Chemical Equilibrium

• Chemical equilibrium is the point at which the concentrations of all species are constant and the rates of the forward reaction and reverse reaction are equal.

Start with NO₂

Start with N₂O₄

Start with NO₂ & N₂O₄

K >> 1

K << 1

Lie to the right

Favor products

Lie to the left

Favor reactants

Equilibrium Will

1. Write a reaction for the following. �2. Write the equilibrium expression. Is the equilibrium constant greater than or less than 1?

1. Write a reaction for each of the following. �2. Calculate Kc. Does the reaction favor reactants or products?

Homogeneous Equilibrium

• Applies to reactions in which all reacting species are in the same phase

​

​

​

​

​

​

K_c - when all species are in solution- concentration (molarity) is used to determine the equilibrium constant

​

K_p- when all species are gases- partial pressure is used to determine the equilibrium constant

​

• K_c does not equal K_p

Converting between K_c and K_p

K_p = K_c (RT)^Δn

​

R = 0.08206 mol/L atm

T= temperature (in K)

Δ n =the change in the number of moles of gas

Δ n = moles of gas products- moles gas reactant

​

If Δn = 0, then K_p= K_c

Example 1

• Carbonyl chloride (COCl₂), also called phosgene, was used in WWI as a poisonous gas. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form carbonyl chloride at 74°C are [CO] = 1.2 x 10^-2M, [Cl₂]= 0.054 M and [COCl₂]= 0.14 M. Calculate the equilibrium constants, K_c and K_p for this reaction. Does this reaction favor reactants or products?

​

CO (g) + Cl₂ (g) COCl₂ (g)

​

​

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.

[CaCO₃] = constant

[CaO] = constant

K_c = [CO₂]

The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

More About K_eq

​

K_eq(forward) = 1/K_eq(reverse)

​

​

For example:

​

At 100 ºC, K_eq(forward) = 6.49

At 100 ºC, K_eq(reverse) = 1/6.49 = 0.154

​

​

​

​

​

​

​

​

The value of K depends on the balanced equation

• The stoichiometry of a reaction that has been multiplied by a number changes the equilibrium constant. K_eq gets raised to the power equal to that number.

​

​

• For example: 2 4

​

• K_eq = (6.49)² = 42.12

​

Multiple equilibrium

• The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

For example: A + B → X + C K_eq(1) = 2.0

X + B → D K_eq(2) = 5.0

A + 2B → C + D K_eq = K₁ x K₂ = 10.0

​

​

Practice Problems

Chapter 15: 1, 4, 19, 21, 23, 27, 29

Warm- up

From the following equilibrium constant expression, write a balanced chemical equation for the gas-phase reaction.

​

​

​

If the value of K_c for the forward reaction is 6, what is the value of the reverse reaction?

​

​

​

WHAT DOES THE EQUILIBRIUM CONSTANT TELL US?

The reaction quotient (Q_c) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (K_c) expression.

IF

• Q_c > K_c system proceeds from right to left to reach equilibrium
• Q_c = K_c the system is at equilibrium
• Q_c < K_c system proceeds from left to right to reach equilibrium

How do we establish equilibrium when all species are present initially?

• http://introchem.chem.okstate.edu/DCICLA/ERGBN.htm

Example

• The equilibrium constant (K_c) for the formation of nitrosyl chloride from nitric oxide and molecular chlorine is 6.5 x 10⁴ at 35°C.

2NO (g) + Cl₂ (g) 2NOCl (g)

​

In a certain experiment, 0.002 mole of NO, 0.0083 mole of Cl₂ and 6.8 moles of NOCl are mixed in a 2.0L flask. In which direction will the reaction proceed to reach equilibrium?

Changes in Equilibrium

• Le Chatelier’s Principle: If a stress is applied to a system that is already at equilibrium, the equilibrium will shift to reduce the effect of the stress.
• The position of the equilibrium will change but the value of the equilibrium constant will remain the same.

​

• Stresses
• Changing concentrations
• Effects of volume and pressure
• Effect of temperature
• Effect of catalyst

​

Concentration Changes�N₂(g) + 3 H₂ (g) ⇌ 2 NH₃ (g)

By looking at the graph, what is the effect on the concentration of N₂ and NH₃ when more H₂ is added to the system? Which direction did the equilibrium shift?

N₂(g) + 3 H₂ (g) ⇌ 2 NH₃ (g)

• What is the effect on if the concentration of NH₃ (g) is increased? Decreased?

​

​

• What is the effect if the concentration of N₂ is increased? Decreased?

Other things to think about

• Solids and liquids are not found in the equilibrium expression and therefore will not shift the equilibrium in either direction.
• There are a lot of “clever” ways to remove things from a solution
• Precipitation
• Addition of a drying agent
• Addition of acid or base

Effect of Volume and Pressure Changes

• Reducing the volume (increasing pressure) of a gaseous equilibrium mixture causes the system to shirt in the direction that reduces the number of moles of gas
• Increasing the volume (decrease pressure) of the system causes a shift in the direction that produces more gas molecules.
• Adding an inert gas does not effect the equilibrium

• What happens to the equilibrium of the following systems if the volume is increased?

​

• 2 SO₂ (g) + O₂ (g) ⇌ 2 SO₃ (g)

​

• H₂ (g) + I₂ (g) ⇌ 2 HI (g)

Effect of temperature

• Endothermic reaction:

Reactants + heat ⇌ products

​

• Exothermic reaction:

Reactants ⇌ heat + products

​

• When the temperature of a system at equilibrium is increases, it is as if we have added a reactant or product (heat) . The equilibrium shifts in the direction that consumes the excess reactant or product (heat)

• The position of the equilibrium and the value of K is changed when the temperature is changes.

​

• Endothermic: Increasing T, increases K
• Exothermic: Increasing T, decreased K

Effect of a catalyst

• A catalyst increases the rate at which equilibrium is achieved but has NO EFFECT on the equilibrium composition.