Equilibrium Mathematics
The rules we have now developed for equilibria systems result in certain mathematical consequences for various procedures we will often perform in chemical equations. Let’s examine four different circumstances.
Reversing Reactions
Since all equilibrium reactions are reversible (<——>), we can represent each equation as the forward or the
backwards way:
Forward: 2 H2 (g) + O2 (g) <——> 2 H2O (g)
Or
Reverse: 2 H2O (g) <——> 2 H2 (g) + O2 (g)
But examine the Kp values for each of them:
Forward:
Kp =
Reverse:
Kp =
P2H2O
2
P H2PO2
P2H2PO2
P2H2O
Notice how the reverse reaction and the forward reaction are the inverse of each other; by reversing a reaction,
the top and bottom of the K expression are flipped. Or to put it another way:
Kforward = K-1reverse
Or
1
Kforward =
Kreverse
Example 1:
For the reaction:
Calculate the Kc for the reaction:
A (aq) + B (aq) <——> C (aq)
C(aq) <——> A (aq) + B (aq)
Since the reactions are reversed, the Kc value is inverted so:
Example 2:
For the reaction:
Calculate the Kp for the reaction:
K = (1.2x10-5)-1 or Kc = 8.3x104
H2 (g) + Br2 (g) <——> 2 HBr (g)
2 HBr (g) <——> H2 + Br2 (g)
Since the reactions are reversed, the Kp value is inverted so:
Kc = 1.2x10-5
Kp = 5.6x106
K = (5.6x106)-1 or Kp = 1.8x10-7
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Changing Stoichiometric Coefficients
In some cases, we may want to change the stoichiometry of an equation to suit our needs. For example:
Equation 1: H2O (g) <——> H2 (g) + 1/2 O2 (g)
Kp =
PH2P0.5O2
PH2O
But what if instead of having the O2 be a coefficient of 1/2, we want whole numbers instead, so we can double
the coefficients of all species to get:
Equation 2: 2 H2O (g) <——> 2 H2 (g) + O2 (g)
Kp =
P2H2PO2
P2H2O
Notice what happened to the Kp value? Each species was raised to the 2nd power because we doubled each of
the coefficients. Consequently,
Knew = Kxold
x = ratio of stoichiometry
Or to put it another way, if you double a reaction, every species in the K equation gets squared (x 2). If you triple a reaction, every species in the equation gets cubed (x3). If you cut an equation in half, every species get
square-rooted (x1/2).
Example 1:
For the reaction:
What is the Kc value for:
1/2 A (aq) + 1/2 B (aq) <——> C (aq)
A (aq) + B (aq) <——> 2 C (aq)
Since you are doubling the equation, the Kc value is squared or
Example 2:
For the reaction:
What is the Kp value for:
Kc = 4.3x10-2
Kc = ???
Kc = (4.3x10-2)2 = 1.85x10-3
P4 (s) + 6 Cl2 (g) <——> 4 PCl3 (g)
1/4 P4 (s) + 3/2 Cl2 (g) <——> PCl3 (g)
Kp = 6.2x105
Kp = ????
Since you are cutting each species by 1/4, the Kp value is raised to the 1/4 power or Kp = (6.2x105)0.25 = 28.1
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Adding Equations Together
Often, you can use more than one equation added together to obtain another overall equation. What happens to
the K values when multiple equations are added together? Let’s look at an example:
Equation 1: S (s) + O2 (g) <——> SO2 (g)
PSO2
PO2
Kp1=
Equation 2: SO2 (g) + 1/2 O2 (g) <——> SO3 (g)
Kp2=
PSO3
PSO2 P0.5O2
Now if we add these two equations together we get the following:
Equation 3: S (s) + SO2 (g) + 3/2 O2 (g) <——> SO2 (g) + SO3 (g)
But we should cancel the SO2 (g) because they appear on both sides so:
Equation 3: S (s) + 3/2 O2 (g) <——> SO3 (g)
Kp =
PSO3
P1.5O2
How do the Kp values of equation 1 and 2 relate to equation 3? If you multiply Kp1 and Kp2 together, you get
Kp3:
Kp1 x Kp2 = Kp3
PSO2
PO2
x
PSO3
PSO2P0.5O2
= Kp3
PSO2
PO2
x
PSO3
PSO2P0.5O2
= Kp3
PSO3
PO2P0.5O2
= Kp3
PSO3
P1.5O2
= Kp3
Which matches the value we
expected from above
Ktot = K1 x K2 x K3….
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Example 1:
Equation 1: H2C2O4 (aq) + H2O (l) <——> HC2O4-1 (aq) + H3O+1 (aq) K1 = 2.3x10-2
Equation 2: HC2O4-1 (aq) + H2O (l) <——> C2O4-2 (aq) + H3O+1 (aq) K2 = 4.7x10-5
What is the K value for the equation:
Equation 3: H2C2O4 (aq) + 2 H2O (l) <——> C2O4-2 (aq) + 2 H3O+1 (aq) K3= ????
To get equation 3, you have to add together equations 1 and 2, combine like terms, and cancel:
H2C2O4 (aq) + H2O (l) + HC2O4-1 (aq) + H2O (l) <——> HC2O4-1 (aq) + H3O+1 (aq) + C2O4-2 (aq) + H3O+1 (aq)
H2C2O4 (aq) + 2 H2O (l) <——> C2O4-2 (aq) + 2 H3O+1 (aq)
So K3 = K1 x K2 = (2.3x10-2)(4.7x10-5) = 1.08x10-6
Example 2:
Equation 1: C (s) + CO2 (g) <——> 2 CO (g) Kp = 2x1013
Equation 2: CO (g) + Cl2 (g) <——> COCl2 (g) Kp = 4x10-4
What is the K value for the equation:
Equation 3: C (s) + CO2 (g) + 2 Cl2 (g) <——> 2 COCl2 (g)
To get equation #3 you have to leave equation #1 the same but you have to double equation #2 and then add
them together:
Equation 1: C (s) + CO2 (g) <——> 2 CO (g) Kp = 2x1013
Equation 2: 2 CO (g) + 2 Cl2 (g) <——> 2 COCl2 (g) Kp = (4x10-4)2 = 1.6x10-7
Remember that when
you double an equation
you have to square (x2)
the K value
Then add them together and cancel terms:
Equation 3: C (s) + CO2 (g) + 2 CO (g) + 2 Cl2 (g) <——> 2 CO (g) + 2 COCl2 (g)
So K3 = K1 x K2 = (2x1013)(1.6x10-7) = 3.2x106
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Kc vs Kp
We have seen problems where we have solved for Kc and other problems where we have used Kp. Are Kc and
Kp interchangeable? Are they the same? If not, is there ever a circumstance where they are the same? Let’s
examine a simple equilibrium example:
CaCO3 (s) <——> CaO (s) + CO2 (g)
Write Kc for this reaction:
Write Kp for this reaction:
Kc = [CO2]
Kp = PCO2
Does Kc = Kp? Does [CO2] = PCO2 ? Clearly not. Pressure does not equal concentration.
But can we turn pressure into concentration? Can we equate them? We can if we remember the ideal gas law:
PV = nRT
We see that there is pressure (P) but where is concentration? Remember that concentration is moles per liter
so:
P = (n/V)RT
or substituting in [X] for (n/V) we get:
P = [X]RT
So pressure does not equal concentration but if we multiply the concentration by RT, then we get pressure:
Moles * Liter atm * K
Liter
mole K
[X]
R
T
So every time you change a concentration [X] into a pressure P, you have to multiply the concentration by RT.
This is going to result in some strange math, however. Examine the following example
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Example 1: The Kc value is 4.2x10-4 for the reaction:
PCl5 (g) <——> PCl3 (g) + Cl2 (g)
Calculate Kp at 273 K.
Kc has the equation
Kc = [PCl3][Cl2]
[PCl5]
Since we need to convert each of those concentration values into pressure, we need to multiply each by RT:
But look what happens. We have
(RT)2 on top and RT on the bottom
so one of those RT values cancels
Kp = ([PCl3]RT)([Cl2]RT)
[PCl5]RT
Which leaves us with
Kc =
But since
Kp = ([PCl3]RT)([Cl2]RT)
[PCl5]RT
Kp = ([PCl3])([Cl2])RT)
[PCl5]
[PCl3][Cl2]
[PCl5]
This equation becomes:
Kp = Kc(RT)
Consequently: Kp = 4.2x10-4 (0.0821 Latm/molK)(273 K) = 9.41x10-3
But that sure seemed like a lot of work, right? Is there a shortcut we could use so we don’t have to do that
long derivation anymore? There is if we look at the big picture of things. Let’s examine a complex example:
Example 2: The Kc value is 8.1x10-2 for the equation:
2 NH3 (g) <——> N2 (g) + 3 H2 (g)
Calculate Kp at 300 K.
Doing the same thing as last time we get that:
Kc = [N2][H2]3
Or to put it another way:
Kc = [N2][H2][H2][H2]
2
[NH3]
[NH3][NH3]
Since each [x] needs to be multiplied by RT we get:
Or: Kc = [N2][H2]3 (RT)4
[NH3]2 (RT)2
Kc = [N2][H2]3 (RT)2
Kc = [N2](RT)[H2](RT)[H2](RT)[H2](RT)
[NH3](RT)[NH3](RT)
[NH3]2
Notice how there is (RT)2 on bottom and (RT)4 on the top? Examine the equation. Notice how there are two
moles of gases on the reactant side and 4 moles of gases on the product side?
2 NH3 (g) <——> N2 (g) + 3 H2 (g)
Since each gas gets modified by RT we can streamline the process by recognizing that every 1 mole of gas on
each side of the equation will cancel out. Only if a side has more gas than the other will RT matter.
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Thus, what matters is the change in the number of moles of gas or to put it another way:
Kp = Kc (RT)gas
Now to finish this example:
The Kc value is 8.1x10-2 for the equation:
2 NH3 (g) <——> N2 (g) + 3 H2 (g)
Calculate Kp at 300 K.
Since there are 2 moles of gas on the left and 4 on the right, the gas = (prod)-(react) = 4-2 = +2 so
Kp = Kc (RT)+2
Kp = (8.1x10-2)[(0.0821)(300)]+2
Kp = 49.1
Example 3:
The Kc value is 2.3x104 for the equation:
2 SO3 (s) <——> 2 SO2 (g) + O2 (g)
Calculate Kp at 500 K
Kp = Kc(RT)gas
Since there are 3 moles of gases on the product side and 2 moles of gases on the reactant side:
gas = 3-2 = +1
So
4
Kp = (2.3x10 )[(0.0821)(500 K)]+1 = 9.4x105
Example 4:
The Kc value is 1.8x10-6 for the equation:
Ti (s) + 2 Cl2 (g) <——> TiCl4 (l)
Calculate Kp at 750 K
Kp = Kc(RT)gas
Since there are 0 moles of gases on the product side and 2 moles of gases on the reactant side:
gas = 0-2 = -2
So
-6
Kp = (1.8x10 )[(0.0821)(750 K)]-2 = 4.75x10-10
Example 5:
The Kp value is 6.7x10-7 for the equation:
3 O2 (g) <——> 2 O3 (g)
Calculate Kc at 250 K
Kp = Kc(RT)gas
Since there are 2 moles of gases on the product side and 3 moles of gases on the reactant side:
gas = 2-3 = -1
So
-7
6.7x10 = Kc[(0.0821)(250 K)]-1 = 1.4x10-5
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So is there ever a circumstance where Kp = Kc ? Yes, if the number of moles of gas on each side are equal.
Examine:
H2 (g) + Cl2 (g) <——> 2 HCl (g)
Kp = Kc(RT)gas
Since there are 2 moles of gases on the product side and 2 moles gas on the reactant side
gas = 2-2 = 0
So
Kp = Kc [RT]0
Since anything raised to the 0 power = 1 we get:
Kp= Kc
So:
If there are the same moles of gas on each side of the
equation:
Kp = Kc
Summary
Kforward = K-1reverse
Knew = Kxold
where x = ratio of stoichiometry (doubled, tripled, halved, etc.)
Ktotal = K1x K2 x K3 …
When adding equations together
Kp = Kc (RT)gas
Where gas is the change in the number of moles of gases (products-reactants)
Kp = Kc
If the number of moles of gases on each side of the equation is the same
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