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Equil Supp sheet Ans S'06

# Equil Supp sheet Ans S'06 - Spring 2006 Solutions to...

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Spring, 2006 Solutions to Chemical Equilibrium Supplementary Sheet 1. Consider the reaction at 700°K CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) for which K c = 5.10. Calculate the equilibrium concentration of all species present if 6.00 moles of CO(g) and 6.00 moles of H 2 O(g) are mixed in a 2.00 liter flask at 700°K. This is a typical equilibrium problem in which a dependent variable is assigned to represent one of the desired concentrations and the solved for from the equilibrium expression. [CO] [H 2 O] [CO 2 ] [ H2 ] 6.00 moles 6.00 moles initial ----------------- = 3.00 M --------------- = 3.00 M 0 0 2.00 L 2 L change - X - X + X + X equil 3.00 - X 3.00 - X X X Let X = the [H 2 ] produced [CO 2 ][H 2 ] (X)(X) x 2 Kc = -------------------= ------------------------- = -------------------- = 5.10 [CO][H 2 O] (3.00 - X)(3.00 - X) (3.00 - X) 2 Taking the square root of both sides of the equation gives X ---------------- = √5.10 = 2.26 (3.00 - X)(2.26) = X (3.00)(2.26) - 2.26X = X 3.00 - x 6.78 = (2.26 + 1)X 6.78 = 3.26 X X = 2.08 [CO] [H 2 O] [CO 2 ] [H 2 ] equil 3.00 - 2.08 = 0.92 3.00 - 2.08 = 0.92 2.08 2.08 Check K c = (2.08)(2.08)/(0.92)(0.92) = 5.11 (Note: If X is rounded off differently, the value of K c may vary slightly.)

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Equilibrium Supplementary Sheet Solutions Page 2 2 . Consider the reaction N 2 O 4 2NO 2 (g) at a temperature where K p = 0.131 atm. A flask initially contains N 2 O 4 (g) at 1.000 atm. pressure. Calculate the pressures of NO 2 (g) and N 2 O 4 (g) at equilibrium. This problem is similar to the last problem except that since K p is given the partial pressures of the gases can be used rather than the concentrations. K p = P NO 2 2 /P N 2 O 4 . There are two routes to the solution of this problem depending on how the dependent variable is defined. Route 1: Let x = the amount of N 2 O 4 that reacts. P N 2 O 4 P NO 2 initial 1.00 0 change - x +2x NOTE: the 2 results from the reaction soichiometry equil 1.00 - x +2x K p = (2x) 2 /(1 - x) = 0.131 Since this is not a perfect square, the quadratic equation must be used to solve for x. 4x 2 = 0.131(1-x) 4x 2 = 0.131 - 0.131 X 4x 2 + 0.131X - 0.131 = 0 x = {-0.131±√[(0.131) 2 - (4)(4)(-0.131)]}/(2)(4) X = {-0.131 + 1.454}/8 = 0.165 (Note: the other route is negative and can be ignored) P N 2 O 4 = 1.00 - 0.165 = 0.835 P NO 2 = 2(0.165) = 0.330 Check K p = (0.33)2/(0.835) = 0.130 Route 2: Let y = the amount of NO 2 produced.
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Equil Supp sheet Ans S'06 - Spring 2006 Solutions to...

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