1. Lecture 25: Calculations with Equilibrium Constants
Standard States, Equilibrium Constants and Reaction Quotients
Developing a Reaction Table: Limiting Reactant, Limiting Product and Everything In Between
Tricks of the Trade

2. Nature of Equilibrium Expressions
Consider some of the equilibrium relations we’ve had so far
For vapor pressure:
For gas-phase dimerization:
The equilibrium expression can be expressed in terms of ratios of partial pressures of the gases to a reference pressure of 1 atm.
The essential properties of a pure liquid do not change as the reaction proceeds at constant temperature. So all the properties of a pure liquid are incorporated into the equilibrium constant.

3. Standard States (a look ahead)
Certain standard states have been agreed upon so that we can consider equilibria in terms of the dimensionless “activity” of a species relative to its agreed-upon standard state.
Species
Standard State
Activity of species
(if ideal)
Gas: i(g)
1 atm
Pi/1 atm
(Dalton’s Law)
Pure liquid: i(l)
Pure liquid 1
Pure solid: i(s)
Pure solid
Solvent:
i(solv) ~ i(l)
Pure solvent
csolvent (mole fraction) ~ 1
(Raoult’s Law)
Solute: i(soln)
1 M = 1 mol/L
(ni/V)/1 M = ci/1 M
(Henry’s Law)

4. Reaction Quotient
A measure of the degree of reaction for comparison with Keq.
For the gas-phase reaction:
Compare Q to the equilibrium constant, Keq:
Q evolves toward Keq as the partial pressures evolve toward their equlibrium values

5. The partial pressure of NO2 is 0. 5 atm and that of N2O4 is also 0
The partial pressure of NO2 is 0.5 atm and that of N2O4 is also 0.5 atm. The equilibrium constant for the dimerization of NO2 is Which of the following is true:
Q = 0.5 and the reaction will proceed backward.
Q = 2.0 and the reaction will proceed backward.
Q = 0.5 and the reaction will proceed forward.
Q = 2.0 and the reaction will proceed forward.

6. Reaction Table
Consider the NO2 on a very smoggy day. If NO2 came to equilibrium with the air, what would it’s partial pressure be?
initial Pi/1atm Qinit = 0.016
What would be the final partial pressures if this were a limiting reactant problem? Assume some constant volume, V.
initial ni (0.8 atm)V/RT (0.2 atm)V/RT (0.05 atm)V/RT
Max # rxns (0.8 atm)V/RT (½)(0.2 atm)V/RT
O2 would be consumed first.
final ni (0.8 atm)V/RT (0.05 atm)V/RT -(½)(0.2 atm)V/RT (½)(0.2 atm)V/RT
final Pi/1atm (½)(0.2) (0.2) = = 0.25

7. How about complete back reaction: “limiting product”?
Reaction Table (con’t)
initial Pi/1atm Qinit = 0.016
Limiting reactant Pi/1atm Qlim react = infinity
How about complete back reaction: “limiting product”?
Limiting product Pi/1atm Qlim prod = 0

8. Reaction Table
So, the best first approximation is the “limiting product” result: PNO2 = 0. Now consider a further change from there.
initial Pi/1atm Qinit = 0.016
Limiting reactant Pi/1atm Qlim react = infinity
Limiting product Pi/1atm Qlim prod = 0
change Pi/1atm - x x x
equil Pi/1atm – x – 2x x Qeq = (2x)2/[(0.825-x)(0.25-2x)2] = Keq = 7.1x10-19
How do we determine x?

9. x must be very small. Try ignoring it in the denominator.
(2x)2/[(0.825)(0.25)2] ~ Keq = 7.1x10-19
x2 ~ (7.1x10-19) (0.825)(0.25)2/4 = 0.92x10-20
x ~ 0.96x10-10 ~ 1x10-10
PNO2 = 2x ~ 2x10-10 atm
How good is this?
Plug into the expression for Q and find out:
So, Q is only 10% too big.
If more accuracy is needed, try a bit less: x = 0.9 or 0.95x10-10