Download (General Equilibrium) Part 1

CHAPTER 14 - CHEMICAL EQUILIBRIUM
Introduction
Most reactions we have written so far go to completion.
e.g. 2 Na(s) + Cl2(g)  2 NaCl(s)
(single right arrow)
What happens when reactants and products are of similar stabilities?
Example:
2 SO2
+
O2
sulfur dioxide
oxygen
Which direction?

2 SO3
sulfur trioxide
______________________
- reaction in which the
product molecules can recombine to form reactant molecules.
1. Substances on the left side of the equation: ___________
2. Substances on the right side of the equation: __________
3. Use a ____________ arrow to indicate a reaction that is
_________________ .
Concentration
As soon as you start the reaction, a little bit of the product goes back to being reactants.
With time, the concentration of reactant decreases and the concentration of product
increases until both concentrations level off at constant, equilibrium values.
Reactants start high and go
___________ .
Products start low and go
____________ .
They reach a __________
value when equilibrium is reached.
The composition of the equilibrium mixture is constant, but
the reaction __________________________
!
You can start with products, but end up with the same mixture either way.
EQUILIBRIUM 14-1
It is not necessary for the concentrations of reactants and products at equilibrium to be
equal.
• The ____________ to which the forward or reverse reaction is favored
over the other is a characteristic property of a reaction under given conditions.
C
o
n
c
e
n
t
r
a
t
i
o
n
Eq
time
Eq
time
Eq
time
Rate
As soon as you start the reaction, a little bit of the product goes back to being reactants.
You start with:
a ___________ forward reaction rate and
a ___________ reverse reaction rate.
At Equilibrium
1. Rates of the forward and reverse reactions are ___________ .
EQUILIBRIUM 14-2
2. This is a _______________ rather than a static process.
The concentrations of products and reactants are constant, but the individual molecules
making up the reactants and products are constantly _____________ .
Reaction Stoichiometry
The balanced chemical equation can help us to determine the composition at equilibrium
if we know the starting concentrations and the equilibrium concentration of at least one
reactant or product.
LP# 1. EXAMPLE:
You place 0.600
mol of N2 and 1.800 mol of H2
into a 2.00L reaction vessel at 450 C and 10.0 atm. The reaction is:
N2(g)
+
3 H2(g)

2 NH3(g)
What is the composition of the equilibrium mixture if you obtain 0.048 mol of ammonia?
N2(g)
3 H2(g)
(I) Initial
(C) Change
(E) Equilib.
Equilib.
EQUILIBRIUM 14-3

2 NH3(g)
The Equilibrium Constant Kc
When substances are in a dynamic equilibrium, a certain ratio of products and reactants
remains a constant, regardless of the absolute value of those concentrations. This ratio is
expressed as follows:
1. For the general reversible reaction: a
c
A + bB

cC + dD
d
C D
Kc 
b
a
A B
All concentrations are equilibrium conc.
Kc = equilibrium constant
 [ ] means the molar concentration of.
 Value is temperature dependant, but is constant at a given temperature.
(Must specify the temperature for which Kc is given.)
 Kc does not depend on initial concentrations.
 Kc is a unitless value.
LP# 2. EXAMPLE:
Determine the value of Kc for the previous reaction between N2 and H2.
N2(g)
+
3 H2(g)

2 NH3(g)
Write the equilibrium constant expression for the reaction and calculate the value of Kc
at 210o C
the equilibrium concentrations are:
[N2] =
[H2] =
[NH3] =
Lets see that ICE tables work in units of molarity as well as moles.
EQUILIBRIUM 14-4
LP# 3. EXAMPLE:
2 H2S(g)  2 H2(g) + S2(g)
For the reaction:
When 0.100
mol H2S was put into a 10.0 L vessel and heated to 1132 C, it gave
an equilibrium mixture containing 0.0285 mol H2. What is the value of Kc at this
temp?
1. Calculate the molar conc. of the H2S =
2. Calculate the molar conc. of the H2 produced =
ICE table:
2 H2S(g)

2 H2(g)
S2(g)
(I) Initial (M)
(C) Change (M)
(E) Equilib.(M)
Equilib.
Kc =
Kc =
Heterogeneous Equilibria
A. Homogeneous equilibria - reactants and products are in a single phase.
B. Heterogeneous equilibria - reactants and products are present in multiple phases.
1. Solids and Liquids- molar concentrations are constants.
a. They can be calculated from the densities and molar masses.
b. This is independent of its amount.
EQUILIBRIUM 14-5
2. water- molar concentration is constant in aqueous solutions. (55.6M)
(This is not true in gas phase reactions that produce water.)
3. Concentrations of pure solids or pure liquids are _________
(their activity is set to “1”.) when writing the equilibrium equation for any
heterogeneous equilibrium.
-The value for the concentrations of solids and liquids are incorporated into the
value of Kc.
C. Units for Kc – Kc is expressed without units. This is because formally, the value
that is put into the Kc expression is not the molar concentration, but the ratio of the
concentration to a reference concentration of 1 M (The 1 is an integer with infinite sig
figs.)
e.g. 1.5M = 1.5 with no units
1M
For the Reaction: 2
NaHCO3(s)  Na2CO3(s) +H2O(g) + CO2(g)
Originally: Kc = [Na2CO3(s)] [H2O(g)] [CO2(g)]
[NaHCO3(s)]2
Assuming activities of 1 for solids and liquids:
Kc =
Manipulating Kc
The equilibrium constant is defined in terms of the balanced chemical equation.
1. If coefficients are multiplied by a constant, x
Knew = (Kold)x
2 SO2 + O2  2 SO3
K=
[SO3]2
= 429
2
[SO2] [ O2]
4 SO2 + 2 O2  4 SO3
K=
EQUILIBRIUM 14-6
2. If the equation is reversed
Knew = 1/Kold
2 SO3  2 SO2 + O2
K=
3. If two equations are added together,
Knew = K1 x K2
CO(g) + 3 H2(g)  CH4(g) + H2O(g);
CH4(g) + 2 H2S(g)  CS2(g) + 4 H2(g);
Sum: CO(g) + 2 H2S(g)  CS2(g) + 4 H2(g) + H2O(g)
K1 = 3.92
K2 = 3.3 x 104
Manually show
cancellations.
Kc= K1K2 = [CH4][H2O] x [CS2][H2]4 = [CS2][H2O][H2]
[CO][H2]3
[CH4][H2S]2
[CO][H2S]2
Kc =
The Equilibrium Constant Kp
A. Kp - equilibrium constant can be defined using partial pressures for gas phase
reactions.(for convenience)
B. Relationship between Kc and Kp.
1.
K p  Kc RT n .
a. n = the sum of the coefficients of the ___________ products
minus the sum of the coefficients of the gaseous reactants
LP# 4. EXAMPLE:
Calculate Kp for the reaction: CO(g) + 2 H2(g)  CH3OH(g) at 210 ºC
(given Kc = 14.5 at 210 ºC)
n =
We can now calculate the value of Kp from the
value of n and the value of Kc.
Kp =
EQUILIBRIUM 14-7
Students, verify the ability to obtain
this value using your calculator.
You need to use the yx button.
LP# 5. EXAMPLE:
Determine the value of Kc and Kp for the reaction:
PCl5 (g)  PCl3 (g) + Cl2 (g)
if [PCl5]eq = 2.75 M
and
[PCl3]eq = [Cl2]eq = 1.26 M
at 500. K.
Step 1: Write the equilibrium equation for Kc and solve for Kc using the molar
concentrations given.
Kc =
Step 2: Determine n.
n =
Step 3: Solve for Kp using the equation that expresses the relationship between
Kc and Kp.
Kp =
Using the Equilibrium Constant
A. Judging the extent of reaction.
1. Large value of Kc ( > 103)
a. In equilibrium mixture: ________________
b. Reaction proceeds almost to completion. (product favored)
c. Position of reaction is _________________
2. Small value of Kc ( < 10-3)
a. In equilibrium mixture: ________________
b. Reaction hardly proceeds at all. (reactant favored)
c. Position of the reaction is ______________
3. 10-3 < Kc < 103
a. appreciable amounts of _________ reactants and products present.
Note: K can never be zero or negative.
EQUILIBRIUM 14-8
B. Predicting the direction of reaction.
1. Reaction quotient, Qc,
 same as the equilibrium constant expression: products
reactants
 but the concentrations are _______ necessarily at equilibrium.
2. Predict direction of reaction by comparing the value of Qc to Kc.
a. Qc < Kc; _________________ products
reaction goes from __________________
b. Qc > Kc; _________________ products
reaction goes from ___________________
c. Qc = Kc; reaction is ____________________
LP# 6. EXAMPLE:
-9
Kc = 4.18  10 at 425o C for:
2 HBr(g)  H2(g) + Br2(g)
If the concentrations of all species present are:
[HBr] = 0.75 M;
[H2] = [Br2] = 2.5  10-4 M
 Is the reaction mixture at equilibrium?
 In which direction will the reaction proceed?
 If not at equilibrium, what will the equilibrium concentrations be?
SOLUTION:
a) To determine if the reaction mixture is at equilibrium, we need to calculate ___
and compare that value to Kc.
b) Direction?
EQUILIBRIUM 14-9
c) What will the equilibrium concentrations be?
2 HBr (g)

I
C
E
H2(g)
Br2(g)
Kc=
2.5  10
 x
 4.18  10 9
2
0.75  2 x 
2.5 10
4
2
4

x
 6.465  10 5
0.75  2 x 
2.5x10-4 – x = (6.465x10-5)(0.75 + 2x)
2.5x10-4 – x =4.849x10-5 + 1.293x10-4x
2.5x10-4 =4.849x10-5 + 1.0001293x
2.5x10-4 - 4.849x10-5 = 1.0001293x
“Skipping a bunch of algebra”
-4
X = 2.0x10
2.015 x10-4 = 1.0001293x
x= 2.015 x10-4 = 2.0x10-4
1.0001293
[HBr] =
[H2] = [Br2] =
Does our answer make sense? Try calculating Kc with these values!
Kc =
EQUILIBRIUM 14-10
Calculating Equilibrium Concentrations using Kc.
(No equilibrium concentrations given)
LP# 7. EXAMPLE: - No Shortcuts - Using the Quadratic
The reaction PCl5 (g)  PCl3 (g) + Cl2 (g) has Kc = 85.0 at a temperature of
760oC. Calculate the equilibrium concentrations of PCl5, PCl3, and Cl2 if the initial
concentration of PCl5 is 5.00 M.
SOLUTION:
I
C
E
PCl5 (g)

PCl3(g)
Cl2(g)
The equilibrium expression for the reaction is:
Kc 
PCl3 Cl 2 
PCl5  .
The equilibrium concentrations from the above table can be substituted into this
expression, and we can solve for x.
Kc =
We need to rearrange the equation (set equal to zero) and use the quadratic
equation to solve for x.
Solving the quadratic equation:
2
 b  b 2  4ac  85.0  85.0  [(4)(1)(425)]  85.0  7.23x103  1.70x103
x


2a
2(1)
2
x =_____________
The mathematical solution that makes chemical sense is _______
Step 4: The equilibrium concentrations can now be calculated.
[PCl5] =
[PCl3] = [Cl2] =
EQUILIBRIUM 14-11
LP# 8. EXAMPLE – Using a Perfect Square:
The value of Kc for the following reaction is 57.0 at 700. K.
H2(g) + I2(g)  2 HI(g)
Determine the equilibrium concentrations if 1.00 mole of hydrogen is allowed to
react with 1.00 mole of iodine in a 10.0 L container.
Initial [H2] = [I2] =
Sometimes
we can avoid
using the
quadratic
equation.

I
C
E
H2(g)
I2(g)

2 HI (g)
[HI]2

Kc  57.0 
[ H 2 ][I 2 ]
 Taking the square root of both sides:
57.0  7.55 
2x
0.100 - x
Solving for both solutions:
+7.55 (0.100 - x) = 2x
x = _________
-7.55 (0.100 – x) = 2 x
x = _________
 Which answer makes sense?
Since the initial conc.s were only 0.1, we could not use up 0.136!

Calculate the equilibrium concentrations using your value for x.
[H2] = [I2] =
[HI] =
Note: If you did not recognize the perfect square, this could be successfully solved using
the quadratic equation.
EQUILIBRIUM 14-12