chapter 15 chemical equilibrium. 15.1 cold temp hot temp 15.1
TRANSCRIPT
Chapter 15
Chemical Equilibrium
15.1
15.1
Cold Temp Hot Temp
15.1
15.1
15.1
15.2: Law of Mass Action
Derived from rate laws by Guldberg andWaage (1864) For a balanced chemical reaction
in equilibrium:
a A + b B ↔ c C + d D Equilibrium constant expression (Keq):
ba
dc
c [B] [A]
[D] [C] K b
Ba
A
dD
cC
p )(P)(P
)(P)(PK
Keq is strictly based on stoichiometry of the reaction (is independent of the mechanism).
Units: Keq is considered dimensionless (no units)
Cato Guldberg Peter Waage
(1836-1902) (1833-1900)
or
15.2 **Relates Kc to Kp
For Example
Write the equilibrium expression Kc for the following reactions:
15.2
Example #2
In the synthesis of ammonia from nitrogen and hydrogen,
Kc = 9.60 at 300°C:
N2(g) + 3H2(g) 2NH3(g)
Calculate Kp for this reaction at this temperature
Hint..you will need to use..
15.2
What does the Equilibrium constant tell us??
15.2
Looking at reversible reactions So far we have only written the expression forwards, but it can also
be written equally backwards! The constants are the recipricals of each other
Kc = 0.212 Kc = 4.72
15.2
15.3: Types of EquilibriaHomogeneous: all components in same phase
(usually g or aq)
N2 (g) + H2 (g) ↔ NH3 (g)
3H
1N
2NH
P )(P)(P
)(PK
22
3
bB
aA
dD
cC
P )(P)(P
)(P)(PK
3 2 1
Fritz Haber(1868 – 1934)
15.3
Heterogeneous: different phases
CaCO3 (s) ↔ CaO (s) + CO2 (g)
Definition: What we use:
][CaCO
)(P [CaO] K
3
COeq
22COp P K
Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium.
Concentrations of pure solids and pure liquids are not included in Keq expression because their concentrations do not vary, and are “already included” in Keq
15.3
For Example
Write equilibrium-constant expressions for Kc and Kp for each of the following reactions:
15.3
Clickfor answers
15.4: Calculating Equilibrium Constants
Steps to use “ICE” table:1. “I” = Tabulate known initial and equilibrium
concentrations of all species in equilibrium expression
2. “C” = Determine the concentration change for the species where initial and equilibrium are known
• Use stoichiometry to calculate concentration changes for all other species involved in equilibrium
3. “E” = Calculate the equilibrium concentrations
Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is 0.0124 M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at 25ºC for the reaction:
NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)
NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)
Initial
Change
Equilibrium
][NH
]][OH[NH K
3
-114
c
0.0124 M
- x
0.0119 M
0 M 0 M
+ x + x
4.64 x 10-4 M 4.64 x 10-4 M
5-2-4
101.81x 0.0119
)10(4.64
NH3 (aq)H2O
(l)NH4
1+ (aq) OH1- (aq)
XXX
x = 4.64 x 10-4 M
Ex: A 5.000-L flask is filled with 5.000 x 10-3 mol of H2 and 1.000 x 10-2 mol of I2 at 448ºC. The value of Keq is 1.33. What are the concentrations of each substance at equilibrium?
H2 (g) + I2 (g) ↔ 2 HI (g)
H2 (g) + I2 (g) ↔ 2 HI (g)
Initial
Change
Equilibrium
]][I[H
[HI] K
22
2
c
1.000x10-3 M
- x M
(1.000x10-3 – x) M
2.000x10-3 M 0 M
- x M + 2x M
(2.000x10-3 – x) M 2x M
33.1x)-10x)(2.000-10(1.000
(2x)3-3-
2
4x2 = 1.33[x2 + (-3.000x10-3)x + 2.000x10-6]
0 = -2.67x2 – 3.99x10-3x + 2.66x10-6
Using quadratic eq’n: x = 5.00x10-4 or –1.99x10-3; x = 5.00x10-4
Then [H2]=5.00x10-4 M; [I2]=1.50x10-3 M; [HI]=1.00x10-3 M
H2 (g) I2 (g) HI (g)
15.6 Le Chatelier’s Principle
Changes that do not affect Keq:
1. Concentration Upon addition of a reactant or product, equilibrium shifts to re-
establish equilibrium by consuming part of the added substance.
Addition of solids/liquids do not appreciably shift the system and can be ignored. But they are still made/consumed!
Upon removal of reactant or product, equilibrium shifts to re-establish equilibrium by producing more of the removed substance.
Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6
H2O (l)
•Add HCl, temporarily inc forward rate
Volume, with a gas present (T is constant)
Upon a decrease in V (thereby increasing P),equilibrium shifts to reduce the number of moles of gas.
Upon an increase in V (thereby decreasing P),equilibrium shifts to produce more moles of gas.
Ex: N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) If V of container is decreased, equilibrium shifts right.
XN2 and XH2 dec
XNH3 inc
3HN
2NH
P )(PP
)(PK
22
33
THTN
2TNH
)P)(XP(X
)P(X
22
3 4T
3HN
2T
2NH
P)(XX
P)(X
22
3 2T
3HN
2NH
P)(XX
)(X
22
3
23HN
2NH
P )(
)(K
22
3
Since PT also inc, KP remains constant.
3. Pressure, but not Volume
Usually addition of a noble gas, p. 560 Avogadro’s law: adding more non-reacting particles “fills in”
the empty space between particles. In the mixture of red and blue gas particles, below, adding
green particles does not stress the system, so there is no Le Châtelier shift.
Catalysts Lower the activation energy of both forward and
reverse rxns, therefore increases both forward and reverse rxn rates.
Increase the rate at which equilibrium is achieved, but does not change the ratio of components of the equilibrium mixture (does not change the Keq)
Energy
Rxn coordinate
Ea, uncatalyzed
Ea, catalyzed
Change that does affect Keq:
Temperature: consider “heat” as a part of the reaction Upon an increase in T, endothermic reaction is favored
(equilibrium shifts to “consume the extra heat”) Upon a decrease in T, equilibrium shifts to produce more
heat.Effect on Keq
1. Exothermic equilibria: Reactants ↔ Products + heat
• Inc T increases reverse reaction rate which decreases Keq
2. Endothermic equilibria: Reactants + heat ↔ Products
• Inc T increases forward reaction rate increases Keq
Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l);
H=+?•Inc T temporarily inc forward rate•Dec T temporarily inc reverse rate
Effect of Various Changes on Equilibrium
Disturbance Net Direction of Rxn
Effect of Value of K
Concentration
Increase (reactant) Towards formation of product
None
Decrease(reactant) Towards formation of reactant
None
Increase (product) Towards formation of reactant
None
Decrease (product) Towards formation of product
None
Effect of Pressure on Equilb.
Pressure
Increase P
(decrease V)
Towards formation of fewer moles of gas
None
Decrease P
(Increase V)
Towards formation of more moles of gas
None
Increase P
( Add inert gas, no change in V)
None, concentrations unchanged
None
Disturbance Direction of Reaction Effect of K
Effect of Temperature on Equilb
Temperature
Increase T
Towards absorption of heat
Increases if endothermic
Decreases if exothermic
Decrease T Towards release of heat
Increases if exothermic
Decreases if endothermic
Catalyst Added None, forward and reverse equilibrium attained sooner
None
Disturbance Direction of Reaction Effect of K
For ExampleFor the following reaction
5 CO(g) + I2O5(s) I2(g) + 5 CO2(g) + 1175 kJ
for each change listed, predict the equilibrium shift and the effect on the indicated quantity.
Change Directionof Shift
( ; ; or no change)
Effect onQuantity
Effect(increase, decrease,
or no change)
(a) decrease in volume Kc No Change
(b) raise temperature amount of CO(g) Increase
(c) addition of I2O5(s) No Change amount of CO(g) No Change(d) addition of CO2(g) amount of I2O5(s) Increase
(e) removal of I2(g) amount of CO2(g) Increase