Chemical Equilibrium

Chemical Equilibrium

In systems that are in equilibrium, reverse processes are happening at the same time and at the same rate.

Rate forward = Rate reverse This is called “dynamic equilibrium,”

because there is no net change, yet individual molecules are constantly changing.

Chemical Equilibrium

Example: evaporation occurs at the surface of a liquid.

Rateevaporation = Ratecondensation

Reverse Reactions

A reaction in which the products can react to re-form the reactants is called a reversible reaction.

Reversible Reactions Example:

N2O4(g) → 2 NO2(g)

And

2 NO2(g) → N2O4(g)

Reversible Reactions Example: N2O4(g) ⇌ 2 NO2(g)

No N2O4

InitiallyNo NO2

InitiallyBoth present

initially

Reversible Reactions Notice that in all cases, there is more N2O4

than NO2 at equilibrium.

We can say N2O4 is favored at equilibrium

Reversible Reactions Notice that in all cases, there is more N2O4

than NO2 at equilibrium.

We can say N2O4 is favored at equilibrium

Equilibrium Constant Expressions Whether reactants or products are favored

at equilibrium is quantified by the equilibrium constant, K.

When K > 1, products are favored. When K < 1, reactants are favored. For N2O4(g) ⇌ 2 NO2(g)

K = 4.6 x 10-3 or 0.0046

4.6 x 10-3 < 1, therefore reactants (N2O4) favored at equilibrium

PracticeFor the following reactions, decide if

reactants or products are favored.

1. H2(g) + I2(g) ⇌ 2 HI(g) K = 0.11 at

425 °C

2. H2(g) + I2(g) ⇌ 2 HI(g) K = 2.40 at

700 °C

3. N2(g) + 3 H2(g) ⇌ 2 NH3(g) K = 0.286 at

500 °C

4. CO2(g) + H2(g) ⇌ CO(g) + H2O(g) K = 4.26 at 650 °C

reactants

reactants

products

products

Equilibrium Constant Expressions The equilibrium constant K is a ratio of

products to reactants raised to their stoichiometric coefficients.

For the reaction aA + bB ⇌ cC + dD

[C]c[D]d

[A]a[B]bK =

Note: • products over reactants• [ ] means molarity

Equilibrium Constant Expressions K is a constant, so it is the same at equilbrium

no matter the starting conditions!

N2O4(g) ⇌ 2 NO2(g) K = 4.6 x 10-3

K = = 4.6 x 10-3

[N2O4]

[NO2]2 K = = 4.6 x 10-3

[N2O4]

[NO2]2

K = = 4.6 x 10-3

[N2O4]

[NO2]2

Equilibrium Constant Expressions When writing equilibrium constant

expressions, ONLY gases and aqueous species appear in the expressions.

Liquids and Solids do NOT appear in the K expressions!

P4(s) + 4 O2(g) ⇌ 2 P2O4(g)

[P2O4]2

[O2]4K =

[P2O4]2

[P4][O2]4K =

Equilibrium Constant Expressions Given a balanced reaction and the

concentrations of all species at equilibrium, one can calculate the value of the equilibrium constant K.

Write the K expression, and substitute the equilibrium concentrations where appropriate.

Equilibrium Constant Expressions 2 SO2(g) + O2(g) ⇌ 2 SO3(g) At 600

°C, the equilibrium concentrations were found to be:– [SO2] = 1.50 M

– [O2] = 1.25 M

– [SO3] = 3.50 M

What is the value of the equilibrium constant K at 600 °C?

Equilibrium Constant Expressions 2 SO2(g) + O2(g) ⇌ 2 SO3(g)

[SO2] = 1.50 M [O2] = 1.25 M [SO3] = 3.50 M

1. Write the equilibrium constant expression.

2. Substitute the equilibrium concentrations and solve.

Practice N2(g) + O2(g) ⇌ 2 NO(g) at 1500 K

At equilibrium:– [N2] = 6.4 x 10-3 M

– [O2] = 1.7 x 10-3 M

– [NO] = 1.1 x 10-5 M

What is the value of the equilibrium constant K at 1500 K?

End of

Day 1

Shifting Equilibria An equilibrium system can be

manipulated to proceed forward or in reverse.

This is summarized by Le Châtelier’s Principle:– If a stress is applied to a system at equilibrium,

the system will react in the direction (forward or in reverse) that relieves the stress.

Shifting Equilibria Possible stresses:

–Changing concentrations of reactants or products

–Changing the pressure–Changing the temperature

Shifting Equilibria Collision Theory:

– In order for a reaction to happen, molecules must collide with enough energy to react.

– In an equilibrium system, both the forward and the reverse reaction can happen depending on the conditions.

– When the forward rate exceeds the reverse rate, the reaction will proceed forward.

– When the reverse rate exceeds the forward rate, the reaction will proceed in reverse.

Changing ConcentrationA + B ⇌ C + D

An increase in [A] will increase the number of collisions of A with B, and the reaction will shift right (forward direction)

Similarly, an increase in [C] will increase the number of collisions of C with D, and the reaction will shift left (reverse direction)

Changing ConcentrationA + B ⇌ C + D

A decrease in [A] will decrease the number of collisions of A with B, and the reaction will shift left (reverse direction)

Similarly, a decrease in [C] will decrease the number of collisions of C with D, and the reaction will shift right (forward direction)

Changing ConcentrationA + B ⇌ C + D

Changing the concentrations do NOT affect the value of the equilibrium constant K

The ratio of products to reactants will be the same at equilibrium, even if the values of concentrations change during the shift.

Changing ConcentrationN2(g) + 3 H2(g) ⇌ 2 NH3(g)

• After adding NH3, the reaction proceeds in reverse to reestablish equilibrium.

• NH3 is consumed.

• H2 and N2 are created.

Changing PressureA(g) ⇌ 2B(g)

A change in pressure will only affect equilibria in which gases are involved.

For a change in pressure to have an effect, the total moles of gas on the reactants side must be different from the total moles of gas on the products side.

Changing PressureA(g) ⇌ 2B(g)

Increasing pressure will create a shift toward the side that has fewer total moles of gas.

By decreasing the total moles of gas, the pressure stress will be relieved.

Changing PressureA(g) ⇌ 2B(g)

Increasing pressure will shift the reaction to the left.

B will be consumed, A will be created, and the pressure will be relieved.

Changing PressureA(g) ⇌ 2B(g)

Decreasing pressure will create a shift toward that side that has more total moles of gas.

By increasing the total moles of gas, the pressure stress will be relieved.

Changing PressureA(g) ⇌ 2B(g)

Decreasing pressure will shift the reaction to the right.

B will be created, A will be consumed, and the pressure will be relieved.

Changing Pressure

Increase pressure, shifts left

Changing liquids and solids Changing the amount of liquid or solid in

an equilibrium system will have NO EFFECT on the equilibrium position, because liquids and solids do NOT appear in the equilibrium constant K expression.

Changing Temperature Reversible reactions are exothermic in

one direction and endothermic in the opposite direction.

That is, they give off heat in one direction and absorb heat in the opposite direction.

Changing Temperature According to Le Châtelier’s principle,

adding heat (increasing the temperature) will create a shift so that heat is absorbed.

This favors the endothermic reaction. Removal of heat favors the exothermic

reaction.

Changing Temperature Treat heat like you would a reactant or

product, and think about it like changing concentration shifts.

Example: CO + 2 H2 ⇌ CH3OH is exothermic (gives off heat)

Rewrite: CO + 2 H2 ⇌ CH3OH + heat– Increasing temperature is adding

heat, will shift the reaction to the left.– Decreasing temperature is removing

heat, will shift the reaction to the right.

Practice Consider the decomposition of

calcium carbonate: CaCO3(s) ⇌ CaO(s) + CO2(g)

endothermic

How would the following stresses shift the equilibrium?

1. Increase [CO2]

2. Increase Pressure3. Increase temperature4. Increase the mass of CaCO3

L R

L RL R

no shift