+
Unit D: Equilibrium
Textbook Reference: Chapters 15 + Equilibrium
1.1K: Define equilibrium and state the criteria that apply to a chemical system in equilibrium; i.e., closed system, constancy of properties, equal rates of forward and reverse reactions
1.2K: Identify, write and interpret chemical equations for systems at equilibrium
1.4K: Define Kc to predict the extent of the reaction and write equilibrium-law expressions for given chemical equations, using lowest whole-number coefficients + Chemical Equilibrium 100% Not all reactions are quantitative (reactants products) Evidence: For many reactions reactants are present even after the reaction appears to have stopped
Recall the conditions necessary for a chemical reaction: Particles must collide with the correct orientation and have sufficient energy If product particles can collide effectively also, a reaction is said to be reversible Rate of reaction depends on temperature, surface area and concentration
+ Chemical Equilibrium Consider the following reversible reaction:
The final state of this chemical system can be explained as a competition between:
The collisions of The collisions of reactants to form products to re-form products reactants
We assume this system is closed (so the reactants and products cannot escape) and will eventually reach a:
DYNAMIC EQUILBRIUM
- Opposing changes occur simultaneously at the same rate
+ Modelling Dynamic Equilibrium
Mini Investigation pg. 678 Assume large straw transfers 5 mL each time Volume of Volume of and the smaller straw Cylinder #1 Cylinder #2 transfers 2 mL each time 25.0 0.0 20.0 5.0 17.0 8.0 14.0 11.0 11.0 14.0 8.0 17.0 5.0 20.0 2.0 23.0 2.0 23.0 2.0 23.0 2.0 23.0 + Chemical Equilibrium
Consider the following hypothetical system:
AB + CD AD + BC forward reaction, therefore
AD + BC AB + CD reverse reaction
• Initially, only AB and CD are present. The forward reaction is occurring exclusively at its highest rate.
• As AB and CD react, their concentration decreases. This causes the reaction rate to decrease as well.
• As AD and BC form, the reverse reaction begins to occur slowly.
• As AD and BC’s concentration increases, the reverse reaction speeds up.
• Eventually, both the forward and reverse reaction occur at the same rate = DYNAMIC EQUILIBRIUM
+ 4 Conditions of Dynamic Equilibrium*
1. Can be achieved in all reversible reactions when the rates of the forward and reverse reaction become equal
Represented by rather than by
2. All observable properties appear constant (colour, pH, etc)
3. Can only be achieved in a closed system (no exchange of matter and must have a constant temperature)
4. Equilibrium can be approached from either direction. This means that the equilibrium concentrations will be the same regardless if you started with all reactants, all products, or a mixture of the two Types of Equilibrium
1. Phase Equilibrium: a single substance existing in more than 1 phase Example: Liquid water in a sealed container with water vapour in the space above it Water evaporates until the concentration of water vapour rises to a maximum and then remains constant
2. Solubility Equilibrium: a saturated solution Rate of dissolving = rate of recrystallization Types of Equilibrium
3. Chemical Equilibrium – reactants and products in a closed system Example: The Hydrogen-Iodine Equilibrium System
The rate of reaction of the reactants decreases as the number of reactant molecules decrease. The rate at which the product turns back to reactants increases as the number of product molecules increases. These two rates become equal at some point, after which the quantity of each will not change. + Describing the Position of Equilibrium
1. Percent Yield- the yield of product measured at equilibrium compared with the maximum possible yield of product. % yield = product eq’m x 100 % product max The equilibrium concentration is determined experimentally, the maximum concentration is determined with stoichiometry
+ Describing the Position of Equilibrium
1. Percent Yield – Example If 2.50 mol of hydrogen gas reacts with 3.0 mol of iodine gas in a 1.00L vessel, what is the percent yield if 3.90 mol of hydrogen iodide is present at equilibrium
% yield = product eq’m x 100 % product max
+ Describing the Position of Equilibrium
2. Using an Equilibrium Constant, (Kc)
This relationship only works if all concentrations are at equilibrium at a constant temperature in a closed system
Think “products over reactants”
If the Kc > 1, the equilibrium favours products
If the Kc < 1, the equilibrium favours reactants
+ Describing the Position of Equilibrium
2. Using an Equilibrium Constant, (Kc)
Example #1: Write the equilibrium law expression for the reaction of nitrogen monoxide gas with oxygen gas to form nitrogen dioxide gas.
+ Describing the Position of Equilibrium
2. Using an Equilibrium Constant, (Kc)
Note: The Kc value describes the extent of the forward reaction.
Kc reverse = 1 . = The reciprocal value
Kc forward
Example #2: The value of Kc for the formation of HI(g) from H2(g) and I2(g) is 40, at a given temperature. What is the value of Kc for the decomposition of HI(g) at the same temperature.
+ Describing the Position of Equilibrium
2. Using an Equilibrium Constant, (Kc) Note: For heterogeneous equilibrium systems, DO NOT include liquids and solids in the expression. (They are assumed to have fixed concentrations)
Example #3: Write the equilibrium law expression for the decomposition of solid ammonium chloride to gaseous ammonia and gaseous hydrogen chloride
Example #4: Write the equilibrium law expression for the reaction of zinc in copper(II) chloride solution.
+ Describing the Position of Equilibrium + PRACTICE
Read Pages 676-680 Do Page 682 #3, 4, 5
Read Pages 684-686 Do Page 688 #1, 4, 6 + Equilibrium Concentrations
2.3K: Calculate equilibrium constants and concentrations for homogeneous when • concentrations at equilibrium are known • initial concentrations and one equilibrium concentration are known • the equilibrium constant and one equilibrium concentration are known. + Calculations in Equilibrium Systems
Using the equilibrium law expression to determine whether a system is at equilibrium:
Substitute in the given concentrations to the equilibrium expression. If the value is the equilibrium constant, the system is at equilibrium
If the value is larger, this means there are more products that reactants. To reach equilibrium, the reaction must proceed to the left (towards the reactants)
If the value is smaller, this means there are more reactants than products. To reach equilibrium, the reaction must proceed to the right (towards the products)
+ Calculations in Equilibrium Systems
Example #1: In the following system:
N2(g) + 3H2(g) ↔ 2NH3(g)
-2 -4 0.249 mol N2(g), 3.21 X 10 mol H2(g) and 6.42 X 10 mol NH3(g) o are combined in a 1.00 L vessel at 375 C, Kc = 1.2
Is the system at equilibrium?
If not, predict the direction in which the reaction must proceed.
+ Calculations in Equilibrium Systems
Example #2: Find the equilibrium concentration of the ions that are formed when solid silver chloride is dissolved in water. The equilibrium constant for this reaction is -4 Kc = 5.4 X 10 .
+ - AgCl(s) Ag (aq) + Cl (aq)
+ ICE Charts and Equilibrium Calculations
STEPS:
Always write out the equilibrium reaction and equilibrium law expression if not given.
Draw an ICE Chart (Initial, Change in and Equilibrium concentrations) (I + C = E)
Substitute values where appropriate
Solve for x
Solve for equilibrium concentrations + ICE Charts and Equilibrium Calculations
Example #1: Consider the following equilibrium at 100 oC:
N2O4(g) ↔ 2 NO2(g)
2.0 mol of N2O4(g) was introduced into an empty 2.0 L bulb. After equilibrium was established, only 1.6 mol of N2O4(g) remained. What is the value of Kc?
N2O4(g) 2NO2(g) I:
C: E:
+ ICE Charts and Equilibrium Calculations Example #2
A 10 L bulb is filled with 4.0 mol of SO2(g), 2.2 mol of O2(g) and 5.6 mol of SO3(g). The gases then reach equilibrium according to the following equation:
2SO2(g) + O2(g) ↔ 2SO3(g) At equilibrium, the bulb was found to contain 2.6 mol of SO2(g). Calculate Kc for this reaction.
2SO2(g) O2(g) 2SO3(g) I:
C:
E: E:
+ PRACTICE
Page 682 # 6
Page 688 # 7-10
+ ICE Charts and Equilibrium Calculations
Example #3: Using a perfect square
Given the following reaction:
N2(g) + O2(g) ↔ 2NO(g) Kc = 0.00250 Determine the equilibrium concentrations for all species present given that the initial concentration of each reactant is 0.200 mol/L.
N O 2NO 2(g) 2(g) (g) I: C: E:
E:
+ ICE Charts and Equilibrium Calculations
Example #4: Using the Approximation Rule
Calculate the concentration of gases produced when 0.100 mol/L COCl2(g) decomposes into carbon monoxide and chlorine gas. The Kc for this reaction is 2.2 X 10-10.
COCl2(g) CO(g) Cl2(g) I:
C:
E: E:
+ PRACTICE
Handout: Practice Problems – Using Stoichiometry to Calculate Kc #21-24
Handout: Practice Problems – Using the Approximation Method #25-29 + Le Chatelier’s Principle
1.3K: Predict, qualitatively, using Le Chatelier’s principle, shifts in equilibrium caused by changes in temperature, pressure, volume, concentration or the addition of a catalyst and describe how these changes affect the equilibrium constant + Le Chatelier’s Principle
Le Chatelier’s Principle is very useful in predicting how a system at equilibrium will respond to change.
It states that when a system at equilibrium is disturbed, the equilibrium shifts in the direction that opposes the change, until a new equilibrium is reached.
There are three common ways an equilibrium may be disturbed: Change in the concentration of one of the reactants or products Change in the temperature Changes in the temperature, will Change in the volume of a container change the Kc value. No other Addition of a catalyst (less common) changes will affect this value. + Le Chatelier’s Principle
Effect of Changes in Concentration If a system at equilibrium is disturbed by the addition of a reactant (or the removal of a product), Le Chatelier’s principle predicts that the equilibrium will shift right.
2N2O(g) + 3O2(g) 4 NO2(g)
If the disturbance is the removal of a reactant (or the addition of a product), Le Chatelier’s principle predicts that the equilibrium will shift left.
2N2O(g) + 3O2(g) 4 NO2(g)
Since concentrations of solids are constants and do not appear in expressions for K, removing or adding some solid does not cause shifts. If you see spike in either a reactant or + product, which causes a gradual change Concentration Change in the other entities – a substance has been added or removed.
Addition of reactant, HF(g) – Removal of product, HCl(g), will shift equilibrium to the will shift the equilibrium to the right right
More products will be More products will be produced, and a new produced, and a new equilibrium is established equilibrium is established + Le Chatelier’s Principle
Effect of Changes in Temperature In endothermic reaction, heat acts like a reactant. Increasing the temperature shifts the reaction right. Decreasing the temperature, shifts the reaction left
Heat + 2N2O(g) + 3O2(g) 4 NO2(g)
In exothermic reactions, heat acts like a product. Increasing the temperature shifts the reaction left. Decreasing the temperature, shifts the reaction right.
4 NO2(g) 2N2O(g) + 3O2(g) + Heat Remember K gets bigger if there are more products being created (i.e. shifts to the right) The equilibrium constant, Kc is temperature dependent
Reaction Type Role of heat Effect of T Effect of T Endothermic Reactants + heat products K K Exothermic Reactants products + heat K K + Temperature Change
The temperature decreases, at the time indicated by the dotted line.
This will cause the equilibrium to shift to the right, creating more products, until a new equilibrium is established.
+ energy
If you see a gradual change in the reactants with an opposite change in the products, you have a temperature change going on + Le Chatelier’s Principle
Effect of Changes in the Volume of the Container
If volume is decreased, pressure increases (Boyle’s Law – Chem 20) So the reaction will shift in the direction which contains the fewest moles of gas
Pressure 2N2O(g) + 3O2(g) 4 NO2(g) 4 moles are fewer than 5
If volume is increased, pressure decreases (Boyle’s Law – Chem 20) So the reaction will shift in the direction which contains the most moles of gas
Pressure 2N2O(g) + 3O2(g) 4 NO2(g) 5 moles are more than 5
If both sides of the equation have the same number of moles of gas, the change in volume of the container has no effect on the equilibrium. + Volume Change
The volume of the container decreased, at the time indicated by the dotted line. This will cause a pressure increase.
This will cause the equilibrium to shift to the right, the side with fewer moles of gas, creating more products, until a new equilibrium is established.
If you see a spike in all of the entities = P increase, V decrease
If you see a drop in all of the entities = P decrease, V increase + Le Chatelier’s Principle
Effect of the Addition of a Catalyst
Catalysts speed up the rate at which equilibrium is obtained, but have no effect on the magnitude of K. They increase both the forward and backward rate of reaction. +
+ Example: Identify the nature of the changes imposed on the following equilibrium system at the four times indicated by coordinates A, B, C and D + Practice
Change Direction of Shift (, , Effect on quantity Effect (increase, decrease, or no change) or no change)
a) Decrease in volume Kc
b) Raise temperature Amount of CO(g)
c) Addition of I2O5(s) Amount of CO(g)
d) Addition of CO2(g) Amount of I2O5(s)
e) Removal of I2(g) Amount of CO2(g) + Practice
Read Pages 690-695
Questions: Page 695 #1-3 Page 699 #1-7