The state of a system at equilibrium is dynamic
Chemical Equilibrium
What are the concentrations of reactants
and products at equilibrium?
How do changes in pressure, volume,
temperature, concentration and the use
of catalysts affect the equilibrium state?
The rate of the forward reaction is equal to the rate of the
reverse reaction. The system hasn’t “stopped” reacting.
In chemical kinetics, the focus is on the portion of the
reaction prior to te (the time at which equilibrium is reached).
In equilibrium, the focus is on time after which te occurs.
The addition of radioactive NaCl to a saturated solution.
Experimental results show
that for a given reaction
equation at a particular
temperature:
Given:
aA + bB + … ↔ gG + hH + …
Kc = [G]g[H]h…/[A]a[B]b…
This is known as the
equilibrium constant
expression and Kc is called
the concentration equilibrium
constant (given as a
dimensionless quantity)
Answers:
Question 1:
Question 1:
Given:
CO(g) + Cl2(g) ↔ COCl2(g)
Given:
Kc = 1.2x103 at 395oC
If [CO] equals [Cl2] at equilibrium for the reaction, is there
just one possible value of [COCl2]? Explain.
Kc = 1.2x103 at 395oC
CO(g) + Cl2(g) ↔ COCl2(g)
If [CO] equals [Cl2] at equilibrium for the reaction, is there just one possible value of [COCl2]? Explain.
No. [COCl2] = Kc[CO][Cl2], but there are many possible values for
[CO]=[Cl2]
Question 2:
Question 2:
Suppose [O2] is fixed at a certain constant value when equilibrium is reached in the following reversible
reaction:
Suppose [O2] is fixed at a certain constant value when
equilibrium is reached in the following reversible reaction:
Do [SO2] and [SO3] have unique values? Does the ratio [SO2]/[SO3] have a unique value? Describe the
equilibrium state when [O2] = 1.00M.
2SO2(g) + O2(g) ↔ 2SO3(g)
Kc = 1.00x102
Do [SO2] and [SO3] have unique values? Does the ratio
[SO2]/[SO3] have a unique value? Describe the equilibrium
state when [O2] = 1.00M.
2SO2(g) + O2(g) ↔ 2SO3(g)
Kc = 1.00x102
[SO2] and [SO3] do not have unique values, but the following ratios do:
[SO3]2/[SO2]2; [SO2]2/[SO3]2; [SO3]/[SO2], [SO2]/[SO3].
If [O2] = 1.00M
100 = [SO3]2/([O2][SO2]2)
100 = ([SO3]/[SO2])2
10 = [SO3]/[SO2]
[SO3] = 10[SO2]
1
Equilibrium From the Perspective of Kinetics and
Thermodynamics
Given
Using the ideal gas law equation PV = nRT and
remembering
2HI(g) ↔ H2(g) + I2(g)
And the rate laws are known to be
Ratef = kf[HI]2
The relationship between Kc (based on concentration)
and Kp (based on partial pressure)
Rater = kr[H2][I2]
At equilibrium Ratef = Rater
If the same rate laws govern the reaction at equilibrium:
kf[HI]2 = kr[H2][I2]
kf/kr = [H2][I2]/[HI]2 = Kc
(the concentration units in kf/kr will match those of Kc only if
they too are based on the stoichiometric coefficients of the
equations)
(Sometimes this requires substitution of values into the rate
determining step)
Molarity (M) = n/V
Solving for the partial pressure of a substance P,
P = [Concentration]RT, which differs by a factor RT
Substituting this into the equilibrium expression gives:
Kp = Kc(RT)∆n
where ∆n is the change in moles of gas
(products – reactants) from the balanced chemical
equation.
Modifying Equilibrium Expressions
Sometimes it is desirable to modify existing equilibrium
expressions to represent a new set of conditions or to
represent a new reaction.
Reversing a reaction:
Adding Chemical Equations:
When equations are added to achieve an overall
equation, the equilibrium constant for each reaction is
multiplied to obtain the overall equilibrium constant.
When a chemical reaction with the equilibrium
constant, Kc, is reversed, the reverse reaction has the
equilibrium constant 1/Kc
Multiplying the coefficients through by some number in a
chemical reaction:
If the coefficients are multiplied through by a common factor,
n, the original Kc is raised to the power n to produce the new
equilibrium constant.
Solids and Pure Liquids:
Since the concentrations of pure solids and liquids are
“constant” during a reaction (i.e. as pure substances,
their concentration does not change with respect to
themselves), those terms are not included in the
equilibrium expression.
expression
(This brings up the necessity of noting the balanced chemical
equation when citing a value for Kc. Kc is also temperature
dependent.)
Question 3:
Given Kc = 1.8x10-6 for the reaction 2NO(g) + O2(g) ↔ 2NO2(g) at 457K,
derive the value of Kp at 457K for the reaction NO2(g) ↔ NO(g) + ½
O2(g)
Question 4:
The equilibrium constant Kc for the reaction ½ N2(g) + 3/2 H2O(g) ↔
NH3(g) + ¾ O2(g) at 900K is 1.97x10-20. Calculate Kc at 900K for the
reaction
4NH3(g) + 3O2(g) ↔ 2N2(g) + 6H2O(g)
Answers
Question 3:
Given Kc = 1.8x10-6 for the reaction 2NO(g) + O2(g) ↔ 2NO2(g) at 457K, derive the value of Kp at
457K for the reaction NO2(g) ↔ NO(g) + ½ O2(g)
Kp = 4564 = 4.6x103
Kp = (1/1.8x10-6).5(.0821x900).5
Question 4:
The equilibrium constant Kc for the reaction ½ N2(g) + 3/2 H2O(g) ↔ NH3(g) + ¾ O2(g) at 900K is
1.97x10-20. Calculate Kc at 900K for the reaction
4NH3(g) + 3O2(g) ↔ 2N2(g) + 6H2O(g)
Question 5:
Kc = 6.64x1078
The reaction of steam with iron is an old method of producing
hydrogen gas:
Question 5:
3Fe(s) + 4H2O(g) ↔ Fe3O4(s) + 4H2(g)
3Fe(s) + 4H2O(g) ↔ Fe3O4(s) + 4H2(g)
Write the equilibrium constant expression Kc and Kp for this
reaction.
Write the equilibrium constant expression Kc and Kp for this reaction.
(1/1.97x10-20)4
The reaction of steam with iron is an old method of producing hydrogen gas:
Kc = [H2]4/[H2O]4;
Kp = PH24/PH2O4
2
ICE Diagram
Equilibrium
Expression:
Using
equilibrium
concentrations
to
experimentally
determine K at
a given
temperature.
0.0175M
Kc = [0.0276]2 / ([0.0037][0.0037]) = 56
(See next slide
for results)
.2076M
This will be true at this temperature
regardless of the initial concentrations
0.0037M
Interpretation of Equilibrium Constants:
A very large Kc or Kp (K value with double digit powers of ten)
signifies the reaction goes to “completion”
A very small Kc or Kp (K value with double digit negative
powers of ten) signifies the forward reaction only occurs to a
slight extent.
The reaction quotient (a.k.a. the massmass-action expression)
expression
gives the current state of a chemical system and
identifies the direction that the reaction will go to
achieve equilibrium.
Given:
aA + bB + … ↔ gG + hH + …
Qc = [G]g[H]h…/[A]a[B]b…
Remember: Kinetics determines when equilibrium is
reached.
Or
QP = PGgPHh…/PAaPBb…
Given: H2(g) + O2(g) ↔ H2O(g)
Although Kp at 298K is very large, it takes an extremely long
time to occur without the addition of energy or a catalyst.
The reaction is said to be thermodynamically favorable, but is
kinetically controlled.
This looks exactly like the equilibrium expression, but
applies to any current state of the system. By comparing
Q with K, the progress of the reaction can be determined
Example 1:
Given:
2HI(g) ↔ H2(g) + I2(g) Kc = 1.84x10-2
In which direction would a net change occur if the initial
conditions in the reaction were [HI] = 1.00M and [H2] = [I2]
= 0.100M?
Example 2:
∞
For the reaction H2S(g) + I2(s) ↔ 2HI(g) + S(s), Kp = 1.34x10-5
at 60oC. Initially, we bring together the following species:
H2S(g) at a partial pressure of 0.010atm, HI(g) at
0.0010atm, I2(s), and S(s). When equilibrium is established,
which gas will show an increase/decrease in its partial
pressure? Which solid will increase/decrease in its
amount?
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Answers:
Example 1:
Le Châtelier’s Principle
Given:
2HI(g) ↔ H2(g) + I2(g)
Kc = 1.84x10-2
In which direction would a net change occur if the initial conditions in the reaction
were [HI] = 1.00M and [H2] = [I2] = 0.100M?
Qc = [0.100][0.100]/[1.00]2 Qc = 1.0x10-2 which is less than
Kc. The reaction will proceed forward towards products.
Example 2:
Henri Le Châtelier 1888: When a change (that is,
a change in concentration, temperature,
pressure or volume) is imposed on a system at
equilibrium, the system responds by attaining a
new equilibrium condition that minimizes the
impact of the imposed change.
For the reaction H2S(g) + I2(s) ↔ 2HI(g) + S(s), Kp = 1.34x10-5 at 60oC. Initially, we
bring together the following species: H2S(g) at a partial pressure of 0.010atm, HI(g)
at 0.0010atm, I2(s), and S(s). When equilibrium is established, which gas will show
an increase/decrease in its partial pressure? Which solid will increase/decrease in
its amount?
Qp = .00102/0.010 = 1.0x10-4 which is larger than Kp. The
pressure of H2S will increase/ HI will decrease, I2 will
increase/ S will decrease.
The use of a catalyst may speed up a reaction,
but it speeds it up in both the forward and
reverse direction, therefore a catalyst has no
effect on the equilibrium state of the system.
Example problems involving establishing the equilibrium
constant, K, and those using K to establish equilibrium
concentrations or partial pressures.
Example 1
2.0 moles of I2 and 4.0 moles of Br2 are placed in a 2.0L
reactor at 150oC, and reaction occurs until equilibrium
is reached:
I2(g) + Br2(g) ↔ 2IBr(g)
Analysis then shows that the reaction contains 3.2
moles of IBr. What is the value of the equilibrium
constant Kc for the reaction?
Answer
Example 1
2.0 moles of I2 and 4.0 moles of Br2 are placed in a 2.0L reactor at 150oC, and reaction
occurs until equilibrium is reached:
I2(g) + Br2(g) ↔ 2IBr(g)
Analysis then shows that the reaction contains 3.2 moles of IBr. What is the value of
the equilibrium constant Kc for the reaction?
3.2 moles of IBr imply that 3.2(1/2) = 1.6mol of I2 and Br2
were used, leaving 2.0 – 1.6 = .40mol I2 and 4.0 – 1.6 =
2.4mol of Br2 left
Kc = [IBr]2/[I2][Br2] = [1.6]2/[.20][1.2] = 1.07x101
Note that you get the same answer in moles or molarity.
4
Example 2
Some hydrogen and iodine are mixed at 400oC in a
1.00 liter container, and when equilibrium is
established the following concentrations are present:
[HI] = 0.490M, [H2] = 0.080M, and [I2] = 0.060M. If an
additional 0.300 moles of HI are added, what
concentrations will be present when the new
equilibrium is established?
Answer:
Example 2
Some hydrogen and iodine are mixed at 400oC in a 1.00 liter container,
and when equilibrium is established the following concentrations are
present: [HI] = 0.490M, [H2] = 0.080M, and [I2] = 0.060M. If an
additional 0.300 moles of HI are added, what concentrations will be
present when the new equilibrium is established?
H2 + I2 ↔ 2HI
50.021
Kc = [0.490]2/[0.080][0.060] =
50.021 = [0.790-2X]2/[0.080+X][.060+X]
X = .0329M
[HI] = .7242M
Example 3
Answer
For the reaction at equilibrium
Example 3
[H2] = .1129M
[I2] = .0929M
For the reaction at equilibrium
2NaHCO3(s) ↔ Na2CO3(s) + H2O(g) + CO2(g)
∆Ho = 128kJ
state the effects (increase, decrease, or no change) of
the following stresses on the number of moles of sodium
carbonate (Na2CO3) at equilibrium in a closed container.
(Note that Na2CO3 is a solid; its concentration will
remain constant, but the amount can change.)
2NaHCO3(s) ↔ Na2CO3(s) + H2O(g) + CO2(g)
∆Ho = 128kJ
state the effects (increase, decrease, or no change) of the following stresses on the number of
moles of sodium carbonate (Na2CO3) at equilibrium in a closed container. (Note that Na2CO3 is a
solid; its concentration will remain constant, but the amount can change.)
a. Removing CO2(g)
b. Adding H2O(g)
c. Raising the temperature
d. Adding NaHCO3(s)
a. Removing CO2(g)
a. Increases moles of Na2CO3
b. Adding H2O(g)
b. Decreases moles of Na2CO3
c. Raising the temperature
c. Increases moles of Na2CO3 (Endothermic)
d. Adding NaHCO3(s)
d. No effect
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