Ch. 13 Notes Mr. Ferwerda, Tecumseh, High School 5/20/08
Chapter 13 Chemical Equilibrium Outline
1. ______- is a dynamic state in which no net change occurs.
- forward reaction rate = reverse reaction rate
- indicated in a chemical reaction by a double arrow(↔)
_____(T/F) All reactions are reversible.
a. Factors determining equilibrium position
- initial concentrations (pressure for gases)
- 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.
_____ (T/F) At equilibrium the forward rate equals the reverse rate.
_____ (T/F) At equilibrium the amounts of reactants and products are always equal.
_____ (T/F) At equilibrium the concentrations of reactants and products are constant.
2. The Equilibrium Constant (K)
a. Based on the law of mass action represented by the following equation for the following reaction :
aA + bB ↔ cC + dD
K =
K is dep. on : - stoichiometry
- temperature
- if T increases
- K increases for endothermic reactions
- K decreases for exothermic reactions
- opposite if T decreases
Units for K are generally ignored.
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] = 0.262 M, and [Cl2] = 0.156 M at equilibrium?
Answer : K =
2. If the above reaction was carried out under the same conditions using 0.250 M CO and 0.356 M Cl2, what would the concentration of COCl be?
Answer : K =
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.
2A(g) + B(g) ↔ 2C(g).
InitialChange
Equilibrium
b. Effects of changing a reaction on K :
i. The equation for the equilibrium constant for the reverse reaction (K') is K' = 1/K (reciprocal of
K)
Ex. If A + B <==> C is changed to C <==> A + B, K(forward) changes to 1/K(reverse)
ii. If the original equation is multiplied by a factor of n to give :
njA + nkB ↔ nlC + nmD
the equilibrium expression becomes :
If an equation at equilibrium is multiplied by a certain factor, then K must be raised to the power of that factor.
e.g. A + 2B <==> 2C to ½A + B <==> C new K =
e.g. A + 2B <==> 2C to 2A + 4B <==> 4C new K =
Your turn : Given : H2 + I2 à 2 HI K = 57 Determine the value of K for the following reactions :
1. 2 H2 + 2 I2 à 4 HI K =
2. 2 HI à H2 + I2 K =
3. HI à ½ H2 + ½ I2 K =
K for a reaction which occurs in two elementary steps equals the product of the K values for the two elementary steps.
Step 1 : A(g) + B(g) <==> 2C(g) K1 = 0.50
Step 2 : 2C(g) <==> 2D(g)_ K2 = 4.0
overall : A(g) + B(g) <==> 2D(g) K’ =
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.
K = KP?
a. 3A(g) + B(g) <==> 4 C(g) b. 2 A(g) + B(g) <==> 2C(g)
c. When the coefficients are not equal, the relationship between K and Kp is
Kp = K(RT)Δn
Where Δn = difference in coefficients : (sum of coeff. of products) - (sum of coeff. of reactants)
Calculate Δn for the following reactions :
2A(g) + 3B(g) <==> 3C(g) Δn =
A(g) + 2B(g) <==> 4C(g) Δn =
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. ______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)
b.
Ex. 2A(g) + B (s) <==> 2C(g)
K = ?
Your turn :
(a) CaO(s) + CO2 (g) <==> K =
(b) NaHCO3(s) <==> K =
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
Write the equilibrium expression for the decomposition of iron(II) sulfite in a closed container.
FeSO3(s) <==> K =
If, at equilibrium, the concentration of sulfur dioxide was found to be 0.25 M, what is the value of K?
For the above reaction K was found to be 1.8 atm at 1100 ºC. If a 20.0 g sample of iron(II) sulfite was placed in a 5.00 L vessel, how many grams of the iron(II) sulfite would decompose at equilibrium?
Day 2 :
Review :
Which of the following factors does not affect the equilibrium position of a reaction?
a. entropies of reactants and products
b. energies of reactants and products
c. temperature
d. concentrations of solid or liquid reactants or products
e. initial concentrations of gaseous or aqueous reactants or products
f. catalysts
Which of the following does affect the value of K?
a. temperature
b. pressure
c. initial concentrations of reactants and products
d. entropies of reactants and products
e. energies of reactants and products
f. stoichiometry
K = 100. for the following reaction : A(g) + B(g) <==> 2 C(g)
What is K for the following reactions?
1. 2A(g) + 2B(g) <==> 4 C(g) K =
2. 2 C(g) <==> A(g) + B(g) K =
3. C(g) <==> ½A(g) + ½B(g) K =
Which reactions favor the products?
A certain reaction at equilibrium at 400. K in a 2.0 L vessel was found to contain 2.0 mol of A, 4.0 mol of B and 2.0 mol of C. Calculate the value of K for this reaction at 400. K.
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 :
Fall into three categories :
1. “perfect square”
2. quadratic formula
3. negligible change – (change can be ignored – usually have a vary small K)
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
A certain reaction taking place in a 1.00 L vessel has a K value of 25.0. If the initial amounts of A and B were each 1.00 mol, what would the equilibrium concentrations be?
A(g) + B(g) ↔ 2 C(g).Initial
Change
Equilibrium
2.0 mol of A and 8.0 moles of C are placed in a 2.0 L vessel. The concentration of B at equilibrium was found to be 0.25 M. Calculate K for this reaction.
A(g) + B(g) ↔ 2 C(g).Initial
Change
Equilibrium
A 2.0 L vessel at 400 K was charged with 1.0 mol of A, 2.0 mol of B and 4.0 mol of C. If K = 0.50, calculate the equilibrium concentrations for this reaction at 400. K.
A(g) + B(g) ↔ 2 C(g).Initial
Change
Equilibrium
Direction of shift? – Need to calculate Q :
Need quadratic to solve :
At 20.0 ºC, K for the following reaction equals 1.2 x 10-6 mol/L. Calculate the equilibrium concentrations if starting with 4.0 mol of gas A in a 2.0 L vessel.
2A(g) ↔ 2B(g) + C(g).Initial
Change
Equilibrium
** When K is very small causing the change to be small you can sometimes assume that x is negligible compared to the initial concentration.
When assuming change is negligible, we must make sure change is less than 5% of original amount (we are really answering the question "Is change smaller than the precision of the least precise value in original amount).
Original amount : 2.0 M ; change was 2x ; x = 0.0106 :
Since it is less than 5%, our assumption of change as negligible stands.
Day :
Review
(1) 2A(g) + B(g) <====> 2C(g)
Equilibrium expression from law of mass action? K =
(2) 2A(g) + B(g) <====> 2C(g)
Equilibrium expression from law of mass action? K =
Calculate the value of K for the following reaction : A(g) + B(g) <==> 2C(g)
Given :
1. E(g) + 2C(g) <==> 3G(g) K1 = 0.020
2. A(g) + B(g) + E(g) <==> 3G(g) K2 = 0.10