (c) K c = [CO 2 ] 2 /[CO] 2 or K p = P CO2 2 /P CO 2 [1 ] [No indication of K c or K p 1 mark only] Give the corresponding concentration terms of CO and CO 2 (e.g. partial pressure ) at a given temperature. [1 ] [Pressure only 1 mark; but should correspond to the equation.] 86 IIA Q.3 (a) (i) (CH 3 COOH) 2 == 2CH 3 COOH 1  2 = degree of dissociation Partial pressure of (CH 3 COOH) 2 = [1  ]P / [1 + ] [ ] Partial pressure of CH 3 COOH = 2 P / [1 + ] [ ] [2 / (1 + )] 2 P 2 K p = [1  ]P / [1 + ] = 4 2 P / [1  2 ] [1] At 298 K and 1500 N m 2 , 72.9 = 4 2 /[1  2 ] x 1500, = 0.11 (0.109 0.111) [1] (ii) K p 72.9 121.1 182.7 302.0 log 10 K p 1.863 2.083 2.262 2.480 T 298 303 308 313 1/T x 10 3 3.36 3.30 3.25 3.19 (Table form : [1]) A plot of log 10 K p against 1/T gives a straight line. The equation is obeyed. [1] 2 Section 6/ Chemical Equilibria / Suggested solutions /...
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This note was uploaded on 08/17/2011 for the course CHEM 100 taught by Professor Fleet during the Fall '08 term at Oxford University.
 Fall '08
 FLEET
 pH

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