Dynamic EquilibriumNotes p. 3-4

2 NO2 (g) N2O4 (g)

Initial concentration N2O4: 0.40 MFinal concentration N2O4: 0.035 MInitial concentration NO2: 0.00 MFinal concentration NO2: 0.01 M

Initial concentration N2O4: 0.00 MFinal concentration N2O4: 0.035 MInitial concentration NO2: 0.08 MFinal concentration NO2: 0.01 M

Equilibrium is reached when concentration stops changing.

Equilibrium is reached when reaction rate stops changing.

Red rover

Red rover

Red rover

Red rover

Chemical Equilibrium

•Reaction: •NH3 + HCl NH4Cl

•Watch this demonstration (Click on the link when the PPT is in “play” mode)•Describe what is happening.

Reversible reaction

• A chemical reaction in which the products can react to re-form the reactants • Symbol used to represent reversible reactions:

• ⇌ or

Example

•Example:•NH3 + HCl NH⇌ 4Cl

• Forward reaction: • NH3 + HCl NH4Cl• Backward reaction: • NH4Cl NH3 + HCl

equilibrium

• Chemical equilibrium exists when: • 1. The rate of the forward reaction () is

equal to the rate of the reverse reaction ()

• 2. The concentration of the products and reactants remains unchanged

LeChatelier’s Principle

When a system atequilibrium is placed understress, the system willundergo a change in sucha way as to relieve thatstress.

Henry Le Chatelier

In other words,

•When you add something to a system at equilibrium, the system shifts in such a way as to use up what you’ve added.•Add to the reactants, shifts towards

the products. • Add to the products, shifts towards

the reactants.

In other words,

•When you take something away from a system at equilibrium, the system shifts in such a way as to replace what you’ve taken away.• Remove from the reactants, shifts towards

the reactants. • Remove from the products, shifts towards

the products.

Easy way TO REMEMBER

•If you ADD, then move AWAY•If you TAKE, then move TOWARDS

LeChatelier Example #1A closed container of ice and water at equilibrium. The temperature is raised.

Ice + Energy Water

The equilibrium of the system shifts to the _______ to use up the added energy.

right

LeChatelier Example #2A closed container of N2O4 and NO2 at equilibrium. NO2 is added to the container.

N2O4 (g) + Energy 2 NO2 (g)

The equilibrium of the system shifts to the _______ to use up the added NO2.left

LeChatelier Example #3A closed container of water and its vapor at equilibrium. Vapor is removed from the system.

water + Energy vapor

The equilibrium of the system shifts to the _______ to replace the vapor.

* Mistake on the notes – should be THIS reaction, not N2O4 (g) + Energy 2 NO2 (g)

right

Equilibrium

• Click to watch an episode of Crash Course about equilibrium • It is very clear and very helpful.

Solutions and Concentrations

Notes Page 5

SOLUTIONS

A homogenous mixture of two or more substances

2 PARTS OF A SOLUTION• SOLUTE• A solute is the dissolved substance in a solution.

Sugar in sodaCarbon dioxide in sodaSalt in salt water

• SOLVENT• A solvent is the dissolving medium in a solution. • Water in salt water

Water in soda

Molarity = concentration

•Amount of substance (mol)•Volume of mixture (L)

•Unit = mol/L or M•1.0 M = 1 molar •12.0 M = 12 molar

solUTE over solVENT

Example 1

•1.5 mol of NaCl are dissolved in 1 L of water. What is the molarity of the solution?

•1.5 mol NaCl = 1.5 mol/L or 1.5 M•1 L H2O

Example 2•100 g of NaCl are dissolved in 1 L of water.

What is the molarity of the solution?

•100 g NaCl |1 mol NaCl = 1.71 mol 58.44 g NaCl

•1.71 mol NaCl = 1.71 M 1 L

Example 3

•100 g of NaCl are dissolved in 250 mL of water. What is the molarity of the solution? •100 g NaCl |1 mol NaCl = 1.71mol• 58.44 g NaCl •250 mL / 103 mL = 0.25 L•1.71 mol NaCl = 6.84 M 0.25 L

Example 4

• How many moles of NaCl are in 500 mL of a 0.25 M solution? • Basic formula: M = mol L • Convert: 500 mL = 0.5 L

• 0.25 M = mol = 0.5 L x 0.25 M = mol 0.5 L

= 0.125 mol NaCl

Example 5

• How many grams of NaCl are in 500 mL of a 2.25 M solution? • Basic formula: M = mol L • Convert: 500 mL = 0.5 L

• 2.25 M = mol = 0.5 L x 2.25 M = mol 0.5 L = 1.125 mol NaCl • 1.125 mol NaCl | 58.44 g NaCl = 65.75 g NaCl 1 mol