Equil Supp sheet Ans S'06

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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.)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 NO2 2 /P N2O4. 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 N2O4 P NO2 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 N2O4 = 1.00 - 0.165 = 0.835 P NO2 = 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. P
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/03/2008 for the course CHEM 114 taught by Professor Martin during the Spring '08 term at Moravian.

Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online