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1
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2
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- Most reactions are capable of going in the forward and the reverse
directions at one time.
- N2O4(g) ¾®
2 NO2(g)
- 2 NO2(g) ¾®
N2O4(g)
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3
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- A reversible reaction will have two reaction rates: a forward reaction rate and a reverse
reaction rate.
- When the forward reaction rate is larger than the reverse reaction rate,
product concentrations increase and reactant concentrations decrease.
- When the reverse reaction rate is larger than the forward reaction rate,
product concentrations decrease and reactant concentrations increase.
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4
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- When the forward reaction rate equals the reverse reaction rate, reactant and
product concentrations do not change with time.
- This condition is called equilibrium.
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5
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6
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7
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- When the forward reaction rate equals the reverse reaction rate for a
reaction.
- At equilibrium, reactant and product concentrations do not change with
time.
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8
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- Consider a reversible reaction in which each direction represents a
unimolecular elementary step.
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9
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- Whether you start with all reactants (for example, 1.0 M N2
and 3.0 M H2) or all products (for example, 2.0 M NH3),
the same relative amounts of reactants and products will be present at
equilibrium at the same temperature.
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10
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11
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- Definition:
- When the forward reaction rate of a reaction is equal to the reverse
reaction rate.
- Consequence of the definition:
- Reactant and product concentrations do not change with time at
equilibrium.
- There are other conditions (namely, steady-state) in which reactant and
product concentrations do not change with time but the reaction is not
at equilibrium.
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12
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- In 1864, Guldberg and Waage postulated the law of mass action:
- For any reaction at equilibrium:
- the equilibrium condition is expressed by an constant, Kc:
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13
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- Find the equilibrium concentrations of all species at equilibrium and
use the law of mass action.
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14
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15
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- Will always be a positive number.
- The exponents on the equilibrium expression are totally dependent on the
reaction’s stoichiometry (unlike kinetics).
- Values can range from very small to very large numbers.
- Values of Kc will change with temperature only.
- For reasons that will not be addressed, we will not worry about the
units on an equilibrium constant.
Kc deals with molar concentrations of reactants and
products.
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16
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- For a reaction in the gas phase, you may evaluate a value of Kc
in terms of molar concentrations or you can evaluate a value of Kp
in terms of the partial pressures of each gas in units of atm’s.
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17
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18
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- Can be derived using the ideal gas law:
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19
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- What is Kp at 1273°C for the reaction:
- If Kc = 2.24 ´ 1022
at the same temperature?
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20
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- Size of the equilibrium constant
- K may be equal to a very large number or a very small number.
- The size of the value of K tells you something about the composition of
the equilibrium mixture.
- Very large value of K: mostly
products at equilibrium.
(Equlibrium lies to the right)
- Very small value of K: mostly
reactants at equilibrium. (Equilibrium lies to the left)
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21
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22
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- If you reverse the direction of a reactions, the value of the
equilibrium constant goes to its inverse.
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23
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- If you multiply a chemical equation by a number, the equilibrium
constant is raised to that number’s power.
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24
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- When two reactions whose equilibrium constants are known are added, the
equilibrium constant for the sum is equal to the product of the
equilibrium constants.
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25
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- The equilibrium constant of a reaction in the reverse direction is the
inverse of the equilibrium constant in the forward direction.
- The equilibrium constant of a reaction that has been multiplied by a
number is the equilibrium constant raised to a power equal to that
number.
- The equilibrium constant for a reaction that is the sum of two equations
whose equilibrium constants are known is equal to the product of the two
equilibrium constants.
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26
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- The following equilibrium constants were determined at 1123 K:
- Use this information to calculate the value of Kp for the
following reaction:
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27
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- Many reactions will have species in different phases. These reactions are called heterogeneous
reactions.
- For a species in the gas-phase, they may be represented in equilibrium
expressions either by molar concentration or partial pressures in
atm’s.
- For a dissolved solute (aqueous), they are represented in equilibrium
expression by molar concentrations only.
- For pure solids, pure liquids, and solvents, they do not appear in
equilibrium expressions because their concentrations remain essentially
constant over the course of the reaction.
- Pure solids ((s)) and liquids ((l)) will not appear in equilibrium
expressions.
- Dissolved solutes ((aq) or (sln)) appear as molar concentrations.
- Gases ((g)) appear as molar concentration or partial pressure in atm’s.
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28
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29
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- If the values of all the equilibrium concentrations or pressures are
known, then the equlibrium constant can be evaluated directly.
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30
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- Consider the reaction:
- If the equilibrium partial pressures of N2, O2,
and NO are 0.15, 0.33, and 0.050 atm, respectively at 2200°C, what is
the value of Kp?
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31
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- The value of Kc is 0.52 at 686°C for the reaction:
- What is the equilibrium concentration of CO if the equilibrium
concentrations of H2O, CO2, and H2 are
equal to 0.040 M, 0.050 M and 0.045 M, respectively.
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32
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- You are given the initial concentrations or pressures of all species.
- Given the value of one equilibrium concentration or pressure, calculate
the value of the equilibrium constant.
- Given the value of the equilibrium constant, calculate the equilibrium
concentrations or pressures.
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33
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- Procedure
- Find the initial pressures or concentrations. (This may require a calculation).
- Make up an “equilibrium table” below the chemical equation with the
rows “Initial”, “Change”, and “Equilibrium”.
- Fill the table with the data given in the problem.
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34
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- Consider the reaction:
- Initially, an evacuated flask was filled with 0.750 atm of PCl5
and allowed to equilibrate. If
the equilibrium partial pressure of Cl2 was equal to 0.505
atm, what is the value of Kp for this reaction.
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35
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36
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37
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- Add the initial pressures and change in pressures to obtain equilibrium
pressures.
- Use these equilibrium pressures to evaluate the value of Kp.
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38
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39
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- In the second type of equilibrium problem, you are given the initial
conditions and the value of the equilibrium constant and asked for the
equilibrium conditions.
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40
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- For the reaction:
- the value of Kc is 0.534 at 700°C. Calculate the equilibrium
concentration of H2 if initially [CO] = [H2O] =
0.0300 M.
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41
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42
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- Once the value of one change is known, all the other changes can be
calculated.
- D[CO] = + x (same side, same coefficient)
- D[CO2] = D[H2O] = –x (opposite side, same coefficient)
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43
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- Add the initial and change to obtain expressions for the equilibrium
concentrations in terms of x.
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44
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- Place these expressions into the equilibrium expression and solve for x.
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45
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46
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- A toxicologist studying mustard gas, S(CH2CH2Cl)2,
a blistering agent, prepares a mixture of 0.675 M SCl2 and
0.974 M C2H4 and allows it to react at room
temperature:
- At equilibrium, [S(CH2CH2Cl)2] = 0.350 M. Calculate the value of Kc
for this reaction.
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47
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- The equilibrium constant, Kp = 5.88 ´ 10-3 for the reaction:
- Suppose an evacuated flask is filled with 0.500 atm of N2O4
and allowed to equilibrate. What
is the equilibrium partial pressures of both NO2 and N2O4?
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48
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- The value of Kp is 0.215 at 400°C for the reaction:
- Suppose some solid NH4I(s) is placed in a flask at 400°C and
allowed to equilibrate. What are
the equilibrium partial pressures of HI and NH3 above the
solid?
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Notes
