CHM 112 - Chapter 16: Equilibrium

Equilibrium

The state where the concentrations of all reactants and products remain constant with time.

Any reaction occurring in a ______ system will achieve an equilibrium.

Closed

Some reactions favor product formation meaning after the reaction...

There are very little reactants remaining. We say this reaction lies far to the right.

Ex. H2(G) 1/2O2(g) → H2O(g)

Some reactions favor the reactant meaning after the reaction...

There is very little formation of product. Sometimes it can be virtually undetectable. We say that this reaction lies far to the left.

Ex. CaO(s) → 2Ca(s) + O2(g)

On a molecular scale for equilibrium...

There is frantic activity. Equilibrium is not static but rather highly dynamic.

Why does equilibrium occur?

Molecules react by collision with each other. More collisions means a faster reaction rate. Reaction rate depends on concentration. There is a greater likelihood of collisions between reactants at higher concentration.

Law of Mass Action

aA + bB → cC + dD

K = [C]^c[D]^d/[A]^a[B]^b or products/reactants

Law of Mass Action Example

4NH3(g) + 7O2 ⇋ 4NO2(g) + 6H2O(g)

K = [NO2]^4[H2O]^6/[NH3]^4[O2]^7

In general, molecules with strong bonds have...

Large activation energies and tend to react very slowly at 25°C.

Another Law of Mass Action Example

T = 127°C
[NH3]eq = 3.1 x 10^-2 mol/L
[N2]eq = 8.5 x 10^-1 mol/L
[H2]eq = 3.1 x 10^-3 mol/L

N2(g) + 3H2(g) ⇋ 2NH3(g)

K = [NH3]eq^2/[N2]eq[H2]eq^3 = (3.1x10^2 mol/L)^2/(8.5x10^-1 mol/L)(3.1x10^-3 mol/L)^3
K = 3.8x10^4 L^2/mol^2

K depends on how equations are...

Written, for example if we were to switch the Haber proccess

2NH3(g) ⇋ N2(g) + 3H2(g)
K' = [N2][H2]^3/[NH3]^2 = 1/K = 1/(3.8x10^4 L^2/mol^2)
K = 2.6 x 10^-5 mol^2/L^2

Example: Multiplying Haber Process by 1/2

1/2 N2(g) + 3/2H2(g) ⇋ NH3(g)
K = [NH3]/([N2]^1/2[H2]^3/2)

Compare with K from the original example
[NH3]/([N2]^1/2[H2]^3/2) = ([NH3]^2/[N2][H2]^3)^1/2
K^1/2 = √(3.8 x 10^4 L^2/mol^2) = 1.9x10^2 L/mol

Since there are an infinite number of initial starting compositions, there are an infinite number of...

Equilibrium positions.

Only one equilibrium ______ for given experimental conditions.

Constant

Equilibrium constants are ______ dependent.

Temperature.

Gas phase reactions can also be expressed as...

Partial pressures (Kp)

C =

P/RT

Partial Pressures (Kp) Example

N2(g) + 3H2(g) ⇋ 2NH3(g)
Kc = CNH3^2/CN2 CH2^3 = (PNH3^2/(RT)^2)/[(PN2/RT)(PH2^3/RT)^3] = PNH3^2/(PN2PH2^3) * (RT)^2

Another Partial Pressures (Kp) Example

H2(g) + F2(g) ⇋ 2HF(g)

Kc = [HF]^2/([H2][F2]) = CHF^2/CH2CF2 = PHF^2/PH2PF2

This time Kp = Kc because there is an equal number of molecules on each side of the equation so RT cancels out.

Kp =

Kc(RT)^Δn

Δn = nfinal-ninitial or products - reactants

Experiments show that position of heterogeneous equilibrium does not depend on the amount of...

Pure solids or liquids present. So previously we would write the lass of mass action as follows:

CaCO3(s) ⇋ CaO(s) + CO2(g)
K = [CO2][CaO]/[CaCo3]

Instead we can generalize as follows:
K = [CO2]C2/C1 = KC2/C1 = [CO2]

If pure solids or liquids are involved in a chemical reaction, their concentrations are not included in the...

Equilibrium expression for the reaction. Liquids and solids are equal to 1.

Pure Solids or Liquids Example

2H2O(l) ⇋ 2H2(g) + O2(g)

K = [H2]^2[O2]
Kp = PO2PH2^2

K equals the

Inherent tendency for a reaction to occur.

K >1

The equilibrium of the reaction system will consist mostly of product. "Equilibrium lies to the right"

K < 1

The value is small, at equilibrium the system will consist mostly of reactant. "Equilibrium lies to the left"

The size of K and the time to reach equilibrium are...

Not directly related. Time deals with reaction rate.

Q =

Reaction Quotient

Reaction Quotient Example

N2(g) + 3H2(g) ⇋ 2NH3(g)
Q = [NH3]0^2/[N2]0[N2]0^3
The 0 means initial concentration
K = 6.0 x 10^-2 L^2/mol^2 @ 500°C

[NH3]0 = 1x10^-3 M, [N2]0 = 1x10^-5 M, [H2]0 = 2x10^-3 M

Q = (1x10^-3mol/L)^2/(1x10^-5mol/L)(2x10^-3mol/L)^3
Q = 1.2 x 10^7 L^2/mol
Q > K
For Q to approach K, the denominator must get larger so
N2+3H2 ← 2NH3 (system shifts left)

Reaction Quotient Example 2

N2(g) + 3H2(g) ⇋ 2NH3(g)

[NH3]0 = 1x10^-4 M, [N2]0 = 5 M, [H2]0 = 1.0x10^-2 M

Q = [NH3]0^2/[N2][H2]0^3 = 2.0 x 10^-3 L^2/mol^2
Q < K
Numerator must get larger to approach K
N2+3H2 → 2NH3 (system shifts right)

Reaction Quotient Example 3

N2(g) + 3H2(g) ⇋ 2NH3(g)

[NH3]0 = 2x10^-4 M, [N2]0 = 1.5 x 10^-5 M, [H2]0 = 3.54x10^-1
Q = [NH3]0^2/[N2][H2]0^3 = 6.01 x 10^-2 L^2/mol^2
K = Q
So the system is at equilibrium

Determining Equilibrium Concentrations (or Pressures) of Reactants and Products Example

N2O4(g) ⇋ 2NO2(g)
Kp = 0.133 atm
At equilibrium PN2O4 = 2.71 atm

Kp = PNO2^2/PN2O4 = 0.133 atm
PNO2^2 = (0.133atm)(2.71atm) = 0.360^2
PNO2 = √(0.360atm^2) = 0.600 atm

LeChatelier's Principle

A "qualitative" approach to predict changes that occur when a system at equilibrium is disturbed. When a change is imposed on a system at equilibrium, the position at equilibrium will shift in a direction that tends to reduce that change. It predicts effects of changes as follows: Pressure, Concentration, and Temperature.

If a reactant or product is added to a system at equilibrium...

The system will shift away from the added component.

If a reactant or product is removed....

The system will shift towards the removed component.

LeChatelier's Principle Example

As4O6(s) + 6C(s) ⇋ As4(g) + 6CO(g)

A. Addition of CO(g)
B. Add or Remove C(s) or As4O6(s)
C. Removal of As4(g)

A. The equilibrium shifts to the left.
B. The amount of pure solid does not affect equilibrium position, so changing these amounts has no effect.
C. The equilibrium position shifts right to form more produce.

Effect of Pressure Change

1. Add or remove gaseous reactants or products.
2. Add inert gas (i.e. not involved in reaction) - there is no effect on equilibrium position
3. Change volume of container (think of a piston)

When container volume changes...

The concentration (and partial pressures) change of all the reactants and products.

When the volume of a container holding a gaseous system is reduced...

The system responds by reducing its own volume. This is done by reducing the total number of gaseous molecules in the system.

Volume Change Example

N2(g) + 3H2(g) ⇋ 2NH3(g)
4 molecules ⇋ 2 molecules
Volume: Decreases

The reaction should shift right, in the direction 4 molecules react to form 2.

The opposite (i.e. volume increases) system shifts so as to...

Increase volume and shift left to make more molecules.

Another Volume Change Example

P4(s) + 6Cl(g) ⇋ 4PCl3(l)

P4 and 4PCl3 are a pure solid and liquid, so only worry about Cl2(g). Kp = 1/[Cl]^6

So if volume decreases, it shifts to less gas molecules on the right.

Volume Change Example with Equal Amounts of Molecules on Both Sides

PCl3(g) + 3NH3(g) ⇋ P(NH2)3(g) + 3HCl(g)
4 gaseous molecules ⇋ 4 gaseous molecules

So change in volume of the system would have no effect, there is no shift in the equilibrium position.

Temperature Change Example

N2(g) + 3H2(g) ⇋ 2NH3(g) ΔH = -92kJ (exothermic reaction)

Write the equation with ΔH as a product or reactant.
N2(g) + 3H2(g) ⇋ 2NH3(g) + 92kJ
Shifts to the left, decreases [NH3]

LeChatelier Principle predicts that the shift will be _____ to the side that consumes energy

Opposite.

Another Temperature Change Example

CaCO3(s) ⇋ CaO(s) + CO2(g) ΔH = +556kJ

556 kJ + CaCO3(s) ⇋ CaO(s) + CO2(g)
Equilibrium position shifts to the right.

What is the value for R? (partial pressure)

0.0821L/atm K