1 / 15

Equilibrium Calculations

Equilibrium Calculations. Law of chemical equilibrium. For an equilibrium a A + b B c C + d D K = [C] c [D] d [A] a [B] b K is the equilibrium constant for that reaction. The [ ] mean concentration in molarity. Problem. 2NH 3 (g) N 2 (g) + 3H 2 (g)

alden
Download Presentation

Equilibrium Calculations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Equilibrium Calculations

  2. Law of chemical equilibrium • For an equilibrium • a A + b B c C + d D • K = [C]c[D]d • [A]a[B]b • K is the equilibrium constant for that reaction. • The [ ] mean concentration in molarity

  3. Problem • 2NH3 (g) N2 (g) + 3H2 (g) • Calculate the equilibrium constant for the above reaction if it comes to equilibrium with the following concentrations: N2 = .18 M, H2 =.35 M, and NH3 = 1.03 M

  4. Answer • K = [N2][H2]3 • [NH3]2 • K = [.18][.35]3 • [1.03]2 • K = .0073

  5. Equilibrium by phase • Equilibrium depends on the concentration of the reactants. • We can calculate the concentration of a gas or of anything dissolved (aqueous). • Insoluble solids or liquids won’t have a concentration. • They in essence are removed from the equilibrium.

  6. So using that • What would the equilibrium expression look like for the following reaction? • 2 H2O2(l) 2 H2O(l) + O2(g) • We ignore the liquids (and solids). • K = [O2]

  7. Another problem • 2 SO2 (g) +O2 (g) 2 SO3 (g) • K = 4.34, for the above reaction. Calculate the concentration of SO3 if the SO2 = .28 M and O2 = .43 M at equilibrium.

  8. Answer • K = [SO3] 2 • [SO2]2 [O2] • 4.34 = [SO3]2 • [.28]2[.43] • [SO3] = .38 M

  9. Solution Equilibrium • All dissociations we have done are equilibriums. • Before we simply stated something was soluble or insoluble. • Actually everything dissolves to some extent, and some dissolved substance fall out of solution. • Higher concentrations force more solute to fall out of solution. • So there is a maximum concentration of solute a solution can hold (saturation)

  10. Solution equilibrium • For Example: • Lead (II) Bromide • PbBr2 (s) Pb2+ (aq)+ 2 Br-(aq) • What would the equilibrium expression look like? • Ksp = [Pb2+ ][Br-]2 • Equilibrium constants for dissociations are called solubility products, and are denoted by Ksp.

  11. Cont. • The Ksp value for PbBr2 is 5.0 x 10-6. • PbBr2 (s) Pb2+ (aq)+ 2 Br-(aq) • I will assume 1 L, and that I start with x moles of PbBr2. • Using the balanced equation I have x mole/L of Pb2+ and 2x mole/L of Br- • So • 5.0 x 10-6= x (2x)2 • 5.0 x 10-6= 4x3 • x=[Pb2+ ] = .011 M 2x = [Br-]= .021M

  12. This means • A solution of PbBr2 would be saturated with [Pb2+ ] = .011 M and [Br-]= .021M • These are both very low concentrations so we say the is compound is insoluble.

  13. Problems • Calculate the saturation concentrations of solutes in aluminum hydroxide Ksp = 5.0 x10 -33, and Barium sulfate Ksp = 1.4 x10 -14

  14. Answer • Al(OH)3 Al3+ (aq)+ 3 OH-(aq) • Ksp = [Al3+ ][OH-]3 • 5.0x10-33 = x (3x)3 • x = [Al3+ ]= 3.7x10-9 M • 3x = [OH-] = 1.1 x10-8 M

  15. Answer • BaSO4 Ba2+ (aq)+ SO42-(aq) • Ksp = [Ba2+ ][SO42-] • 1.4x10-14 = x (x) • x = [Ba2+]= [SO42-] = 1.2 x10-7 M

More Related