1. Chapter 14 Chemical Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA

2. Arrow Conventions
Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions
A single arrow indicates all the reactant molecules are converted to product molecules at the end
A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products
 in these notes
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3. Reaction Dynamics
When a reaction starts, the reactants are consumed and products are made
forward reaction = reactants  products
therefore the reactant concentrations decrease and the product concentrations increase
as reactant concentration decreases, the forward reaction rate decreases
Eventually, the products can react to re-form some of the reactants
reverse reaction = products  reactants
assuming the products are not allowed to escape
as product concentration increases, the reverse reaction rate increases
Processes that proceed in both the forward and reverse direction are said to be reversible
reactants Û products
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4. Hypothetical Reaction 2 Red Û Blue
The reaction slows over time,
but the Red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the Red and
the Blue molecules no longer change –
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of Red molecules turning into Blue is the
same as the rate of Blue molecules turning
into Red
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5. Hypothetical Reaction 2 Red Û Blue
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6. Reaction Dynamics Once equilibrium is established,
the forward and reverse reactions
proceed at the same rate, so the
concentrations of all materials
stay constant
Eventually, the reaction proceeds
in the reverse direction as fast as
it proceeds in the forward direction.
At this time equilibrium is established.
Because the reactant concentration
decreases, the forward reaction slows.
As the products accumulate, the
reverse reaction speeds up.
As the forward reaction proceeds
it makes products and uses reactants
Initially, only the forward
reaction takes place
Rate
Rate Forward
Rate Reverse
Time
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7. Dynamic Equilibrium
As the forward reaction slows and the reverse reaction accelerates, eventually they reach the same rate
Dynamic equilibrium is the condition wherein the rates of the forward and reverse reactions are equal
Once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant
because the chemicals are being consumed and made at the same rate
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8. Equilibrium  Equal
The rates of the forward and reverse reactions are equal at equilibrium
But that does not mean the concentrations of reactants and products are equal
Some reactions reach equilibrium only after almost all the reactant molecules are consumed – we say the position of equilibrium favors the products
Other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed – we say the position of equilibrium favors the reactants
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9. An Analogy: Population Changes
However, after a time, emigration will occur in
both directions at the same rate, leading to
populations in Country A and Country B that are
constant, though not necessarily equal
When Country A citizens feel overcrowded, some will emigrate to Country B
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10. Equilibrium Constant
Even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them
The relationship between the chemical equation and the concentrations of reactants and products is called the Law of Mass Action
For the general equation aA + bB Û cC + dD, the Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the balanced chemical equation
always products over reactants
K is called the equilibrium constant
unitless
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11. Writing Equilibrium Constant Expressions
For the reaction aA(aq) + bB(aq) Û cC(aq) + dD(aq) the equilibrium constant expression is
So for the reaction
2 N2O5(g) Û 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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12. What Does the Value of Keq Imply?
When the value of Keq >> 1, we know that when the reaction reaches equilibrium there will be many more product molecules present than reactant molecules
the position of equilibrium favors products
When the value of Keq << 1, we know that when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules
the position of equilibrium favors reactants
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13. A Large Equilibrium Constant
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14. A Small Equilibrium Constant
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15. Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following
2 SO2(g) + O2(g) Û 2 SO3(g) K = 8 x 1025
N2(g) + 2 O2(g) Û 2 NO2(g) K = 3 x 10−17
favors products
favors reactants
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16. Relationships between K and Chemical Equations
When the reaction is written backward, the equilibrium constant is inverted
For the reaction aA + bB Û cC + dD
the equilibrium constant expression is
For the reaction cC + dD Û aA + bB
the equilibrium constant expression is
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17. Relationships between K and Chemical Equations
When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor
For the reaction aA + bB Û cC
the equilibrium constant expression is
For the reaction 2aA + 2bB Û 2cC
the equilibrium constant expression is
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18. Relationships between K and Chemical Equations
When you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations
For the reactions (1) aA Û bB and (2) bB Û cC the equilibrium constant expressions are
For the reaction aA Û cC
the equilibrium constant expression is
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19. Example 14.2: Compute the equilibrium constant at 25 °C for the reaction NH3(g)  0.5 N2(g) H2(g)
Given:
Find:
for N2(g) + 3 H2(g) Û 2 NH3(g), K = 3.7 x 108 at 25 °C
K for NH3(g)  0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual Plan:
Relationships:
Kbackward = 1/Kforward, Knew = Koldn K’ K
Solution:
N2(g) + 3 H2(g) Û 2 NH3(g) K1 = 3.7 x 108
2 NH3(g) Û N2(g) + 3 H2(g)
NH3(g) Û 0.5 N2(g) H2(g)
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20. Practice – When the reaction A(aq) Û 2 B(aq) reaches equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M. When the reaction 2 B(aq) Û Z(aq) reaches equilibrium [B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M. Calculate the equilibrium constant for each of these reactions and the equilibrium constant for the reaction 3 Z(aq) Û 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2 B Û A K3 = 1/Krxn 1 = 6.25 x 10−5
Z Û 2 B K4 = 1/Krxn 2 = 2
Z Û A K3K4 = 1.25 x 10−4
3 Z Û 3 A (K3K4)3 = 2.0 x 10−12
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21. Equilibrium Constants for Reactions Involving Gases
The concentration of a gas in a mixture is proportional to its partial pressure
Therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases
For aA(g) + bB(g) Û cC(g) + dD(g) the equilibrium constant expressions are or
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22. Kc and Kp In calculating Kp, the partial pressures are always in atm
The values of Kp and Kc are not necessarily the same
because of the difference in units
Kp = Kc when Dn = 0
The relationship between them is
Dn is the difference between the number of moles of reactants and moles of products
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23. Deriving the Relationship between Kp and Kc
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24. Deriving the Relationship Between Kp and Kc
for aA(g) + bB(g)  cC(g) + dD(g)
substituting
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25. Example 14.3: Find Kc for the reaction 2 NO(g) + O2(g) Û 2 NO2(g), given Kp = 2.2 x 1012 @ 25 °C
Conceptual Plan:
Relationships: Kp Kc
Solution:
2 NO(g) + O2(g)  2 NO2(g)
Dn = 2  3 = −1
Check:
K is a unitless number
because there are more moles of reactant than product, Kc should be larger than Kp, and it is
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26. 2 SO2(g) + O2(g) Û 2 SO3(g) Kc = 8 x 1025
Practice – Calculate the value of Kp or Kc for each of the following at 27 °C
2 SO2(g) + O2(g) Û 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) Û 2 NO2(g) Kp = 3 x 10−17
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27. Heterogeneous Equilibria
Pure solids and pure liquids are materials whose concentration doesn’t change during the course of a reaction
its amount can change, but the amount of it in solution doesn’t
because it isn’t in solution
Because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant expression
For the reaction aA(s) + bB(aq) Û cC(l) + dD(aq) the equilibrium constant expression is
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28. Heterogeneous Equilibria
The amount of C is different, but the amounts of CO and CO2 remain the same. Therefore the amount of C has no effect on the position of equilibrium.
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29. Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following
HNO2(aq) + H2O(l) Û H3O+(aq) + NO2−(aq) K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) Û CaSO4(s) + 2 HNO3(aq) K = 1 x 104
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30. Calculating Equilibrium Constants from Measured Equilibrium Concentrations
The most direct way of finding the equilibrium constant is to measure the amounts of reactants and products in a mixture at equilibrium
The equilibrium mixture may have different amounts of reactants and products, but the value of the equilibrium constant will always be the same
as long as the temperature is kept constant
the value of the equilibrium constant is independent of the initial amounts of reactants and products
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31. Initial and Equilibrium Concentrations for H2(g) + I2(g) Û 2HI(g) @ 445 °C
Constant
[H2]
[I2]
[HI]
0.50
0.0
0.11
0.78
0.055
0.39
0.165
1.17
1.0
0.5
0.53
0.033
0.934
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32. Calculating Equilibrium Concentrations
Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know initial concentrations and one equilibrium concentration
Use the change in the concentration of the material that you know to determine the change in the other chemicals in the reaction
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33. Calculating Equilibrium Concentrations: An Example
2 A(aq) + B(aq) Û 4 C(aq)
Suppose you have a reaction 2 A(aq) + B(aq) Û 4 C(aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M.
Product C increased by 0.50 mol/L
Because Product C increased by 0.50 mol/L, you must have used ½ that amount of Reactant A, according to the stoichiometry.
Because Product C increased by 0.50 mol/L, you must have used ¼ that amount of Reactant B, according to the stoichiometry.
-½(0.50)
-¼(0.50)
+0.50
0.75
0.88
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34. Example 14.6: Find the value of Kc for the reaction 2 CH4(g) Û C2H2(g) + 3 H2(g) at 1700 °C if the initial [CH4] = M and the equilibrium [C2H2] = M
+0.035
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35. Example 14.6: Find the value of Kc for the reaction 2 CH4(g) Û C2H2(g) + 3 H2(g) at 1700 °C if the initial [CH4] = M and the equilibrium [C2H2] = M
−2(0.035)
+0.035
+3(0.035)
0.045
0.105
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36. Practice –The following data were collected for the reaction 2 NO2(g) Û N2O4(g) at 100 °C. Complete the table and determine values of Kp and Kc.
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37. Practice – A second experiment was done and the following data were collected for the reaction 2 NO2(g) Û N2O4(g) at 100 °C. Complete the table and determine values of Kp and Kc. Compare them to the first experiment.
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38. The Reaction Quotient
If a reaction mixture, containing both reactants and products, is not at equilibrium; how can we determine in which direction it will proceed?
The answer is to compare the current concentration ratios to the equilibrium constant
The concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q
for the gas phase reaction aA + bB Û cC + dD
the reaction quotient is:
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39. The Reaction Quotient: Predicting the Direction of Change
If Q > K, the reaction will proceed fastest in the reverse direction
the [products] will decrease and [reactants] will increase
If Q < K, the reaction will proceed fastest in the forward direction
the [products] will increase and [reactants] will decrease
If Q = K, the reaction is at equilibrium
the [products] and [reactants] will not change
If a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction
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40. Q, K, and the Direction of Reaction
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41. If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse Q
Example 14.7: For the reaction below, which direction will it proceed if PI2 = atm, PCl2 = atm. & PICl = atm
Given:
Find:
for I2(g) + Cl2(g) Û 2 ICl(g), Kp = 81.9
direction reaction will proceed
Conceptual Plan:
Relationships:
If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse Q
PI2, PCl2, PICl
Solution:
I2(g) + Cl2(g) Û 2 ICl(g) Kp = 81.9
because Q (10.8) < K (81.9), the reaction will proceed to the right
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42. Practice – For the reaction CH4(g) + 2 H2S(g) Û CS2(g) + 4 H2(g) Kc = 3.59 at 900 °C. For each of the following measured concentrations determine whether the reaction is at equilibrium. If not at equilibrium, in which direction will the reaction proceed to get to equilibrium?
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43. Example 14. 8: If [COF2]eq = 0. 255 M and [CF4]eq = 0
Example 14.8: If [COF2]eq = M and [CF4]eq = M, and Kc = 1000 °C, find the [CO2]eq for the reaction given.
Given:
Find:
Sort: You’re given the reaction and Kc. You’re also given the [X]eq of all but one of the chemicals
2 COF2 Û CO2 + CF4
[COF2]eq = M, [CF4]eq = M
[CO2]eq
Conceptual Plan:
Relationships:
Strategize: You can calculate the missing concentration by using the equilibrium constant expression
K, [COF2], [CF4]
[CO2]
Solution:
Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amounts
Check: substitute approximations back in to see if the value agrees
Check: 
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44. Practice – A sample of PCl5(g) is placed in a 0
Practice – A sample of PCl5(g) is placed in a L container and heated to 160 °C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of PCl5 if Kc =
PCl5 Û PCl3 + Cl2
if in addition you calculate Kp from Kc you find that it is 2.26, slightly greater than 1 – showing that the equilibrium constant may be unreliable for predicting the position of equilibrium when it is close to 1
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45. Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures
Step 1: decide which direction the reaction will proceed
compare Q to K
Step 2: define the changes of all materials in terms of x
use the coefficient from the chemical equation as the coefficient of x
the x change is + for materials on the side the reaction is proceeding toward
the x change is  for materials on the side the reaction is proceeding away from
Step 3: solve for x
for 2nd order equations, take square roots of both sides or use the quadratic formula
may be able to simplify and approximate answer for very large or small equilibrium constants
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46. Example 14.11: For the reaction I2(g) + Cl2(g) Û 2 25 °C, Kp = If the initial partial pressures are all atm, find the equilibrium concentrations
Qp(1) < Kp(81.9), so the reaction is proceeding forward
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47. Example 14.11: For the reaction I2(g) + Cl2(g) Û 2 25 °C, Kp = If the initial partial pressures are all atm, find the equilibrium concentrations x x
+2x
0.100x
0.100x x
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48. Example 14.11: For the reaction I2(g) + Cl2(g) Û 2 25 °C, Kp = If the initial partial pressures are all atm, find the equilibrium concentrations x x
+2x
0.100x
0.100x x
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49. Example 14.11: For the reaction I2(g) + Cl2(g) Û 2 25°C, Kp = If the initial partial pressures are all atm, find the equilibrium concentrations
−0.0729 x
−0.0729 x
+2(0.0729)
+2x
0.100x
0.027
0.100x
0.027 x
0.246
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50. Example 14.11: For the reaction I2(g) + Cl2(g) Û 2 25 °C, Kp = If the initial partial pressures are all atm, find the equilibrium concentrations
−0.0729
−0.0729
2(0.0729)
0.027
0.027
0.246
Kp(calculated) = Kp(given) within significant figures
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51. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]? (Hint: you will need to use the quadratic formula to solve for x)
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52. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0 and the reaction must proceed forward
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53. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
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54. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
0.500  = 0.498
[I2] = M
2( ) =
[I] = M
x =  
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55. Approximations to Simplify the Math
When the equilibrium constant is very small, the position of equilibrium favors the reactants
For relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium
the [X]equilibrium = ([X]initial  ax)  [X]initial
we are approximating the equilibrium concentration of reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
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56. Checking the Approximation and Refining as Necessary
We can check our approximation by comparing the approximate value of x to the initial concentration
If the approximate value of x is less than 5% of the initial concentration, the approximation is valid
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57. Example14.13: For the reaction 2 H2S(g) Û 2 H2(g) °C, Kc = 1.67 x 10−7. If a L flask initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations.
because no products initially, Qc = 0, and the reaction is
proceeding forward
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58. Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) °C, Kc = 1.67 x 10−7. If a L flask initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations.
2x
+2x +x
2.50E−4
2x 2x x
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59. Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) °C, Kc = 1.67 x 10−7. If a L flask initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations.
2x
+2x +x
2.50E−4
2x 2x
2.50E−4 x
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60. the approximation is not valid!!
Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) °C, Kc = 1.67 x 10−7. If a L flask initially containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations.
2x
+2x +x 2x x
2.50E–4
the approximation is not valid!!
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61. Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) °C, Kc = 1.67 x 10−7. If a L flask initially containing 1.25 x 10−4 mol H2S is heated to °C, find the equilibrium concentrations.
2x
+2x +x
2.50E−4
2x 2x x
xcurrent = 1.38 x 10−5
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62. Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) °C, Kc = 1.67 x 10−7. If a L flask initially containing 1.25 x 10−4 mol H2S is heated to °C, find the equilibrium concentrations.
2x
+2x +x
2.50E−4
2x 2x x
xcurrent = 1.27 x 10−5
since xcurrent = xnew, approx. OK
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63. Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) °C, Kc = 1.67 x 10−7. If a L flask initially containing 1.25 x 10−4 mol H2S is heated to °C, find the equilibrium concentrations.
2x
+2x +x
2.50E−4
2x 2x x
2.24E−4
2.56E−5
1.28E−5
xcurrent = 1.28 x 10−5
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64. Kc(calculated) = Kc(given) within significant figures
Example 14.13: For the reaction 2 H2S(g) Û 2 H2(g) °C, Kc = 1.67 x 10−7. If a L flask initially containing 1.25 x 10−4 mol H2S is heated to °C, find the equilibrium concentrations.
2x
+2x +x
2.24E−4
2.56E−5
1.28E−5
Kc(calculated) = Kc(given) within significant figures
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65. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]? (use the simplifying assumption to solve for x)
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66. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
because [I]initial = 0, Q = 0 and the reaction must proceed forward
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67. the approximation is valid!!
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
the approximation is valid!!
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68. Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3
Practice – For the reaction I2(g) Û 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
0.500  = 0.498
[I2] = M
2( ) =
[I] = M
x =  
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69. Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1
Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is M, what will be the equilibrium concentrations of [N2O4] and [NO2]?
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70. Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1
Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is M, what will be the equilibrium concentrations of [N2O4] and [NO2]?
because [NO2]initial = 0, Q = 0 and the reaction must proceed forward
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71. the approximation is valid!!
Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is M, what will be the equilibrium concentrations of [N2O4] and [NO2]?
without approximation
with approximation
x = 1.82 x 10−4
the approximation is valid!!
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72. Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1
Practice – For the reaction N2O4(g) Û 2 NO2(g) the value of Kc = 1.07 x 10−5 . If the initial concentration of N2O4 is M, what will be the equilibrium concentrations of [N2O4] and [NO2]?
[N2O4]=  1.83 x 10−4
[N2O4] = M
[NO2]=2(1.83 x 10−4)
[NO2] = 3.66 x 10−4 M
x = 1.83 x 10−4  
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73. Disturbing and Restoring Equilibrium
Once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same
However if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is restored
The new concentrations will be different, but the equilibrium constant will be the same
unless you change the temperature
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74. Le Châtelier’s Principle
Le Châtelier's Principle guides us in predicting the effect various changes in conditions have on the position of equilibrium
It says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance
disturbances all involve making the system open
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75. An Analogy: Population Changes
When an influx of population enters Country B
from somewhere outside Country A, it disturbs the
equilibrium established between Country A and Country B.
The result will be people moving from Country B into Country A faster than people moving from Country A into Country B. This will continue until a new equilibrium between the populations is established; the new populations will have different numbers of people than the old ones.
When the populations of Country A and Country B
are in equilibrium, the emigration rates between the
two countries are equal so the populations stay constant.
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76. Disturbing Equilibrium: Adding or Removing Reactants
After equilibrium is established, a reactant is added
as long as the added reactant is included in the equilibrium constant expression
i.e., not a solid or liquid
How will this affect the rate of the forward reaction?
How will it affect the rate of the reverse reaction?
How will it affect the value of K?
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77. Disturbing Equilibrium: Adding Reactants
Adding a reactant initially increases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction
The reaction proceeds to the right until equilibrium is restored
At the new equilibrium position, you will have more of the products than before, less of the non-added reactants than before, and less of the added reactant
but not as little of the added reactant as you had before the addition
At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same
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78. Disturbing Equilibrium: Adding or Removing Reactants
After equilibrium is established, a reactant is removed
as long as the added reactant is included in the equilibrium constant expression
i.e., not a solid or liquid
How will this affect the rate of the forward reaction?
How will it affect the rate of the reverse reaction?
How will it affect the value of K?
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79. Disturbing Equilibrium: Removing Reactants
Removing a reactant initially decreases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction.
so the reaction is going faster in reverse
The reaction proceeds to the left until equilibrium is restored
At the new equilibrium position, you will have less of the products than before, more of the non- removed reactants than before, and more of the removed reactant
but not as much of the removed reactant as you had before the removal
At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same
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80. The Effect of Concentration Changes on Equilibrium
Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found
that has the same K
Removing a product will increase the amounts of the other products and decrease the amounts of the reactants.
you can use this to drive a reaction to completion!
Equilibrium shifts away from side with added chemicals or toward side with removed chemicals
Remember, adding more of a solid or liquid does not change its concentration – and therefore has no effect on the equilibrium
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81. The Effect of Concentration Changes on Equilibrium
When NO2 is added, some of it combines to make more N2O4
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82. The Effect of Concentration Changes on Equilibrium
When N2O4 is added, some of it
decomposes to make more NO2
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83. The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium
Adding a gaseous reactant increases its partial pressure, causing the equilibrium to shift to the right
increasing its partial pressure increases its concentration
does not increase the partial pressure of the other gases in the mixture
Adding an inert gas to the mixture has no effect on the position of equilibrium
does not affect the partial pressures of the gases in the reaction
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84. Effect of Volume Change on Equilibrium
Decreasing the volume of the container increases the concentration of all the gases in the container
increases their partial pressures
It does not change the concentrations of solutions!!
If their partial pressures increase, then the total pressure in the container will increase
According to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure
The way the system reduces the pressure is to reduce the number of gas molecules in the container
When the volume decreases, the equilibrium shifts to the side with fewer gas molecules
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85. Disturbing Equilibrium: Changing the Volume
After equilibrium is established, the container volume is decreased
How will it affect the concentration of solids, liquid, solutions, and gases?
How will this affect the total pressure of solids, liquid, and gases?
How will it affect the value of K?
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86. Disturbing Equilibrium: Reducing the Volume
Decreasing the container volume increases the concentration of all gases, but not solids, liquid, or solutions
same number of moles, but different number of liters, resulting in a different molarity
Decreasing the container volume will increase the total pressure
Boyle’s Law
if the total pressure increases, the partial pressures of all the gases will increase – Dalton’s Law of Partial Pressures
Because the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules
shift toward the side with fewer gas molecules
At the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same
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87. The Effect of Volume Changes on Equilibrium
Because there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the side with fewer molecules to decrease the pressure
When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant side
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88. The Effect of Temperature Changes on Equilibrium Position
Exothermic reactions release energy and endothermic reactions absorb energy
If we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changes
even though heat is not matter and not written in a proper equation
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89. The Effect of Temperature Changes on Equilibrium for Exothermic Reactions
For an exothermic reaction, heat is a product
Increasing the temperature is like adding heat
According to Le Châtelier’s Principle, the equilibrium will shift away from the added heat
Adding heat to an exothermic reaction will decrease the concentrations of products and increase the concentrations of reactants
Adding heat to an exothermic reaction will decrease the value of K
How will decreasing the temperature affect the system?
aA + bB  cC + dD + Heat
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90. The Effect of Temperature Changes on Equilibrium for Endothermic Reactions
For an endothermic reaction, heat is a reactant
Increasing the temperature is like adding heat
According to Le Châtelier’s Principle, the equilibrium will shift away from the added heat
Adding heat to an endothermic reaction will decrease the concentrations of reactants and increase the concentrations of products
Adding heat to an endothermic reaction will increase the value of K
How will decreasing the temperature affect the system?
Heat + aA + bB  cC + dD
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91. The Effect of Temperature Changes on Equilibrium
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92. Not Changing the Position of Equilibrium: The Effect of Catalysts
Catalysts provide an alternative, more efficient mechanism
Catalysts work for both forward and reverse reactions
Catalysts affect the rate of the forward and reverse reactions by the same factor
Therefore, catalysts do not affect the position of equilibrium
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93. Practice – Le Châtelier’s Principle
The reaction 2 SO2(g) + O2(g) Û 2 SO3(g) with DH° = −198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored?
adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
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94. Practice – Le Châtelier’s Principle
The reaction 2 SO2(g) + O2(g) Û 2 SO3(g) with DH° = −198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored?
adding more O2 to the container shift to SO3
condensing and removing SO3 shift to SO3
compressing the gases shift to SO3
cooling the container shift to SO3
doubling the volume of the container shift to SO2
warming the mixture shift to SO2
adding helium to the container no effect
adding a catalyst to the mixture no effect
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