1. aA + bB  dD + eE that there is an equilibrium constant
Calculations With Equilibrium Constants
In the last lecture we showed that for a reaction
aA + bB  dD + eE that there is an equilibrium constant
K 𝑇 = 𝐷 𝑑 𝐸 𝑒 𝐴 𝑎 𝐵 𝑏
And for the reverse reaction
dD + eE  aA + bB
𝐾 ′ 𝑇 = 1 𝐾

2. Calculations With Equilibrium Constants
We also showed that for two reactions
aA + bB  dD + eE K1
fF + gG  hH + jJ K2
The equilibrium constant for the sum of these reactions
aA+bB+fF+gG  dD+eE+hH+jJ K=K1K2

3. Concentration = 𝑛 𝑉 = 𝑃 𝑅𝑇
Calculations With Equilibrium Constants
And finally since concentration can be related to pressure using the ideal gas law
Concentration = 𝑛 𝑉 = 𝑃 𝑅𝑇
we showed that
Kp=Kc(RT)Dn
where n is the difference between the number of moles of gas produced and consumed in a reaction and
𝐾 𝑝 = 𝑃 𝑐 𝑃 𝑑 𝑃 𝑎 𝑃 𝑏
for the gas phase reaction aA +bB = cC +dD

4. Rules for Equilibrium Constant Problems
We learned how to calculate the equilibrium constants given equilibrium concentrations of the products and reactants.
Now we want to learn how to calculate equilibrium concentrations given the equilibrium constant and starting concentrations of reactants and products.

5. Rules for Equilibrium Constant Problems
To organize the problem we use a chart with the initial concentrations, the changes needed to reach equilibrium and the final equilibrium concentrations
Initial
Change
Final

6. Rules for Equilibrium Constant Problems
We write the BALANCED chemical reaction across the top of the chart and the concentrations of reactants and products in the first row
aA bB = cC dD
[A]
[B]
[C]
[D]
Initial
Change
Final
We then fill in the chart with the given information, write out the equilibrium constant expression, formulate the necessary algebraic equations and solve them

7. Rules for Equilibrium Constant Problems
Write the balanced chemical equation
Write the equilibrium constant equation
Fill out the table using the stoichiometric relations from the balanced equation
Use the information in the table to evaluate the equilibrium constant.
Solve any remaining algebraic equations.

8. An Example For H2(g) + I2(g)  2HI(g)
What is the equilibrium constant in terms of the concentrations of the product and reactants?
K= 𝐻𝐼 𝑔 𝐻 2 𝑔 𝐼 2 𝑔

9. Example (cont.)
In an experiment at 490 oC the following concentrations were measured at equilibrium H2 I2 HI
Initial
Change
Final M M M
What is the equilibrium constant?

10. Example (cont.) What is the equilibrium constant? Substitute intoH2 I2 HI
Initial
Change
Final M M M
Substitute into
K= 𝐻𝐼 𝑔 𝐻 2 𝑔 𝐼 2 𝑔 = ( 𝑀) 2 ( 𝑀 ( 𝑀 =45.9

11. Example (cont.)
We can now use the equilibrium constant to describe any equilibrium mixture of these three molecules at 490 C
1.00 moles of H2 and 1.00 mole of I2 are introduced into a 2.00 L box at 490 C. What is the final concentration of all the molecules at equilibrium
First we calculate the concentration of all of the reactants and products at the start
𝐻 2 = 𝐼 2 = 1.00 𝑚𝑜𝑙 2.00 𝐿 =0.500 𝑀 𝐻𝐼 =0
And substitute into the ICE table

12. Example (cont.) H2 I2 HI
Initial
 0.500
0.500
0.00
Change
Final
The change in [H2] and [I2] will be the same, -x, and from the stoichiometry the change in [HI] will be +2x
H2(g) + I2(g)  2HI(g) H2 I2 HI
Initial
 0.500
0.500
0.00
Change
 -x -x
+2x
Final
We can add these to get the algebraic amounts at equilibrium

13. Example (cont.) H2 I2 HI
Initial
 0.500
0.500
0.00
Change
 -x -x
+2x
Final
0.500-x
And substitute these expressions into the equilibrium constant expression
K= 𝐻𝐼 𝑔 𝐻 2 𝑔 𝐼 2 𝑔 = (2𝑥) −𝑥 −𝑥 =45.9
K= 𝐻𝐼 𝑔 𝐻 2 𝑔 𝐼 2 𝑔 = (2𝑥) −x 2 =45.9

14. Example (cont.) (2𝑥) 0.500−x = 45.9 =6.77 2𝑥= 0.500−x 6.77
K= 𝐻𝐼 𝑔 𝐻 2 𝑔 𝐼 2 𝑔 = (2𝑥) −𝑥 −𝑥 =45.9
We can simplify this expression
K= 𝐻𝐼 𝑔 𝐻 2 𝑔 𝐼 2 𝑔 = (2𝑥) −x 2 =45.9
taking the square root of both sides
(2𝑥) −x = =6.77
Then multiplying by (0.500-x) to get
2𝑥= 0.500−x 6.77

15. Example (cont.) 2𝑥= 0.500−x 6.77 8.77𝑥=3.39 𝑥=0.387
Which can be solved for x
8.77𝑥= 𝑥=0.387
So the equilibrium concentrations are
𝐻 2 = 𝐼 2 = 0.500−.387 =0.113𝑀
𝐻𝐼 =2𝑥=0.773 𝑀
To check use these concentrations to calculate the equilibrium constant
K= 𝐻𝐼 𝑔 𝐻 2 𝑔 𝐼 2 𝑔 = [ 𝑀 ] 2 (0.113 𝑀 (0.113 𝑀 =46.4 ~ 45.9

16. Example (cont.) K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔 =45.9
It doesn’t matter whether we start with hydrogen and iodine, or with HI. If two moles of HI are injected into a 1 L box at 490 C what will be the concentration of each species in the box at equilibrium
K= 𝐻𝐼 𝑔 𝐻 2 𝑔 𝐼 2 𝑔 =45.9 H2 I2 HI
Initial
Change
Final

17. Example (cont.) K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔 =45.9
Or even if we start with a mixture of all three molecules. One mole of H2, two moles of I2 and three moles of HI are injected into a 1 L box. What will be the concentration of each species at equilibrium at 490 C?
K= 𝐻𝐼 𝑔 𝐻 2 𝑔 𝐼 2 𝑔 =45.9 H2 I2 HI
Initial
Change
Final

18. Yet Another Example
A mol sample of phosphorus pentachloride, PCl5 dissociates at 160 C and 1.00 atm to give mol of phosphorus trichloride at equilibrium. What is the composition of the final mixture? What is the equilibrium constant? H2 I2 HI
Initial
Change
Final

19. And Yet Another Example
Methanol is manufactured commercially by
CO(g) + 2H2(g)  CH3OH(g)
A L vessel was filled with mol CO and mol H2. When this mixture came to equilibrium at 500K the vessel contained mol CO. How many moles of each substance were in the vessel? What is the equilibrium constant?
Initial
Change
Final