Reaction Rates and Equilibrium CHAPTER 12 Reaction Rates and Equilibrium 12.2 Chemical Equilibrium

CITY traffic in traffic out There is a balance or equilibrium between cars that go into the city and cars that leave the city

Concept of equilibrium Physical equilibrium There is a balance or equilibrium between cars that go into the city and cars that leave the city Physical equilibrium

Concept of equilibrium Physical equilibrium Chemical equilibrium There is also a “balance” in a chemical system Reactants Products N2(g) + 3H2(g) 2NH3(g) Chemical equilibrium Physical equilibrium

Concept of equilibrium Physical equilibrium Chemical equilibrium Le Châtelier’s principle Effect of temperature Effect of concentration Effect of pressure / volume Reactants Products N2(g) + 3H2(g) 2NH3(g) Chemical equilibrium Products Reactants

Concept of equilibrium Physical equilibrium Chemical equilibrium Le Châtelier’s principle Effect of temperature Effect of concentration Effect of pressure / volume The equilibrium expression Calculating the equilibrium constant, K Determining where the equilibrium lies Reactants Products N2(g) + 3H2(g) 2NH3(g) Chemical equilibrium

Physical equilibrium H2O(l) H2O(g) An equilibrium between two phases of water: H2O(l) H2O(g)

Physical equilibrium H2O(l) H2O(g) An equilibrium between two phases of water: H2O(l) H2O(g)

H2O(l) H2O(g) An equilibrium between two phases of water: physical equilibrium: a “balance” in a physical system where there is an equilibrium between two or more phases at the same time.

Physical equilibrium Pouring 50 mL from A into B Pouring 50 mL from B into A It looks like nothing is happening, but there is in fact a dynamic equilibrium 50 mL 50 mL 50 mL 50 mL

and the rate of transfer are the same from both sides Physical equilibrium Equilibrium does not mean that there must be equal amounts on each side It is an equilibrium when the level of neither tank changes over time. This occurs when the amount transferred and the rate of transfer are the same from both sides 50 mL 50 mL 50 mL 50 mL

Physical equilibrium The illustration below shows an example of . non-equilibrium 25 mL 50 mL 50 mL 25 mL

Physical equilibrium The illustration below shows an example of . non-equilibrium 25 mL 50 mL 50 mL 25 mL

Chemical equilibrium N2O4(g) 2NO2(g) a closed system: clear red-brown a closed system: no exchange of matter with the surroundings; both reactant and product are present as a mixture Depending on the color of the gas, we can tell whether there is more reactant (clear) or more product (red-brown)

Chemical equilibrium N2O4(g) 2NO2(g) clear red-brown more reactant more product

Chemical equilibrium In each case, an equilibrium has been established. (Remember: amounts of reactants and products don’t have to be equal.) N2O4(g) 2NO2(g) clear red-brown more reactant more product

Chemical equilibrium N2O4(g) 2NO2(g) Nitrogen oxide (NO2) is a poisonous gas responsible for respiratory problems! N2O4(g) 2NO2(g) clear red-brown The smoggy skyline of Los Angeles

an equilibrium is reached Chemical equilibrium N2O4(g) 2NO2(g) clear red-brown Start: Only N2O4 is present Over time, an equilibrium is reached End: The concentrations of N2O4 and NO2 are constant

an equilibrium is reached Chemical equilibrium N2O4(g) 2NO2(g) clear red-brown Start: Only N2O4 is present Over time, an equilibrium is reached End: The concentrations of N2O4 and NO2 are constant

Notice that the equilibrium concentrations of N2O4 and NO2 are the same in cases A and B N2O4(g) 2NO2(g) N2O4 NO2 N2O4 NO2

The equilibrium favors the reverse reaction At equilibrium, there is more N2O4 (reactant) than NO2 (product) The equilibrium favors the reverse reaction N2O4(g) 2NO2(g) N2O4 NO2 equilibrium position: the favored direction of a reversible reaction, determined by each set of concentrations for the reactant(s) and product(s) at equilibrium.

From this graph of concentrations versus time, which side of the reaction is favored? 2SO2(g) + O2(g) 2SO3(g)

The forward reaction is favored From this graph of concentrations versus time, which side of the reaction is favored? 2SO2(g) + O2(g) 2SO3(g) The forward reaction is favored Higher concentration

Temperature has an effect on the equilibrium Concept of equilibrium Physical equilibrium Chemical equilibrium Le Châtelier’s principle Effect of temperature Effect of concentration Effect of pressure/ volume N2O4(g) 2NO2(g) more product more reactant Temperature has an effect on the equilibrium

Le Châtelier’s principle If an equilibrium has been established… N2(g) + 3H2(g) 2NH3(g) You no longer have an equilibrium! … what happens when two NH3 molecules are removed?

Le Châtelier’s principle If an equilibrium has been established… N2(g) + 3H2(g) 2NH3(g) … what happens when two NH3 molecules are removed? The forward reaction is favored so there is more product You no longer have an equilibrium! The equilibrium is re-established

Favored reaction If removed N2(g) + 3H2(g) 2NH3(g) Le Châtelier’s principle: principle that states that when a “change” is made to a system at equilibrium, the system will shift in a direction that partially offsets the “change.”

Temperature and equilibrium In an endothermic reaction, energy is considered a reactant N2O4(g) + 58.0 kJ/mole 2NO2(g) clear red-brown Raising the temperature is equivalent to adding more reactants The equilibrium shifts to the right More NO2 (red-brown gas) is produced

Temperature and equilibrium In an endothermic reaction, energy is considered a reactant N2O4(g) + 58.0 kJ/mole 2NO2(g) clear red-brown Lowering the temperature is equivalent to having less reactants The equilibrium shifts to the left More N2O4 (clear gas) is produced

Concentration and equilibrium less of it 2SO2(g) + O2(g) 2SO3(g) reactants product In a reversible equilibrium reaction, forming product means “using up” or consuming reactants More product means less of the reactants.

Concentration and equilibrium 2SO2(g) + O2(g) 2SO3(g) reactants product shifts left In a reversible equilibrium reaction, forming product means “using up” or consuming reactants

Forming reactants means consuming some of the products Concentration and equilibrium 2SO2(g) + O2(g) 2SO3(g) reactants product In a reversible equilibrium reaction, forming product means “using up” or consuming reactants shifts left less of it More of the reactants means less of the product. 2SO2(g) + O2(g) 2SO3(g) reactants product Forming reactants means consuming some of the products

Forming reactants means consuming some of the products Concentration and equilibrium 2SO2(g) + O2(g) 2SO3(g) reactants product In a reversible equilibrium reaction, forming product means “using up” or consuming reactants shifts left 2SO2(g) + O2(g) 2SO3(g) reactants product shifts right Forming reactants means consuming some of the products

Concentration and equilibrium Using the equilibrium system predict the direction in which the system will shift as a result of an increase in the concentration of Cl2. PCl5(g) PCl3(g) + Cl2(g)

Concentration and equilibrium Using the equilibrium system predict the direction in which the system will shift as a result of an increase in the concentration of Cl2. PCl5(g) PCl3(g) + Cl2(g) Asked: Direction that the system will shift Given: Increase in Cl2 Relationships: Note that Cl2 is a product on the right side of the equation.

Concentration and equilibrium Using the equilibrium system predict the direction in which the system will shift as a result of an increase in the concentration of Cl2. PCl5(g) PCl3(g) + Cl2(g) Asked: Direction that the system will shift Given: Increase in Cl2 Relationships: Note that Cl2 is a product on the right side of the equation. Solve: The system will shift toward the reactant to consume some of the added Cl2.

Concentration and equilibrium Using the equilibrium system predict the direction in which the system will shift as a result of an increase in the concentration of Cl2. PCl5(g) PCl3(g) + Cl2(g) Asked: Direction that the system will shift Given: Increase in Cl2 Relationships: Note that Cl2 is a product on the right side of the equation. Solve: The system will shift toward the reactant to consume some of the added Cl2. Answer: Shifts left to produce some PCl5 Discussion: The shift in direction partially offsets the increase in Cl2.

The number of particles in the system is unchanged Pressure and equilibrium Pressure or volume only affect gaseous equilibrium systems Higher pressure Smaller volume Lower pressure Larger volume The number of particles in the system is unchanged

Pressure and equilibrium 2SO2(g) + O2(g) 2SO3(g) 2 moles 1 mole 2 moles 3 moles 2 moles Reactants take up more space (larger volume) than the product If we increase the pressure, the volume will decrease The equilibrium will shift to the right

Pressure and equilibrium 2SO2(g) + O2(g) 2SO3(g) 2 moles 1 mole 2 moles 3 moles 2 moles Reactants take up more space (larger volume) than the product If we decrease the pressure, the volume will increase The equilibrium will shift to the left

Le Châtelier’s principle helps to determine where the equilibrium lies when the system undergoes a change in: Temperature Concentration Pressure or volume (for gaseous systems) Le Châtelier’s principle: principle that states that when a “change” is made to a system at equilibrium, the system will shift in a direction that partially offsets the “change.”

The equilibrium position depends on a ratio between products and reactants, called the equilibrium expression: aA + bB cC + dD raise to the power of coefficients Equilibrium constant [A] means “molarity of A”

aA + bB cC + dD Write the equilibrium expression for the following reaction: CH4(g) + 2H2S(g) CS2(g) + 4H2(g) Answer: Molarities of products Molarities of reactants

Experimental results for: N2(g) + 3H2(g) 2NH3(g) at 500oC

Experimental results for: N2(g) + 3H2(g) 2NH3(g) at 500oC

Experimental results for: N2(g) + 3H2(g) 2NH3(g) at 500oC K is constant at the same temperature

2H2(g) + O2(g) 2H2O(l) K = 1.4 x 1083 at 25oC Large K favors product CaCO3(s) CaO(s) + CO2(g) K = 1.9 x 10–23 at 25oC Small K favors reactant

Calculate the equilibrium concentration of hydrogen iodide [HI], given the following information: K = 50 at 450oC [H2] = 0.22 M, and [I2] = 0.22 M H2(g) + I2(g) 2HI(g)

Calculate the equilibrium concentration of hydrogen iodide [HI], given the following information: K = 50 at 450oC [H2] = 0.22 M, and [I2] = 0.22 M H2(g) + I2(g) 2HI(g) Answer: Calculate the concentration of HI at equilibrium Given: K = 50 at 450oC, [H2] = [I2] = 0.22 M Relationships:

Calculate the equilibrium concentration of hydrogen iodide [HI], given the following information: K = 50 at 450oC [H2] = 0.22 M, and [I2] = 0.22 M H2(g) + I2(g) 2HI(g) Answer: Calculate the concentration of HI at equilibrium Given: K = 50 at 450oC, [H2] = [I2] = 0.22 M Relationships: Solve:

Calculate the equilibrium concentration of hydrogen iodide [HI], given the following information: K = 50 at 450oC [H2] = 0.22 M, and [I2] = 0.22 M H2(g) + I2(g) 2HI(g) Answer: Calculate the concentration of HI at equilibrium Given: K = 50 at 450oC, [H2] = [I2] = 0.22 M Relationships: Solve: Answer: The equilibrium concentration of [HI] is 1.56 M

Calculate the equilibrium concentration of hydrogen iodide [HI], given the following information: K = 50 at 450oC [H2] = 0.22 M, and [I2] = 0.22 M H2(g) + I2(g) 2HI(g) Answer: Calculate the concentration of HI at equilibrium Relationships: Solve: Answer: The equilibrium concentration of [HI] is 1.56 M Discussion: Here we used a known equilibrium constant (K) to solve for the unknown equilibrium concentration of the product.

Le Châtelier’s principle helps to determine where the equilibrium lies when the system undergoes a change in: Temperature Concentration Pressure or volume (for gaseous systems) The equilibrium constant helps to determine where the equilibrium lies: Large K favors products Small K favors reactants aA + bB cC + dD raise to the power of coefficients Equilibrium constant [A] means “molarity of A”