Question

(b) Isopropyl alchohol (C3H7OH) (the main ingredient in rubbing alchohol) can decompose into acetone (C3H60) and hydrogen according to C3H2OH(g) = C3H6O(g)+H2(g) At 180°C this has the equilibrium constant K=0.45. (i) 1.5 mol of isopropyl alcohol is placed in a 5 L container at 180 C and allowed to come to equilibrium. What are the concentrations of isopropyl alcohol, acetone, and hydrogen in the resulting mixture? [30 marks] (ii) Determine the mole fractions of isopropyl alcohol, acetone, and hydrogen in the equilibrium mixture (10 marks) de the reactants?

Answer 1

The given reaction is :- Cathy OH (9) - C3460 (9) + H₂ (9) Ke = 0.45. No. of moles of Caljon taken = 1.5 moles volume of the container = 5L. un concentration of C3H OH - no. of moles = 1.5 volume 6. [C3H₂ 047 = 0.3M.. & Catty OH(g) - CB HOO (9) + H₂ (9) Initial 0.3M Change tatu. equilibrium 0.3-1 WALI - -u ta ta 7 0.45 = i. Equilibrium constant, Ke = 2(31.60] [2] [Cathy OH] 6) (n) (0.3-2) 7 0.135 - 0.45n = n² q?+ 0.46% - 0.136 = 0 e x = 0.2058 6. At equilibuium [C3 M J OH] = 0.3-4 = 0.3-0.2068 = 0.0942m and [L3M60] = [th] = n = 0.2068 M. y using the relation, no. of moles concentuation & volume. No. of moles of C3 HjOH at equilibrium = NCHIOH = 0.0942x5 = 0.471 moles. Similarly, nato = 0.2058 x 5 and N = 0. 2068 x 5 0 = 1.029 mores. 1.029 miles. i. More fraction of C3HH OH NEGHAOH = Meston MCH OH th C3 460 the 0.47 1 0.471 + 1.029+1.029

. Исои он = 0. 18Ь Ам по но Similarly, more fraction of Cat60 = Kesho I am MegnyoH & Resto this 2 —-— 0 — .004 О. АН+). 024 + 1.0 29 - р. 401 42 and more faction of H₂ = No = Пари +по 10 +ИИ. - 1.024 0.4 + 1.024 + 1-0 29 2 0.40 3