Wednesday, March 19 th : “A” Day Thursday, March 20 th : “B” Day Agenda

Wednesday, March 19th: “A” DayThursday, March 20th: “B” Day

AgendaHomework questions/problems?Section 14.1 QuizBegin Section 14.2: “Systems at Equilibrium”Homework:

Practice pg. 504: #1, 2Practice pg. 506: #1, 2Concept Review: “Systems at Equilibrium”: #1-14

We will finish section 14.2 next time…

Homework Questions/Problems?Section 14.1 review

Pg. 501: #1-5

Quiz14.1: “Reversible Reactions and Equilibrium”

You may use your book and your notes to complete the quiz…

Good Luck!(Pick up the notes fortoday when you turn inyour quiz)

The Equilibrium Constant, Keq

There is a mathematical relationship between product and reactant concentrations at equilibrium.

Equilibrium constant, Keq: a number that relates the concentrations of starting materials and products of a reversible chemical reaction to one another at a given temperature.

The Equilibrium Constant, Keq

Equilibrium constants:

do NOT have any units depend on temperatureapply only to systems in equilibrium must be found experimentally or from tables

** Rules for determining Keq **1. Write a balanced chemical equation.

Make sure that the reaction is at equilibrium before you write a chemical equation.

2. Write an equilibrium expression. To write the expression, put the product

concentrations in the numerator and the reactant concentrations in the denominator.

The concentration of any solid or a pure liquid that takes part in the reaction is left out.

For a reaction occurring in aqueous solution, water is omitted.

** Rules for determining Keq **

3. Complete the equilibrium expression Raise each substance’s concentration to the

power equal to the substance’s coefficient in the balanced chemical equation.

The Equilibrium Constant, KeqLimestone caverns form as rainwater, slightly

acidified by H3O+, dissolves calcium carbonate. The reverse reaction also takes place, depositing

calcium carbonate and forming stalactites and stalagmites.

CaCO3(s) + 2 H3O+(aq) Ca2+(aq) + CO2(g) + 3 H2O(l)

When the rates of the forward and reverse reactions become equal, the reaction reaches chemical equilibrium.

The Equilibrium Constant, KeqWrite the equilibrium constant expression for

limestone reacting with acidified water at 25°C according to the balanced equation:CaCO3(s) + 2 H3O+(aq) Ca2+(aq) + CO2(g) + 3 H2O(l)

(left out) (left out)

Keq = [Ca2+] [CO2] [H3O+]2

Keq for this reaction at 25˚C is 1.4 X 10 -9The [ ] means “concentration” in M, moles/liter.Equilibrium constants do not have any units and apply only to systems at equilibrium.

Calculating Keq from Concentrations of Reactants and Products ( Sample Problem)

Calculate Keq for the following reaction:

Br2 (g) 2 Br (g)

At equilibrium [Br2] = 0.99 M, [Br] = 0.020 M

Write Equilibrium expression: [Br] 2 [Br2]

Plug in values = (0.020) 2 and Solve… 0.99

Keq = 4.0 X 10-4

Calculating Keq from Concentrations of Reactants and Products

Sample Problem A, pg. 504

An aqueous solution of carbonic acid reacts to reach equilibrium as described below::

The solution contains the following concentrations: - carbonic acid, 3.3 × 10−2 mol/L- bicarbonate ion, 1.19 × 10−4 mol/L- hydronium ion, 1.19 × 10−4 mol/L

Determine the Keq

Sample Problem A, pg. 504, cont.

3.3 X 10-2 M 1.19 X 10-4 M 1.19 X 10-4 M

K eq = [HCO3-] [H3O+]

[H2CO3]

Plug in values: K eq = (1.19 X 10-4 ) (1.19 X 10-4 ) (3.3 X 10-2 )

= 1.42 X 10-8

3.3 X 10-2

Keq = 4.3 X 10-7

** Keq Shows if the Reaction is Favorable **If Keq is small, reaction favors reactants. This is

called an “unfavorable” reaction.If Keq is large, reaction favors products. This is

called a “favorable” reaction.If Keq=1, products and reactants are equal.

Keq Shows if the Reaction is Favorable

The synthesis of ammonia is very favorable at 25°C and has a large Keq value.

Keq Shows if the Reaction is Favorable

However, the reaction of oxygen and nitrogen to give nitrogen monoxide is not favorable at 25°C.

It’s a good thing that this reaction is not favorable or we would have trouble breathing!

K eq = [NO]2 = 4.5 X 10-31 at 25˚C [N2] [O2]

Calculating Concentrations of Products and Reactants from Keq; Sample Problem B, pg. 506

Keq for the equilibrium below is 1.8 × 10−5 at a temperature of 25°C. Calculate [NH4

+] when [NH3] = 6.82 × 10−3

Write equilibrium expression: K eq = [NH4 +] [OH-] [NH3] NH4

+ and OH- ions are produced in equal numbers, so [NH4

+] = [OH-] = X1.8 X 10-5 = ___X2 ___

6.82 X 10-3

[NH4+] = 3.5 X 10-4

Find [H2] equilibrium at 700K when [CH3OH] = 0.25, [CO] = 0.0098, Keq = 290

CO (g) + 2 H2 (g) CH3OH (g)Write equilibrium expression: Keq = [CH3OH]

[CO] [H2]2

Plug in values: 290 = 0.25__ (.0098) [H2]2

Cross multiply: 2.8 [H2]2 = 0.25

Divide each side by 2.8: [H2]2 = .089 [H2] = 0.30

Calculating Concentrations of Products and Reactants from Keq; Additional Example

Homework

Practice pg. 504: #1, 2Practice pg. 506: #1, 2Concept Review: “Systems at Equilibrium”: #1-14

Use your time wisely….the concept review will be due before you know it!

We will finish section 14.2 next time…