Dynamic Equilibrium Notes p. 3-4. 2 NO 2 (g) N 2 O 4 (g) Initial concentration N 2 O 4 : 0.40 M...
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Transcript of Dynamic Equilibrium Notes p. 3-4. 2 NO 2 (g) N 2 O 4 (g) Initial concentration N 2 O 4 : 0.40 M...
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Dynamic EquilibriumNotes p. 3-4
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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
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Equilibrium is reached when concentration stops changing.
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Equilibrium is reached when reaction rate stops changing.
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Red rover
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Red rover
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Red rover
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Red rover
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Chemical Equilibrium
•Reaction: •NH3 + HCl NH4Cl
•Watch this demonstration (Click on the link when the PPT is in “play” mode)•Describe what is happening.
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Reversible reaction
• A chemical reaction in which the products can react to re-form the reactants • Symbol used to represent reversible reactions:
• ⇌ or
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Example
•Example:•NH3 + HCl NH⇌ 4Cl
• Forward reaction: • NH3 + HCl NH4Cl• Backward reaction: • NH4Cl NH3 + HCl
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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
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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
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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.
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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.
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Easy way TO REMEMBER
•If you ADD, then move AWAY•If you TAKE, then move TOWARDS
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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
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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
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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
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Equilibrium
• Click to watch an episode of Crash Course about equilibrium • It is very clear and very helpful.
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Solutions and Concentrations
Notes Page 5
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SOLUTIONS
A homogenous mixture of two or more substances
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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
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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
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solUTE over solVENT
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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
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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
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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
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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
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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