Ch 14 Chemical Equilibrium
Chemical Equilibrium
- At equilibrium, the reactant and product concentrations are no longer changing because:
forward rate = reverse rate
- This process is called dynamic equilibrium, because the forward and reverse reactions
have not stopped completely, they are still occurring, simultaneously.
- Equilibrium requires a reversible reaction system. Catalytic methanation is reversible and it produces
methane (natural gas) from incomplete combustion products that contain carbon monoxide.
CO(g) + 3H2(g)  CH4(g) + H2O(g)
- The reverse reaction is called steam reforming, which produces H2 and CO from methane and steam.
Hydrogen and carbon monoxide are typically used as manufacturing feedstocks or for fuel.
CH4(g) + H2O(g)  CO(g) + 3H2 (g)
- Symbolize both sides reacting simultaneously with a reversible double arrow.
CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g)
- Conditions of the reaction can be changed to favor one side or the other.
Equilibrium Table
- If all initial concentrations are known, and at least one equilibrium concentration can be measured,
then remaining concentrations can be determined from stoichiometry using an Equilibrium Table.
- The starting concentrations and the change in concentrations are combined algebraically
to find the equilibrium concentrations for each reactant and product.
Ex 14.1 Creating an Equilibrium Table
CO
+
3H2 ⇌
CH4
+
H2O
Initial (starting)
0.100 M
0.300 M
0
0
Change (with variable)
-x
-3x
+x
+x
-------------------------------------------------------------------------Equilibrium (sum)
0.100 - x
0.300 - 3x
x
x
0.387 moles
10.00 L
[CH4] = [H2O] = x
=
[CO] =
[H2] =
= 0.100 – 0.0387
= 0.300 – 3(0.0387)
0.100 - x
0.300 - 3x
= 0.0387 M
= 0.0613 M
= 0.1839 M
Equilibrium Constant Expression
- For reversible reaction aA + bB ⇌ cC + dD at equilibrium,
the concentrations are no longer changing.
- So the relationship between concentrations can be defined by an expression, which is
obtained by multiplying product concentrations, dividing reactant concentrations, and
raising each concentration to its stoichiometric coefficient as its exponent.
[C]c [D]d
Kc =
[A]a [B]b
Law of Mass Action
- The Kc numerical value is constant at a particular set of circumstances (e.g., constant temperature).
This numerical value is known as the equilibrium constant.
Ex 14.2.
-
-
[CH4 ][H2 O]
a.
For CO + 3H2 ⇌ CH4 + H2O, the expression is Kc =
b.
c./d.
The same reaction written in reverse has the reciprocal Kc.
If stoichiometric coefficients are divided by 2, then exponents are as well.
[CO][H2 ]3
The kinetics argument for the law of mass action is that Kc is a constant because it is related to rate
constants for elementary reactions, which have a rate law that is obtained from stoichiometry.
For N2O4 ⇌ 2NO2 at equilibrium, the forward rate equals reverse rate.
kf[N2O4] = kr[NO2]2
and
Kc = kf / kr = [NO2]2/[N2O4]
Since both rate constants (k’s) have constant values, Kc must as well.
Since the rate-determining step for all reaction mechanisms is an elementary reaction,
a constant Kc is found for all net chemical equations.
The Equilibrium Constant’s Value
- The value of the equilibrium constant can be determined from concentration data.
- We can use the data from Example 14.1 on the previous page for catalytic methanation.
Kc =
[CH4 ][H2 O]
[CO][H2 ]3
=
(0.0387)(0.0387)
(0.0613)(0.1839)3
= 3.93 at 1200 K (See Figure 14.5)
Ex 14.3 Value for an Equilibrium Constant
2HI ⇌ H2 + I2 [HI]o
=
0.800
0
0
x = [I2]eqm
=
−2x
x
x
[HI]eqm
=
0.800-2x x
x
Kc = [H2][I2] / [HI]2 =
4.00 mol / 5.00 L
0.442 mol / 5.00 L
0.800 − 2x
(0.0884)2 / (0.623)2
= 0.800 M
= 0.0884 M
= 0.623 M
= 0.0201
Equilibrium Constant of Gaseous Reactions using Partial Pressures (Kp)
- The ideal gas law, PV = nRT, rearranges to P = (n/V)RT = MRT, where M is Molarity in mol/L.
- So, equilibrium constants for gases can be expressed using partial pressures of reactants and products.
- We can use partial pressures to get Kp for CO + 3H2 ⇌ CH4 + H2O (catalytic methanation) at 1200 K.
Kp =
-
PCH4 PH2O
PCO P3H2
=
[CH4 ][H2 O]
[CO][H2 ]3
× (RT)(1+1-1-3) = Kc × (RT)-2
Kp = (3.93)(0.0821 × 1200)-2 = 4.04 × 10-4
So, Kp = Kc(RT)n, where n = (sum of product coefficients) – (sum of reactant coefficients).
The n is –2 for the catalytic methanation reaction because it produces two moles from four moles.
Exercise 14.5:
Kp = Kc × (RT)2-1 = (3.26 × 10-2)(0.0821 × 464)1 = (0.0326)(38.1) = 1.24
Sum of Chemical Equations
- If a net chemical equation is the sum of other net chemical equations with known eqm constants,
then the eqm constant for the sum reaction is the product of the component eqm constants.
- CO + 3H2 ⇌ CH4 + H2O
Kc1 = 3.93
CH4 + 2H2S ⇌ CS2 + 4H2
Kc2 = 3.3 × 104
---------------------------------------CO + 2H2S ⇌ CS2 + H2O + H2
Kc3 = (Kc1)(Kc2) = (3.93)(3.3 × 104) = 1.3 × 105
Homogeneous Equilibrium (single phase)
- All reactants and products are in the same phase, contained in a single liquid or gas solution.
- For example, in catalytic methanation, all of the substances are in the same gaseous mixture.
- Homogeneous reactions can also occur completely dissolved in a single aqueous solution.
Cu2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)42+](aq)
Heterogeneous Equilibrium (involves more than one phase)
- Solids have surface concentrations that are small, difficult to measure, and approximately constant.
g
L
g
18.0
mol
1000
-
Pure liquids also have constant concentrations: [H2O] =
-
Omit solids and pure liquids from Kc to simplify the expression by leaving out constants.
Consider 3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g).
We have Kc′ =
[Fe3 O4 ][H2 ]4
[Fe]3 [H2 O]4
= 55.6 M
as the full equilibrium constant, but it is not necessary.
We simplify the expression by cancelling the solids.
Kc = (Kc′) [
[Fe]3
Fe3 O4 ]
and
[H ]4
Kc = [H 2
2 O]
4
-
Equilibrium is not affected by pure solids/liquids as long as they are simply present
Example 14.4
CaCO3(s) ⇌ CaO(s) + CO2(g)
Kc = [CO2]
-
Exercise 14.6
Ni(s) + 4CO(g) ⇌ Ni(CO)4(g)
Kc =
[Ni(CO)4 ]
[CO]4
The equilibrium constant can be used to answer important questions about chemical reactions.
1) The magnitude of the constant’s value can tell you whether reactants or products are favored.
2) Kc can predict direction of reaction for a mixture that is not at equilibrium.
3) Kc can be used to calculate equilibrium concentrations using an equilibrium table.
Qualitatively Interpreting the Magnitude of the Equilibrium Constant
- Large value means products (in numerator) are favored (Kc >> 1)
- Small value means reactants (in denominator) are favored (Kc << 1)
- If Kc ≈ 1, then equilibrium mixture contains both reactants and products
- N2 + 3H2 ⇌ 2NH3 Kc = 4.1 × 108 at 25 oC
If [N2] = [H2] = 0.01M, then [NH3] = 2.0 M, which is more than 200X each reactant’s concentration.
Predicting the Direction of Reaction for a Mixture that is Not Necessarily at Equilibrium
- Use the reaction quotient (Qc), which has same form as Kc,
but uses interim (not equilibrium) concentrations.
[C]ci [D]d
i
where “i” stands for interim
-
Qc =
-
If Qc > Kc, the reaction will go to the left, towards the reactants.
Shifting left decreases the numerator and increases the denominator.
If Qc < Kc, the reaction will go to the right, towards the products.
Shifting right increases the numerator and decreases the denominator.
If Qc = Kc, then the reaction is at equilibrium.
No shift occurs.
-
[A]ai [B]b
i
Example 14.5 Evaluating Qc and Comparing it with Kc
Qc =
[NH3 ]i 2
[N2 ]i [H2 ]i 3
=
(0.0100)2
(0.0200)(0.0600)3
= 23.1
Kc = 0.500 at 400 oC, so Qc > Kc and reaction goes left (reverse)
Equilibrium Concentrations can be determined by the following steps with an equilibrium table.
1) Set up a table of concentrations with three lines:
initial, change (including the variable), and equilibrium (as their sum).
2) Substitute the equilibrium line (with the variable) into the Kc expression.
3) Use the value for Kc to solve for the variable, and then find the concentrations.
x2
Example 14.7 uses a perfect square:
Kc = (0.02
Example 14.8 uses the quadratic formula:
x=
− x)2
−b ± √b2 −4ac
2a
Effect of Changing Conditions on Equilibrium Concentrations
- The equilibrium concentrations can be changed by:
o adding or removing either reactants or products
o changing partial pressures (Pi) of reactants and products (by changing the volume)
o changing the temperature
- The effect can be predicted from LeChatelier’s principle of dynamic equilibrium.
When a system at chemical eqm is disturbed by a change in [concentration], Pi , or T,
then the equilibrium concentrations shift to counteract that change.
- For instance, if you remove or add a substance in the reaction,
the equilibrium constant will not change, but Qc does.
The reaction will proceed to restore the balance, until Qc = Kc again.
- If a reactant is added or a product is removed, then Qc < Kc and the equilibrium shifts right (fwd).
- If a reactant is removed or a product is added, then Qc > Kc and the equilibrium shifts left (rev).
- Example 14.9 has H2 + I2 ⇌ 2HI where H2 reactant is removed. The equilibrium shifts left.
- Reactions involving gases can shift their equilibrium as a result of changes to pressure and volume.
Suppose the partial pressures of the gases are all increased proportionally by decreasing the volume.
The reaction will counteract the change by shifting the equilibrium towards fewer moles of gas.
This occurs because moles are proportional to pressure PA = (nA/V)RT.
Also, increasing the volume will shift the equilibrium towards more moles of gas.
- See example 14.10.
- CO + 3H2 ⇌ CH4 + H2O
If volume becomes V/2 (reduced by half), then each PA doubles.
Qc =
-
(2[CH4 ]0 )(2[H2 O]0 )
(2[CO]0 )(2[H2 ]0
)3
=
4Kc
16
=
Kc
4
Qc < Kc
So, the reaction goes to the right.
Adding a nonparticipant (like He), though, only changes the total P.
The nonparticipant does not change any other PA.
So, the Qc and the equilibrium concentrations do not change.
Temperature affects Equilibrium Concentrations by Changing Kc
- For an endothermic reaction (H > 0), more heat favors products, so increasing T will increase Kc.
Higher T makes Qc < Kc, and equilibrium shifts right (forward).
Adding heat to an endothermic reaction is like adding a reactant.
- For an exothermic reaction (H < 0), less heat favors products, so decreasing T will increase Kc.
Higher T shifts the exothermic reaction to the left (reverse).
Lower T is needed for Qc < Kc, so that equilibrium shifts to right.
Removing heat from an exothermic reaction is like removing a product.
- See Table 14.2 for Kc vs. T of an exothermic reaction (catalytic methanation).
Kc decreases as T increases.
Optimizing an Exothermic Reaction
-
CO + 3H2 ⇌ CH4 + H2O
H = − 206
kJ
mol CO
Increase T just enough to increase the reaction rate, but
not enough to reduce Kc significantly (230 – 450 oC).
Also, use moderately high partial pressures (Pi’s) to shift
the equilibrium to the right, towards fewer moles of gas.
See example 14.11.
Catalysts
- Catalysts do not affect equilibrium compositions.
However, they decrease the time needed to reach equilibrium.
- 2SO2 + O2 ⇌ 2SO3
Kc = 1.7 × 1026
The reaction has a very large Kc, and is used to produce H2SO4 from industrially from sulfur.
However, combustion of sulfur produces mostly SO2 and little SO3.
This is because the rate for SO3 is very slow, and its equilibrium is not reached.
Pt and V2O5 increase the rates (both fwd/rev) for SO3 so that equilibrium can be reached.
- N2 + O2 ⇌ 2NO
Kc = 4.6 × 10-31 at 25 oC
This reaction has a small K at 25 oC, so catalysts do not help at room T,
because they do not change equilibrium concentrations.
Kc increases with T, so that air can be used with a catalyst at 2000 oC to form 0.4 % NO in air.