General Equilibrium: A Review.

From thermodynamics:

All chemical process proceed to an equilibrium at which point there exists an equilibrium constant:

K = P a(i)m(i)

(K has no units: activities are defined as ratios without units)

General Equilibrium: A Review.

From thermodynamics:

All chemical process proceed to an equilibrium at which point there exists an equilibrium constant:

K = P a(i)m(i)

(K has no units: activities are defined as ratios without units)

In practice:

Experimentally, the equilibrium constant is defined only when the activities can be measured. When all components are ideal gases or all components are ideal solutions two other K values can be defined.

Kc is defined as P [i]m(i) MSm(i)

(note that Kc has units which depend on the process.)

Kp is defined as P P(i)m(i) kPaSm(i)

(note that Kp has units which depend on the process and on the pressure units chosen.)

These K values must be converted to the activities K for use in the equation: DT = -RT ln K.

Units of K values.

For the process

6 H2O(g) + 4 NO2(g) ľ 7 O2(g) + 4 NH3(g)

define K, Kp, and Kc. Interconversion of Kp and Kc values.

If all components are ideal gases, then there is a relationship between Kp and Kc:

Kp = P P(i)m(i) kPaSm(i)

For an ideal gas: PV = nRT

or

P(i) = {n/V}RT = [i]RT

Substituting:

Kp = P {[i]RT}Sm(i) MSm(i)

= Kc {RT}Sm(i) MSm(i)

Note: when Sm(i) = 0, all K values are identical, and there are no units for any of them.

Example: for the above reaction:

6 H2O(g) + 4 NO2(g) ľ 7 O2(g) + 4 NH3(g)

Kp = Kc {RT}Sm(i) kPaSm(i)

Kp = Kc*{RT}+1 kPa

Manipulating Equilibrium Constants.

In thermodynamics, we have learned that whatever you do to a chemical equation, you do the same to the intensive thermodynamic parameters.

 Reverse an equation - negate the thermodynamic parameter Double an equation - double the thermodynamic parameter. Add two equations - add thermodynamic parameters.

Thermodynamic parameters for which these rules apply:

DE, DH, DS, DG.

For equilibrium purposes, simply remember that the thermodynamic parameter is ln K which is related to DG.

If an equation is reversed, ln K is negated, K is inverted.

If an equation is doubled, ln K is doubled, K is squared.

If equations are added, ln K values are added, the K values are multiplied.

Examples:

At a certain temperature,

for N2(g) + 2 O2(g) ľ 2 NO2(g) K = 1 x 10-5

and for 2 NO2(g) ľ N2O4(g) K = 1 x 104

Calculate K for the following:

• ˝ N2(g) + O2(g) ľ NO2(g)

• N2O4(g) ľ 2 NO2(g)

• NO2(g) ľ ˝ N2(g) + O2(g)

• N2O4(g) ľ N2(g) + 2 O2(g)

a 3 x 10-3

b 1 x 10-4

c 3 x 103

d 10

Equilibrium Problem Solving - Review.

Problem solving involves the five step process:

• Analyze the problem.

• Brainstorm for a solution.

• Evaluate the process used.

(Dave has a handout on problem solving if you would like one. Or you could visit my web site:

http://www.sci.ouc.bc.ca/chem/probsol/ps_intro.html.)

In stoichiometry and many other aspects of chemistry, routine analysis of a problem involves writing a balanced equation and analyzing the data given on the basis of the equation.

In equilibrium problems, the same applies: write a balanced equation and analyze the data given in the problem using the I.C.E. method:

1. Balanced equation.

2. Initial conditions.

3. Change occurring to move to equilibrium, use mix.

4. Equilibrium conditions.

5. Write down a suitable expression for K.

6. Substitute any values from the ICE into the expression for K.

Example:

For the process

H2(g) + I2(g) ľ 2 HI(g),

Kc is 46.0. If the initial concentrations of H2 and I2 are equal at 0.120 M, calculate the concentrations of all species at equilibrium.

Analysis step:

 1 Equation H2(g) + I2(g) ľ 2 HI(g) 2 Initial (M) 0.120 0.120 0 3 Change (mix) (-1)x (-1)x (+2)x 4 Equilibrium 0.12-x 0.12-x 2x

5 Kc = [HI]2 / [H2]*[I2]

6 46.0 = (2x)2 / (0.12-x)(0.12-x) = { 2x / (0.12-x)}2

Brainstorm step:

Unknown are equilibrium concentrations: need to find x.

Calculation step:

x = 0.0927

At equilibrium, [HI] = 0.185 M, [H2] = [I2] = 0.0273 M

(Check the answer by substituting x back into the expression for Kc)

Example.

For the reaction

N2O4(g) ľ 2 NO2(g),

at 416 K the equilibrium constant, Kp, is 9850 kPa.

If the initial pressure of N2O4(g) is 177 kPa, and that of NO2(g) is 280 kPa, find the pressure of N2O4(g) at equilibrium.

Problem Solving Step A: Analysis:

 1 Equation N2O4 ľ 2 NO2 2 Initial (kPa) 177 280 3 Change (mix) (-1)x (+2)x 4 Equilibrium 177 - x 280 + 2x

5 Kp = P(NO2)2 / P(N2O4)

6 9850 = (280 + 2x)2 / (177 - x)

Problem Solving Step B: Brainstorm for a solution.

P(N2O4) at equilibrium is required. This is (177 - x) kPa. We need to find x.

This can be done by solving the quadratic.

Problem Solving Step C: Calculate.

x = 144.2

Problem Solving Step D: Defend and present the solution.

The pressure of N2O4(g) at equilibrium is 32.8 kPa

Example:

At a certain temperature Kp = 46.0 kPa for the process

PCl5(g) ľ PCl3(g) + Cl2(g).

What initial pressure of PCl5 must be placed in a reaction vessel if the final total pressure is to be 100 kPa?

Analysis:

 1 Equation PCl5(g) ľ PCl3(g) + Cl2(g). 2 Initial (kPa ) x 0 0 3 Change (-1)y (+1)y (+1)y 4 Equilibrium x - y y y

5 Kp = P(PCl3) * P(Cl2) / P(PCl5)

6 46.0 = y * y / (x - y)

Brainstorm:

Unknown required is x, the initial pressure of PCl5. To find this we need to solve the quadratic.....but that involves two unknowns (x and y), so we need a second equation.

Total pressure = 100 kPa, so (x - y) + y + y = 100

100 = x + y

This gives us a second equation and the problem is now solvable.

Calculate:

substitute y = 100 - x

46 = (100 - x)2 / (x - (100 - x))

46 = (100 - x)2 / (2x - 100) Solving x = 228 or x = 64.0

Note that the physical limits of this problem (100 = x + y), preclude x > 100 (since y cannot be negative!)

Thus x = 64.0

the initial pressure of PCl5 in the container was 64.0 kPa.

The Le Châtellier Principle

When a reaction at equilibrium is disturbed, the reaction will move in the direction that will minimize the disturbance.

A disturbance is defined as a change in the activity of at least one component in the reaction.

(If no activities are changed, then no disturbance occurs.)

Example:

For the reaction:

6 H2O(g) + 4 NO2(g) ľ 7 O2(g) + 4 NH3(g)

(endothermic)

fill in the following table:

 Change Equilibrium movement Change in Equilibrium Constant, K Addition of O2(g) Removal of NO2(g) Increased pressure Increased temperature Add He (inert) at constant P Add catalyst

Example:

For the process:

H2(g) + I2(g) ľ 2 HI(g), (exothermic)

fill in the following table:

 Change Equilibrium movement Change in Equilibrium Constant, K Addition of H2(g) Removal of HI(g) Increased volume Increased temperature Add He (inert) at constant P Add catalyst

Example :

For the process:

CaCO3(s) ľ CaO(s) + CO2(g) (endothermic)

fill in the following table:

 Change Equilibrium movement Change in Equilibrium Constant, K Addition of CaCO3(s) Removal of CO2(g) Increased pressure Increased temperature Add He (inert) at constant volume Add catalyst