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Chemical Equilibrium
Chapter 14
Materials are referenced to Chemistry
7th ED. by Chang, 2001
Rate Laws and Reverse Reactions
 | Reactions don’t always occur in a single direction. |
| Consider the decomposition of ozone |
|
k1 |
| O3(g)
<====> O2(g) + O(g) |
|
k-1 |
| Forward Rate = k1[O3] |
| Reverse Rate = k-1[O2][O] |
Chemical Equilibrium
| If the rate of the forward and reverse reactions are equal , a dynamic
equilibrium is achieved. |
| Forwar Rate = Reverse
Rate |
| No change in Products or Reactants is observed |
Chemical Equilibrium
| If the rate of the forward and reverse reactions are equal, a dynamic
equilibrium is achieved. (Draw graph from Board) |
Change in Rate to Achieve Equilibrium
| At equilibrium the rates of the forward and reverse reactions are the
same. (Draw graph from Board) |
Change in Concentration To Achieve Equilibrium
| At equilibrium both forward and reverse reactions continue! (Draw graph from Board) |
Change in Concentration To Reach Equilibrium
| At equilibrium both forward and reverse reactions continue! But the
overall concentration doesn’t change. (Draw graph from Board) |
An Equilibrium Constant
| Similar to the rate law, we can determine the equilibrium constant that
describes the final concentrations achieved by a system at equilibrium. |
| jA
+ kB <=====> lC + mD |
The Law of Mass Action
| Similar to the rate law, we can determine the equilibrium constant that
describes the final concentrations achieved by a system at equilibrium. |
|
K
= Products / Reactants =
[ C]l
[ D ]m / [
A]j [ B]k |
Example: Writing an Equilibrium-Constant
Expression
| CO(g) + 3H2(g) CH4(g) + H2O(g) |
| Kc = [CH4][H2O] / [CO][H2]3 |
Example: continued
| b) What if the reaction is reversed? |
| CH4(g) + H2O(g) CO(g) + 3H2(g)
|
| Kc = [CO][H2]3 / [CH4][H2O] |
Example: c)
| N2 (g) + 3H2 (g) 2NH3(g) |
| Kc = [NH3]2 / [N2][H2]3 |
Example: d)
| What if the equation in c) where written : |
| 1/2N2 (g) + 3/2H2 (g) NH3(g) |
| Kc = [NH3] / [N2]1/2[H2]3/2 |
Kc for
a Heterogeneous Reaction
lCH4(g) +
H2O(l)
= CO(g) +
3H2(g)
lKc
= [CO][H2]3/l
[CH4]
lNote
that since the concentration of a pure liquid is not going to change
significantly, its effect is included as part of the Kc value.
Equilibrium Constants from Reaction Compositions
| 2HI(g) H2(g) + I2(g) |
| When 4.00 mol HI was placed in a 5.00 L vessel at 458oC, the
equilibrium mixture was found to contain 0.442 mol I2 . What is the value of Kc
for the decomposition of HI at this temperature ? |
Solution
| First, use the mole values and volume given to find the concentrations. |
| [HI] = 4.00mol = 0.800 M |
| 5.00 L |
| [I2] = 0.442 mol = 0.0884 M |
| 5.00 L |
Solution Setup
| 2HI(g)
H2(g)
+ I2(g) |
| S 0.800
0
0 |
| C -2x
x
x |
| E 0.800-2(0.884)
0.0884
0.0884 |
|
(0.623) |
| Next write the Kc equation |
Kc Equation
| Kc = [H2][I2] / [HI]2 |
| Kc = (0.0884)(0.0884) = 0.0201 |
| (0.623)2 |
The Equilibrium Constant, Kp
| Remember: M(n/V) is related to P by the Ideal gas law; PV = nRT
Therefore, while Kc and Kp are not necessarily equal, they are
proportional and the equations are similar. |
| For the equation: |
| CO(g) + 3H2(g) CH4(g) + H2O(g) |
Kp
| Kp = PCH4PH2O / PCOPH23 |
| Kp = Kc(RT) n |
| /\n = sum of gaseous products - sum of gaseous reactants |
Direction of Equilibria from K
| We can use the current conditions/concentrations of a system to determine
the direction that the system will respond. |
| Q = K : System is at equilibrium |
| Q > K : System shifts left |
| Q < K : System shifts right |
Working a Problem
| The formation of hydrogen iodide occurs according to the equilibrium
shown below. H2(g) + I2(g) 2HI(g) KC=60 (350 oC) |
| Is the following set of concentrations at equilibrium? Which way will the
reaction shift? |
| [H2] = [I2] = [HI] = 0.010 M |
The Solution
| After making sure the equation is balanced, determine the equilibrium
quotient. |
The Solution
| After making sure the equation is balanced, determine the equilibrium
quotient. |
The Solution
| After making sure the equation is balanced, determine the equilibrium
quotient. |
Equilibrium Constant
The following reaction has an equilibrium constant Kc equal to 3.07 x
10-4 at 24oC.
2NOBr(g) 2NO(g) + Br2(g)
For the following composition, decide whether
the reaction mixture is at equilibrium. If it is
not, decide which direction the reaction
should go:
Equilibrium Constant
Equilibrium Constant
Equilibrium Constant
Equilibrium Composition
| The equilibrium constant Kc for the reactionPCl3(g)
+ Cl2 (g) PCl5 (g)equals 49 at 230 oC. If 0.500 mol of
PCl3 and Cl2 are added to a 5.0 L vessel, what is the equilibrium
composition of the mixture at 230 oC? |
| Convert moles to concentration:0.500 mol/ 5.0 L = 0.10 M = [PCl3]
= [Cl2] |
Equilibrium Composition
| Assemble the table for change in concentration. PCl3(g) + Cl2
(g) PCl5 (g) |
Equilibrium Composition
| Assemble the table for change in concentration. PCl3(g) + Cl2
(g) PCl5 (g) |
Equilibrium Composition
| Assemble the table for change in concentration. PCl3(g) + Cl2
(g) PCl5 (g) |
| Rearranging and solving for X yields: |
Equilibrium Composition
| Assemble the table for change in concentration. PCl3(g) + Cl2
(g) PCl5 (g) |
| Rearranging and solving for X yields: |
Equilibrium Composition
| Assemble the table for change in concentration. PCl3(g) + Cl2
(g) PCl5 (g) |
| Rearranging and solving for X yields: |
| Thus, [PCl3] = [Cl2] = 0.036 M and [PCl5]
= 0.064 M |
| OR, PCl3 = Cl2 = 0.18 mol and PCl5 =
0.32 mol |
Le Chatelier’s Principle
"A system at equilibrium reacts to a disturbance in T, P, or M by shifting the
equilibrium in a way that tends to counteract this change in variable."
A Problem to Review
| In the reaction |
6H2O (g) + 6 CO2 (g) C6H12O6
(s) + 6 O2 (g)
A Problem to Review
| In the reaction |
6H2O (g) + 6 CO2 (g) C6H12O6
(s) + 6 O2 (g)
A Problem to Review
In the reaction
6H2O (g) + 6 CO2 (g) C6H12O6
(s) + 6 O2 (g)
A Problem to Review
| In the reaction |
6H2O (g) + 6 CO2 (g) C6H12O6
(s) + 6 O2 (g)
A Problem to Review
| In the reaction |
6H2O (g) + 6 CO2 (g) C6H12O6
(s) + 6 O2 (g)
A Problem to Review
| In the reaction |
6H2O (g) + 6 CO2 (g) C6H12O6
(s) + 6 O2 (g)
A Problem to Review
| In the reaction |
6H2O (g) + 6 CO2 (g) C6H12O6
(s) + 6 O2 (g)
A Problem to Review
| In the reaction |
6H2O (g) + 6 CO2 (g) C6H12O6
(s) + 6 O2 (g)
A Problem to Review
In the reaction
6H2O (g) + 6 CO2 (g) C6H12O6
(s) + 6 O2 (g)
A Problem to Review
| In the endothermic reaction |
6H2O (g) + 6 CO2 (g) C6H12O6
(s) + 6 O2 (g)
A Problem to Review
| In the endothermic reaction |
6H2O (g) + 6 CO2 (g) C6H12O6
(s) + 6 O2 (g)
A Problem to Review
| In the endothermic reaction |
6H2O (g) + 6 CO2 (g) C6H12O6
(s) + 6 O2 (g)
Temperature and Equilibrium
Equilibrium constant changes with T
The terms endothermic and exothermic are used in reference to the REACTANTS.
With increasing temperature :
Exothermic reaction shifts LEFT
Endothermic reaction shifts RIGHT
Catalyst
lSince
a catalyst changes the rate of reaction by lowering activation energy, it
changes both the forward and reverse rates equally.
lThe
catalyst does not shift the position of equilibrium for a system, it merely
allows the equilibrium to be achieved sooner.
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