Tutorial 6 Solutions

# Tutorial 6 Solutions - CHEMISTRY 1AO3 TUTORIAL#6 Oct 19-23...

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CHEMISTRY 1AO3 TUTORIAL #6 Oct. 19-23, 2009 Solutions – Chapters 15 and 5 ______________________________________________________________________________ Page 1 of 7 The following problems (1-7) are based on material from Chapters 15 and class notes. 1. Write the expression for the equilibrium constant, K c for each of the following reactions: SOLUTION: (i) 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) K c = [SO 3 ] 2 /[SO 2 ] 2 [O 2 ] (ii) CH 3 COOH(aq) + H 2 O(l) CH 3 COO (aq) + H 3 O + (aq) K c = [CH 3 COO ][H 3 O + ]/[CH 3 COOH] (iii) CH 4 (g) + H 2 O (g) CO (g) + 3H 2 (g) Kc = [CO][H2]3/[CH4][H2O] (iv) 4 NO 2 (g) 2 N 2 O (g) + 3 O 2 (g) K c = [N 2 O] 2 [O 2 ] 3 /[NO 2 ] 4 (v) PCl 3 (l) + Cl 2 (g) PCl 5 (s) K c = 1/[Cl 2 ] 2. For the gas phase reaction H 2 + I 2 2HI at 490 o C, the equilibrium concentrations (in mol litre -1 ) are: [H 2 ] = 8.62 10 -4 M, [I 2 ] = 2.63 10 -3 M, [HI] = 1.02 10 -2 M. (a) Calculate K c for the equilibrium K c = [HI] 2 /[H 2 ][I 2 ] = (1.02 10 -2 ) 2 /(8.62 10 -4 )(2.63 10 -3 ) = 45.9 (no units) (b) What is K c for the reverse reaction, such that: HI 1/2 H 2 + 1/2I 2 ? For the reverse process, K c ' = 1/ K c and also the stoichiometry is divided in half, so take the square root: K c ' = (1/ K c ) 1/2 = (1/45.9) 1/2 = 0.148 (c) When one mole of H 2 and one mole of I 2 in a 1.00 litre container are heated to 490 o C, what are the equilibrium concentrations of H 2 , I 2 and HI ? (mol/L) [H 2 ] [I 2 ] [HI] initial 1.00 1.00 0 change - x - x + 2x at equilibrium (1.00 - x) (1.00 - x) (2x)

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CHEMISTRY 1AO3 TUTORIAL #6 Oct. 19-23, 2009 Solutions – Chapters 15 and 5 ______________________________________________________________________________ Page 2 of 7 K c = (2x) 2 /(1 - x) 2 = 45.9 (this expression is a perfect square – take square root of both sides) therefore, (2x)/(1 - x) = 45.9 = 6.77, and solving for x gives x = 0.772 mol/L [H 2 ] = [I 2 ] = (1.00 - x) = 0.23 mol/L [HI] = 2x = 1.54 mol/L (d) Under the same conditions as for (c), an extra 0.20 mole of H 2 are added; what are the concentrations of H 2 , I 2 and HI when equilibrium has been reached? One could start with the equilibrium conditions for (c), but it is easier to assume the additional moles were added at the beginning. The final equilibrium will be the same either way. (mol/L) [H 2 ] [I 2 ] [HI] initial 1.20 1.00 0 change - x - x + 2x at equilibrium (1.20 - x) (1.00 - x) (2x) K c = (2x) 2 /(1.20 - x)(1.00 - x) = 45.9 which gives a quadratic: 41.9x 2 – 100.98x + 55.08 = 0 Find a solution, with the quadratic formula: x = [ - b  (b 2 - 4ac) ] / 2a, in this case a = 41.9, b = -100.98 and c = 55.08 Solve the quadratic to see that: x = 1.58 or 0.834 (the value of 1.58 is not possible, since it would give a negative value for [H 2 ] and [I 2 ] ) therefore, x = 0.834 is the correct solution [H 2 ] = (1.20 - x) = 0.37 mol/L [I 2 ] = (1.00 - x) = 0.17 mol/L; [HI] = 2x = 1.67 mol/L (e) Explain the impact of performing the original reaction in (a) at higher pressure in terms
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Tutorial 6 Solutions - CHEMISTRY 1AO3 TUTORIAL#6 Oct 19-23...

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