Transcript Document 7931448

I.
The Reaction Quotient
1.
2.
We can write K only for a reaction at equilibrium
How can we describe a reaction at other times? N2 + 3H2
2NH3
2
Q
3.
4.
5.
[NH3 ]0
3
[N 2 ]0 [H 2 ]0
Describes the reaction when not at equilibrium
Q = reaction quotient has the same form as K, but is at non-equilibrium times.
Q can tell us how a reaction will change to get to equilibrium
1) If Q = K, we are at equilibrium
2) If Q > K, then products > reactants and the reaction will shift left
3) If Q < K, then reactants > products and the reaction will shift right
Q
K
Predicted Direction of Reaction
0.55
1.45
To the right (towards products)
2.55
1.45
To the left (towards reactants)
1.45
1.45
No change (at equilibrium)
Example: For the above reaction, K = 6 x 10-2. Predict change:
1) [N2]0 = 1 x 10-5 M, [H2]0 = 0.002 M, [NH3]0 = 0.001M
2) [N2]0 = 1.5 x 10-5 M, [H2]0 = 0.354 M, [NH3]0 = 0.0002M
3) [N2]0 = 5 M, [H2]0 = 0.01 M, [NH3]0 = 0.0001 M
II.
Solving Equilibrium Problems
A.
Problems where equilibrium concentrations (or pressures) are given
1. Example: N2O4(g)
2NO2(g)
a) KP = 0.133
b) P(N2O4) at equilibrium = 2.71 atm
c) Find P(NO2) at equilibrium
2.
B.
Example: PCl5(g)
PCl3(g) + Cl2(g)
a) [PCl3]0 = 0.298 M, [PCl5]0 = 0.0087 M
b) At equilibrium, [Cl2]0 = 0.002 M
c) Calculate all concentrations and K
Procedure when not given equilibrium concentrations
1. Write the balanced equation
2. Write the equilibrium expression
3. List the initial conditions
4. Calculate Q to decide which way the reaction will shift
5. Write equilibrium conditions using x as unknown changes in concentration
6. Substitute the equilibrium conditions into the equilibrium expression and
solve for x
7. Check your answer to see if it gives the correct value of K
A.
Example: 3 mol of hydrogen and 6 mol of fluorine are mixed in a 3 L flask to
give hydrogen fluoride gas. K = 115. Find the equilibrium concentrations.
1. Write the balanced equation: H2 + F2
2HF
[HF] 2
K
 115
[H 2 ][F2 ]
2.
Write the equilibrium expression:
3.
List the initial conditions: [H2]0 = 1 M, [F2]0 = 2 M, [HF]0 = 0
4.
Calculate Q to decide which way the reaction will shift
2
[HF]0
02
Q

0
[H 2 ]0 [F2 ]0 1 x 2
5.
Write equilibrium conditions using x as unknown changes in
concentration
Initial
Change
Equilibrium
[H2] = 1
-x
1-x
[F2] = 2
-x
2-x
[HF] = 0
+2x
2x
Shorthand:
H2
Initial:
1
Change
-x
Equilibrium 1-x
•
+
F2
2
-x
2-x
2HF
0
+2x
2x
Substitute and solve for x
[HF] 2
[2x] 2
4x 2
K

 2
 115
[H 2 ][F2 ] [1 - x][2 - x] x - 3x  2
4x 2  115x 2  345x  230
111x 2  345x  230  0
- b  b 2 - 4ac
Quadratic Equation : x 
2a
a = 111, b = -345, c = 230
x = 2.14 M or x = 0.968 M
Reject x = 2.14 M since [H2] = 1 – x = 1 – 2.14 = -1.14 M
•
x = 0.968M
Check your answer to see if it gives the correct value of K
[HF]2
[2(0.968)] 2
K

 113
[H 2 ][F2 ] [1 - 0.968][2 - 0.968]
B.
Example: CO(g) + H2O(g)
CO2 + H2
K = 5.1
Find equilibrium conditions if 1 mol of each is mixed in a 1L flask.
C.
Example: 3 mol of hydrogen and 3 mol of fluorine are mixed in a
with 3 mol hydrogen fluoride gas. K = 115. Find the equilibrium
concentrations. H2 + F2
2HF
D.
Example: H2 + I2
2HI KP = 100. If we mix HI (P = 0.5 atm), H2
(P = 0.01 atm), and I2 (P = 0.005) in a 5 L flask, what are eq. conc.?
E.
Approximations when x is small
2NOCl(g)
2NO + Cl2(g)
Initial
Equil.
0.5 M
0.5 – 2x
0
2x
0
x
1.5 L flask
K = 1.6 x 10-5
Since K is small, x will be very small.
[NO]2 [Cl2 ] (2x) 2 (x) (2x) 2 (x) 4x 3
5
K




1.6x10
[NOCl]2
(0.5  x) 2
(0.5) 2
0.25
x = 1 x 10-2
[NO]2 [Cl 2 ] (2 x10 -2 ) 2 (1 x10 -2 )
5
K


1.6x10
[NOCl]2
(0.5) 2
II.
Le Chatelier’s Principle
A.
We can change the position of an equilibrium by changing conditions
1. N2 + 3H2
2NH3
2. Low temperature and high pressure favor the forward reaction
Temp. (oC)
300 atm
400 atm
500 atm
400
48% NH3
55% NH3
61% NH3
500
26% NH3
32% NH3
38% NH3
600
13% NH3
17% NH3
21% NH3
Le Chatelier’s Principle: When a chemical system at equilibrium is
disturbed, the system shifts in a direction that minimizes the disturbance.
Effect of change in concentrations N2 + 3H2
2NH3 K = 0.0596
1) At equilibrium, [N2] = 0.399 M, [H2] = 1.197 M, [NH3] = 0.202 M
2) Let’s add 1 M N2. How does the equilibrium shift?
3.
B.
[NH3 ]2
(0.202) 2
Q

 0.017  K
3
2
[N 2 ][H 2 ]
(1.399)(1.197)
a)
b)
c)
Too much N2 for equilibrium
Reaction must adjust to get back to equilibrium
Reaction must shift to the right
Eq. shifts to the right
3.
We could do the x calculation and find the new concentrations if we wanted, but
sometimes you just want to know which way the reaction will shift. Le
Chatelier’s principle lets us do that simply.
4.
Another Example:
What would happen if we add N2O4?
5.
C.
Example: As4O6(s) + 6C(s)
As4(g) + 6CO(g)
a) Add CO
b) Add or remove As4O6(s)
c) Remove As4(g)
Effect of Change in Pressure/Volume
1. If we add/remove reactant/product, we change concentration (see above)
2. Adding an inert gas, doesn’t change concentrations, so no effect on equilibrium
3. Changing the size of the flask:
V↓ = P↑
V↑ = P↓
4.
a)
b)
The volume changes, so the concentrations must change
To re-establish equilibrium, the concentrations will adjust
i. Fewer gas particles are favored at higher pressures
ii. This reduces the pressure (Le Chat’s principle)
c)
N2(g) + 3H2(g)
2NH3(g) Increase the pressure
i. There are 4 particles on the left and 2 on the right
ii. To reduce the pressure, the equilibrium will shift right
d)
If we increased the volume and decreased the pressure, the reaction
will shift to increase the number of particles (shift left)
Example: Reduce the volume on the following reactions:
a) P4(s) + 6Cl2(g)
4PCl3(l)
b)
PCl3(g) + Cl2(g)
c)
PCl3(g) + 3NH3(g)
PCl5(g)
P(NH2)3(g) + 3HCl(g)
D.
Effect of change in Temperature
1. The above changes (P, Concentration) effect the equilibrium position, but not the
equilibrium constant K
2. Changing Temperature changes the equilibrium constant K
3. We can predict the changes by treating heat as a product or reactant of every
reaction
4.
5.
6.
Exothermic reactions produce heat as a product
a) Adding heat (increasing temp.) shifts away from heat, to left, K decreases
b) Removing heat (decreasing temp.) shifts towards heat, to right, K increases
c) N2 + 3H2
2NH3 + 93 kJ/mol (DH = - 93 kJ/mol)
d) Temperature (oC): 500
600
700
800
K: 90
3
0.3
0.04
Endothermic reactions require heat as a reactant
a) Add heat = shift to right, K increases
b) Remove heat = shift to left, K decreases
c) 556 kJ/mol + CaCO3(s)
CaO(s) + CO2(g) (DH = + 556 kJ/mol)
Examples:
N2(g) + O2(g)
2SO2(g) + O2(g)
2NO(g) DH = 181 kJ/mol
2SO3 DH = - 198 kJ/mol)