Ch. 13 Notes Mr. Niedenzu Providence High School
Chapter 13 Chemical Equilibrium Outline
1. Chemical equilibrium - a dynamic state in which no net change occurs
- forward reaction rate = reverse reaction rate
- indicated in a chemical reaction by a double arrow()
a. Factors determining equilibrium position
- initial concentrations
- relative energies of reactants and products
- entropies of reactants and products
- temperature
For the reaction 3A(aq) + 2B(aq) 4C(g) + D(aq) under 1 atm of pressure would have the following equilibrium position (drawing 1). Under 2 atm of pressure it might have the equilibrium position in the second diagram (an increase in pressure shifts the equilibrium position away from gaseous reactants or products).
1. 2.
2. The Equilibrium Constant (K)
a. Based on the law of mass action represented by the following equation for the following reaction :
jA + kB lC + mD
Sample problems :
1. What is the value of K for the equilibrium reaction COCl(g) CO(g) + Cl2(g) at 1000 K if [COCl] = .125 M, [CO] = .262 M, and [Cl2] = .156 M ?
Answer : K = [CO][Cl2]/[COCl] = (.262 M)(.156 M)/(.125 M) = .327 M or .327 mol/L
2. If the above reaction was carried out under the same conditions using .250 M CO and .356 M Cl2, what would the concentration of COCl be?
Answer : K = [CO][Cl2]/[COCl], rearranging to solve for [COCl] : [COCl] = [CO][Cl2]/K = (.250 M x .356 M)/.327 M = .272 M COCl
3. A certain reaction takes place by the equation 2A(g) + B(g) 2C(g). A 1.00 L vessel was filled with 2.00 moles of A and 2.00 moles of B ([A] = [B] = 2.00 mol/1L = 2.00M). At equilibrium [C] was found to be .50M. Calculate K for this reaction.
Sample problems, cont'd
Answer : Use [C] and the stoichiometry of the balanced equation to calculate the concentrations of reactants. [C] = .50 M and since the mole ratio of [A] to [C] is 1:1 then .50 M of A has converted to C so the [A] at equilibrium equals [A]0 - .50 M = 1.50 M. The mole ratio of B : C is 1 : 2 so only .25 M of B has been changed to C so at equilibrium [B] = [B]0 - .25 = 1.75 M
Solving for K : K = [C]2/([A]2[B]) = (.50 M)2/(1.50 M2 x 1.75 M) = .063 M-1
b. The equation for the equilibrium constant for the reverse reaction (K') is K' = 1/K (reciprocal of K)
c. If the original equation is multiplied by a factor of n to give :
njA + nkB nlC + nmD
the equilibrium expression becomes :
Sample problem : At a particular set of conditions K= 62.5 for the reaction
2H2(g) + S2(g) 2H2S(g)
Calculate the value of K for the following reactions at the same conditions.
1. H2(g) + ½S2(g) H2S(g)
2. 2H2S(g) 2H2(g) + S2(g)
3. H2S(g) H2(g) + ½S2(g)
4. 6H2(g) + 3S2(g) 6H2S(g)
Answers :
1. K = 7.91 The reaction is cut in half (n = .5, see 2c above) Knew = Kn = 62.5.5 = 7.91
2. K = .0160 The reaction is reversed so Knew = 1/K= 1/62.5 = .0160
3. K = .126 This reaction is no. 2 cut in half (n = .5) so Knew = .0160.5 = .126
4. K = 2.44 x 105 This reaction is the original equation multiplied by 3, so Knew = K3 = 62.53
d. The units for K depend on powers of the various concentration terms in the equilibrium expression
e. K is constant for each set of conditions for a given reaction in equilibrium.
f. Each set of equilibrium concentrations will result in a specific equilibrium position, meaning that although there is only one value of K for a particular system at a particular temperature, there is an infinite number of equilibrium positions. (e.g. For the above reaction, there are an infinite number of concentrations of A and B that can be combined, but at a certain temperature K will always have the same value).
3. Equilibrium expressions Involving Pressures (all gaseous reactants and products)
a. For the following reaction :
N2(g) + 3 H2(g) 2 NH3(g)
The equilibrium expression would be :
Since the partial pressure of a gas is directly proportional to its concentration we can rewrite the above expression as:
b. K will equal Kp when the sum of the coefficients on either side of an equation are equal
c. When the coefficients are not equal, the relationship between K and Kp is
Kp = K(RT)n
Where n = difference in coefficients : (sum of coeff. of products) - (sum of coeff. of reactants)
Sample problems :
1. At a certain set of conditions the equilibrium pressures for the reaction 2NOCl(g) 2 NO(g) + Cl2(g) are PNOCl = 1.4 atm, PNO = 1.00 x 10-2 atm and PCl2 = 2.75 x 10-1 atm. Calculate KP for this reaction.
Answer : KP = (PNO)2(PCl2)/PNOCl2 = (1.00 x 10-2 atm)2(2.75 x 10-1 atm)/(1.4 atm)2 = 1.4 x 10-5 atm
2. Calculate K for the above reaction if T = 350. K.
Answer : Since the coefficients are not balanced on both sides of the equation K does not equal KP. Rearranging the equation Kp = K(RT)n to K = KP/(RT) n :
n = 3-2 = 1
K = 1.45 x 10-5 atm/(.08206 L • atm/K • mol x 350K)1 = 5.04 x 10-7 mol/L
4. Heterogeneous equilibria : not all reactants and products in the same state
a. The equilibrium position of a reaction is independent of pure solids and pure liquids (concentrations of pure solids and liquids do not change and are therefore not included in the equilibrium expression)
Sample problems :
1. Write the expressions for K and KP for the following reactions :
a. CO2(g) + H2(g) CO(g) + H2O(l)
b. SnO2(s) + 2CO(g) Sn(s) + 2 CO2(g)
Answers : Note solids and liquids do not appear in equilibrium expressions.
a. and
b. and
5. Applications of the Equilibrium Constant
a. From the equilibrium constant we can predict :
- the tendency of a reaction to occur (but not speed)
- whether or not the given concentrations represent an equilibrium condition
- the equilibrium position from a given set of initial concentrations
b. The extent of a reaction (how far it proceeds to form products) is indicated by the magnitude of K
- a K value much larger than one indicates that at the equilibrium position the system will consist of
mostly products
- a K value much smaller than one indicates that at the equilibrium position the system will consist
of mostly reactants
- the time required to reach equilibrium from a set of initial concentrations is independent of K
c. The Reaction Quotient (Q) - useful in determining if a reaction is at equilibrium or not (and if not, which way it will shift to attain equilibrium) by comparing Q to K
- For the following reaction :
N2(g) + 3 H2(g) 2 NH3(g)
The reaction quotient would be :
Where the concentrations used are the initial concentrations given instead of concentrations at equilibrium conditions.
- if Q is equal to K the system is at equilibrium
- if Q is greater than K (more products than at equilibrium) the system will shift to the left
- if Q is less than K ( fewer products than at equilibrium) the system will shift towards the right
d. Calculation of Equilibrium Pressure and Concentrations
Sample problems :
At 721 K the equilibrium constant for the reaction H2(g) + I2(g) 2 HI(g) is 51. For which of the following sets of initial conditions at 721 K will the reaction be at equilibrium and if not at equilibrium, which way will the reaction shift to attain equilibrium?
a. [H2] = 5.0 x 10-3 M, [I2] = 1.5 x 10-2 M, [HI] = 1.0 x 10-3
b. [H2] = 4.0 x 10-3 M, [I2] = 3.5 x 10-2 M, [HI] = 2.0 x 10-1
c. PH2 = 3.0 x 10-3 torr, PI2 = 2.5 x 10-2 torr, PHI = 6.18 x 10-3 torr
Answers :
a. Q = (1.0 x 10-3)2/(5.0 x 10-3 x 1.5 x 10-2) = .013
Since Q < K, the reaction will shift to the right ([HI] needs to increase and [H2] and [I2] need to decrease).
b. Q = 286 Since Q > K, the reaction will shift to the left.
c. Q = 51 = KP = K so the reaction is at equilibrium. KP = K because n = 0, units of torr can be used
because K is does not have any units.
6. Solving Equilibrium Problems
a. Procedure :
- write balanced chemical equation
- write the equilibrium expression using the law of mass action
- list initial concentrations
- calculate Q and determine which way equilibrium lies
- define change needed to reach equilibrium and define the equilibrium concentrations
- substitute in the equilibrium concentrations into the equilibrium expression and solve for the
unknown
- check your answers by using them to solve for K to see if it agrees with the accepted value of K (5% rule - if an answer is within 5% it is considered valid)
b. if K is small approximations can be used to simplify the math - must be checked for accuracy
Sample Problem :
At a certain temperature K = 75.0 for the reaction H2(g) + I2(g) 2 HI(g). If an experiment started with 2.00 moles of each species in a 4.00 L container, what would the equilibrium concentrations be?
Answer : Initial concentrations equal mol/L. Calculate Q to determine which way the reaction will shift. (Q = 1, so the reaction will shift to the right). Set up an ICE (Initial, Change, Equilibrium) table as follows:
H2(g) + I2(g) 2 HI(g).Initial / .500 M .500 M .500 M
Change / -x -x +2x
Equilibrium / .500 M - x .500 M -x .500 M +2x
The variable x is the amount of change. It is negative on the reactant side and positive on the product side because the reaction will shift to the right. The stoichiometry of the balanced equation is used to determine the coefficient of x.
Solving : ,
Solving for x, x = .359 so at equilibrium : [H2] = .500 - x = .500 -.359 = .141 M
[I2] = .500 - x = .500 -.359 = .141 M
[HI] = .500 + 2x = .500 + (2 x .359) = 1.22 M
To check your answer substitute equilibrium values into the equilibrium expression :
The value of K is reasonably close (within 5%) for confidence in the answer.
7. Le Châtelier's Principle - if a change is imposed on a system at equilibrium, the position of the
equilibrium will change in a direction that reduces that change.
a. Changes in concentration - an increase (or decrease) in the concentration of a reactant or product
will cause the system to shift away (toward) from the added reactant or product.
b. Changes in pressure :
- can be caused by :
- the addition or removal of a gaseous reactant or product-system will change as described above
- the addition of an inert gas- will not affect equilibrium position (changes total pressure, but nor partial pressures (concentrations) of gaseous reactants or products( e.g. if you have 2.0 moles of a reactant gas in a one liter vessel, the concentration of the reactant will be 2.0 M. Adding helium to the one liter vessel will not change the concentration of the reactant gas - there is still 2.0 moles in a one liter vessel).
- a change in the volume of a container - will compensate for an increase in volume by shifting toward the side with the larger volume and vice versa
c. Changes in temperature - different from above because a change in temperature will change the value of K as well as the equilibrium position
- energy can be treated as a reactant or product when using Le Châtelier's principle
- ex. For the reaction :
N2(g) + 3 H2(g) 2 NH3(g) + 92 kJ
an increase in temperature will add energy and the reaction will shift to the left. This will decrease the concentrations of the products and reduce the value of K.
Sample Problem :
Nitrogen gas will combine with chlorine gas to form nitrogen trichloride according to the equation N2(g) + 3 Cl2(g) 2 NCl3(g). If the reaction is at equilibrium and the following changes are made, which way will the reaction shift?
a. Additional chlorine gas is added.
b. Nitrogen trichloride is removed from the reaction vessel.
c. The pressure is increased by decreasing the size of the vessel.
d. The pressure is increased by the addition of helium.
e. The temperature is increased (the reaction is exothermic).
Answers :
a. The reaction will shift to the right to reduce stress of added chlorine.
b. The reaction will shift to the right to replace removed nitrogen trichloride.
c. The reaction will shift to the right - fewer moles of gas.
d. No change, the addition of an inert gas does not change the partial pressures of reactive gases.
e. The reaction will shift to the left. If the reaction is exothermic heat is a product and will shift away
from added heat.