Chemical Equilibrium
DETERMINING HOW MUCH TO ADD TO OBTAIN A GIVEN EQUILIBRIUM CONCENTRATION:
1. Put the given information into the reaction table. The unknown goes into the initial line.
2. Determine the entry on the Δ line for the substance whose final concentration is given.
3. Use stoichiometry and the result of Step 2 to complete the Δ line.
4. Add the initial and Δ lines to obtain the equilibrium line.
5. Substitute the equilibrium line entries into the equilibrium constant expression and solve for the unknown.
PRACTICE EXAMPLE 5.8
0.80 mol N2 and 0.80 mol O2 are mixed and allowed to react in a 10-L vessel at 2500 oC. How many moles of NO would be present at equilibrium?
N2(g) + O2(g) U 2NO(g) K = 2.1 x10-3
Reaction Table (let x = change in O2 concentration) N2(g) + O2(g) U 2NO(g)
in Δ eq
Equilibrium constant expression in terms of x
2.1x10-3 =
moles of NO = _____________ mol
Example 5.15
How many moles of H2 would have to be added to 1.00 mol I2 in a 1.00-L flask to produce 1.80 mol HI at equilibrium? H2 + I2 U 2HI K = 49.0
ng = 0, so K = Kp = Kc. Thus, we can use concentrations with the given K.
1/2. Put the given information into the reaction table with x in the initial line and then determine the entry on the Δ line for the known substance.
H2(g) + I2(g) U 2HI(g)
Initial x 1.00 0 mol.L-1
Δ +1.80 mol.L-1
Eq 1.80 mol.L-1
3/4. Use stoichiometry and the Δ line entry above to complete the Δ line then add the initial and Δ lines to obtain the equilibrium line.
H2(g) + I2(g) U 2HI(g)
Initial x 1.00 0 mol.L-1
Δ -0.90 -0.90 +1.80 mol.L-1
Eq x-0.90 0.10 1.80 mol.L-1
5. Setup the equilibrium constant expression and solve for x.
Equilibrium constant expression: 2 2
2 2
[HI] 1.80
49.0 = =
[H ][I ] (x-0.90)(0.10) Solve for x:
2 -1
2
x = 0.90 + 1.80 = 0.90 + 0.66 = 1.56 mol L H
(49.0)(0.10) ⋅
Simple stoichiometry predicts that only 0.90 mol H2 are required to react with 0.90 mol I2; but, because the equilibrium constant is not very large, an excess of H2 is required to drive the reaction to just 90% completion.
Example 5.16
What are the equilibrium concentrations in a solution made by mixing 100. mL each of 0.100 M KF and 0.100 M HNO2? The chemical equation and its K are given below.
Mixing two solutions of equal volume results in a 1:2 dilution because the volume of solution is doubled from 100 to 200 mL. Set up the reaction table.
F1-(aq) + HNO2(aq) U HF(aq) + NO21-(aq) K = 0.56
initial 0.0500 0.0500 0 0 M
Δ -x -x +x +x M
eq 0.0500 - x 0.0500 - x x x M Solve the equilibrium expression for x.
PRACTICE EXAMPLE 5.9
What are the concentrations of ClF3 and ClF in an equilibrium mixture produced by the decomposition of 1.84 M ClF3 at a temperature where Kc = 2.76 M for the decomposition?
Construct the reaction table to establish partial pressures ClF3(g) U ClF(g) + F2(g) Kc = 2.76 M in
Δ eq
Equilibrium constant expression in terms of x:
K = 2.76 =
Eliminate the denominator:
Write the equation in the form of a quadratic equation:
Solve for x
[ClF3] = _________ M [ClF] = [F2] = _________ M
1- 2 2
1- 2
2
[HF][NO ] (x)(x) x
K = 0.56 = = =
(0.050 -x)(0.050 - x)
[F ][HNO ] (0.050 - x)
The numerator and denominator are perfect squares, so the algebra can be simplified by taking the square root of both sides to eliminate squared terms,
0.56 = 0.75 = x
0.050 - x
Multiply both sides of the equation by 0.050-x to obtain 0.037-0.75 x = x, and solve.
0.037
x = = 0.021 M 1.75
The equilibrium concentrations are:
[F1-] = [HNO2] = 0.050 - x = 0.050 - 0.021 = 0.029 M [HF] = [NO21-] = x = 0.021 M.
Example 5.17
How many moles per liter of fluoride ion should be added to the equilibrium solution discussed in Example 5.16 to increase the HF concentration to 0.040 M?
First, recognize that we are asked for the amount to be added, so the unknown is in the initial line. We start with the equilibrium concentrations from Example 5.16.
F1- (aq) + HNO2(aq) U HF(aq) + NO21-(aq)
eq 0.029 0.029 0.021 0.021 M
Add the unknown amount of fluoride ion to convert the equilibrium line to an initial line.
F1- (aq) + HNO2(aq) U HF(aq) + NO21-(aq) initial 0.029 + x 0.029 0.021 0.021 M Enter the given equilibrium concentration.
F1- (aq) + HNO2(aq) U HF(aq) + NO21-(aq)
initial 0.029 + x 0.029 0.021 0.021 M
Δ
eq 0.040 M
The concentration of HF increases by 0.019 M. The other Δ-line entries are then determined by applying stoichiometry to this entry. The equilibrium line is found by summing the initial and Δ lines.
F1- (aq) + HNO2(aq) U HF(aq) + NO21-(aq) initial 0.029 + x 0.029 0.021 0.021 M Δ -0.019 -0.019 0.019 0.019 M eq 0.010 + x 0.010 0.040 0.040 M
PRACTICE EXAMPLE 5.10
The following equilibrium mixture was found in a 1.00-L flask at some temperature.
2SO2(g) + O2(g) U 2SO3(g)
eq 0.0500 0.0250 0.0180 mol
How many moles of O2 must be added to double the number of moles of SO3 at equilibrium?
Evaluate the equilibrium constant (Kc).
K =
On which line of the reaction table is the unknown?
Reaction Table
2SO2(g) + O2(g) U 2SO3(g)
in Δ eq
Solve the equilibrium constant expression for the number of moles added.
The unknown is determined by evaluation of the equilibrium constant expression with these equilibrium concentrations.
2
1-1- 2
[HF][NO ] (0.040)(0.040) 0.0016 0.16
K = 0.56 = = = =
(0.010 +x)(0.010) (0.010 +x)(0.010) 0.010 + x [F ][HNO ]
0.010 + x = 0.16 = 0.29 x = 0.29 - 0.01 = 0.28 M
0.56 ⇒
The molarity of the fluoride ion must be increased by 0.28 mol.L-1, which is over 1.5 times the amount that reacts (0.19 mol.L-1). In other words, only about 68% of the added fluoride ion reacts!
Hopefully, you have noticed that equilibrium problems are done in the same way whether they deal with gas-phase or aqueous equilibria. These same methods are applied in Chapters 6 - 8 in dealing with acid-base and metal ion equilibria. In this chapter, the equilibrium constants are neither very large nor very small; but, in the following chapters, the equilibrium constants are typically very small. The difference in magnitude allows us to make approximations to simplify the algebra, but it does not affect the way the problem is done.
5.4 CHAPTER SUMMARY AND OBJECTIVES
The activity of gases in equilibrium constants derived from standard free energies of formation are numerically equal to the pressures expressed in atmospheres. However, we can define the equilibrium constant in terms of concentrations, as well. The two values are designated as Kp (activities are in pressures) and Kc (activities are in concentrations), and they are related by the expression
p c ng
K = K (RT)Δ
The equilibrium constant for a reaction depends upon how the reaction is expressed. If a reaction is multiplied by a number n, then the resulting equilibrium constant is the original K raised to the nth power. If the reaction is written in the reverse direction, the equilibrium constant is the reciprocal of the original. Finally, the equilibrium constant of a reaction that is the sum of two other reactions is the product of the equilibrium constants of the summed reactions.
Le Châtelier’s principle states that when a stress is placed on a reaction at equilibrium, the reaction proceeds so as to minimize the effects of the stress. The shift can be
PRACTICE EXAMPLE 5.11
How many moles/liter of HCN should be added to 0.100 M NH3
to prepare a solution in which [NH41+] = 0.065 M?
Reaction table (K = 0.71)
NH3(aq) + HCN(aq) U NH41+ (aq) + CN1-(aq) initial
Δ eq
Equilibrium constant expression in terms of x.
0.71 =
Solve the equilibrium constant expression for the number of moles/liter added.
x = _________ mol/L HCN
understood in terms of changing the reaction quotient and the effect that change has on the free energy of the system. The addition of a solid that is involved in the equilibrium does not result in any change because the solid is not included in the reaction quotient.
Three types of equilibrium problems involving reaction tables were discussed:
i. Determining K from the initial concentrations and one final concentration.
ii. Determining concentrations from K and the initial conditions.
iii. Determining how much of one substance must be added or removed to produce a desired equilibrium concentration.
After studying the material presented in this chapter, you should be able to:
1. convert between Kp and Kc (Section 5.1);
2. determine the value of K for a reaction given the value of K for a related reaction that differs only by a multiple and/or direction (Section 5.1);
3. determine the equilibrium constant of a reaction that is the sum of several other reactions (Section 5.1);
4. relate Le Châtelier’s principle to the relationship between Q and K (Section 5.2);
5. predict the direction of the shift in equilibrium caused by stress placed on the equilibrium (Section 5.2);
6. calculate the concentrations of one substance in an equilibrium mixture from the other concentrations and the equilibrium constant (Example 5.8);
7. determine the equilibrium constant for a reaction given the initial amounts and one equilibrium amount (Examples 5.9 and 5.10);
8. determine the equilibrium composition from the initial composition and the equilibrium constant (Example 5.11);
9. calculate the amount of one reactant required to react with a given amount of another reactant to produce a given amount of product (Example 5.12);
10. determine the extent of the change caused by the addition of known amounts of reactants or products to an equilibrium mixture, given K and the initial equilibrium concentrations (Example 5.13);
11. calculate the concentrations resulting from a volume change (Example 5.14); and 12. calculate the amount of one substance that would have to be added to an equilibrium
mixture to change the concentration of another substance in the equilibrium by a given amount (Example 5.16).
ANSWERS TO PRACTICE EXAMPLES
5.1 ΔGo = +3.79 kJ; K = 0.272; Kp = 0.272 atm-1; Kc = 7.81 M-1 5,2 ΔGo = -1.89 kJ; K = 1.92; K = 1.92
5.3 K = 7.5x10-5
5.4 a) [HSO41-] decreases; K unchanged b) [HSO41-] increases; K unchanged c) [HSO41-] increases; K decreases 5.5 [Ca2+] = 1.9x10-5 M
5.6 PO2 = 2.5 atm; PSO2 = 5.0 atm; PSO3 = 7.5 atm; K = 1.1 5.7 K = 1.3
5.8 0.036 mol NO
5.9 x2 + 2.76x - 5.08 = 0; x = 1.26 M [ClF3] = 0.58 M; [ClF] = [F2] = 1.26 M 5.10 0.228 mol O2
5.11 0.24 mol.L-1 HCN
5.5 EXERCISES
6. If equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products?
a) HCN(aq) + H2O(l) U CN1-(aq) + H3O1+(aq) K = 6.2x10-10 10. Given the following information at 1000 K:
CaCO3(s) U CaO(s) + CO2(g) K1 = 0.039 C(s) + CO2(g) U 2 CO(g) K2 = 1.9
Determine the equilibrium constant at 1000 K for:
CaCO3(s) + C(s) U CaO(s) + 2 CO(g)
11. Lead fluoride dissolves in strong acid by the following reaction:
PbF2(s) + 2H3O1+(aq) U Pb2+(aq) + 2HF(aq) +2H2O(l) a) What is the equilibrium constant expression for the reaction?
b) Use the following equilibrium constants to determine the value of the equilibrium constant of the above reaction:
PbF2(s) U Pb2+(aq) + 2F1-(aq) K1 = 3.7x10-8 HF(aq) + H2O(l) U H3O1+ + F1-(aq) K2 = 7.2x10-4
12. Aluminum hydroxide dissolves in strong acid by the following reaction:
Al(OH)3(s) + 3H3O1+(aq) U Al3+ (aq) + 6H2O(l)
a) What is the equilibrium constant expression for the reaction?
b) Use the following equilibrium constants to calculate the value of K for the above reaction
H3O1+(aq) + OH1-(aq) U 2H2O(l) K1 = 1.0x10+14 Al(OH)3(s) U Al3+(aq) + 3OH1-(aq) K2 = 1.9x10-33
13. Equal numbers of moles of Cl2 and NO are placed in a vessel at some temperature where they reach the following equilibrium:
2 NO(g) + Cl2(g) U 2 ClNO(g)
Indicate whether each of the following statements about the resulting equilibrium mixture is true, false, or depends upon the value of the equilibrium constant.
a) [NO] > [ClNO] b) [Cl2] < [NO] c) [Cl2] > [ClNO]
14. Equal number of moles of NH3 and N2 are added to a flask where they equilibrate according to: 2NH3(g) U N2(g) + 3H2(g). Indicate whether each of the following statements about the resulting equilibrium mixture is true, false, or depends upon the value of the equilibrium constant.
a) [NH3] < [H2] b) [H2] > [N2] c) [NH3] > [N2] LE CHÂTELIER’S PRINCIPLE
15. What effect (increase, decrease, or no effect) does increasing the volume of the following equilibrium mixtures at constant temperature have on Q?
What effect does each have on K?
a) H2(g) + I2(g) U 2HI(g) ΔHo = +53 kJ b) 3H2(g) + N2(g) U 2NH3(g) ΔHo = -92 kJ c) N2O4(g) U 2NO2(g) ΔHo = +58 kJ
16. What effect does increasing the temperature at constant volume of each of the equilibrium mixtures in Exercise 15 have on Q and K?
17. Does increasing the volume of each of the equilibria in Exercise 15 increase the number of moles of reactant, product, or neither?
18. Does increasing the temperature of each of the equilibria in Exercise 15 increase the number of moles of reactant or product?
19. Consider the equilibrium, NH3(g) + H2S(g) U NH4HS(s) , ΔHo < 0. Which of the following would increase the number of moles of ammonia in the equilibrium mixture?
a) increasing the temperature
b) increasing the volume of the container c) adding more H2S gas
d) adding more NH4HS solid
20. Methanol is manufactured by the following reaction:
CO(g) + 2 H2(g) U CH3OH(g) ΔH° = -91 kJ
Does the amount of methanol increase, decrease, or remain the same when an equilibrium mixture is subjected to the following changes?
a) the temperature is increased
b) the volume of the container is decreased c) CO is added
d) CH3OH is added
21. Consider the following: P4(s) + 6Cl2(g) U 4PCl3(l) ΔH < 0 What happens to the mass of phosphorus in each of the following?
a) the volume is increased b) chlorine is removed c) phosphorus trichloride is added d) the mixture is cooled
22. Predict how an increase in temperature will change K for the following chemical reactions:
a) N2(g) + O2(g) U 2NO(g) ΔH° = 181 kJ b) 2SO2(g) + O2(g) U 2SO3(g) ΔH° = -198 kJ
SOLVING FOR AN UNKNOWN EQUILIBRIUM CONCENTRATION