Chapter 14: Chemical
Equilibrium
Sections 14.4 - 14.7
Sarah Rodriguez
14.4 Expressing the Equilibrium Constant in Terms of
Pressure
Previously, expressed equilibrium constant w/ concentrations of reactants and products
●
○ in gaseous reactions the partial pressure of a particular gas is proportional to its
concentration
○ *Reminder* Partial pressure (Pn) = the pressure due to any individual component in a gas
mixture
■ equilibrium constant can be expressed in terms of partial pressures of reactants
and products
●
Consider
○ 2SO3 (g) ⇌ 2SO2(g) + O2 (g)
○ Kc → equilibrium constant in terms of concentration (M)
■ Kc = [SO2]^2 [O2]
[SO3]^2
● Kp → equilibrium constant in terms of partial pressures (atm)
○ Kp = (PSO2)^2 PO2
(PSO3)^2
Formulas below can be found in reference tables:
...Continued
●
Kp not necessarily = to Kc (partial pressure of a gas in atm not same as its concentration in
molarity)
○ BUT, if gases behaving ideally, a relationship b/w Kp and Kc can be derived
○ **Derivation is beyond scope of AP Exam**
■ → ∆n = c + d - (a + b)
● (sum of stoichiometric coefficients of gaseous products) - (sum of
stoichiometric coefficients of gaseous reactants)
○ ∆n represents difference b/w # moles of gaseous products + gaseous
reactants
R= constant on reference table (.08206 L atm/ mol K)
● n = # moles
■ → Kp = Kc (RT) ^∆n
○ IF the total # moles of gas is the same after the reaction as before →
■ ∆n = 0
■ Kp = Kc
Let’s try a practice question!
○ Q: Nitrogen monoxide, a pollutant in automobile exhaust, is oxidized to nitrogen dioxide
in the atmosphere according to the equation:
■ 2NO (g) + O2 (g) ⇌ 2NO2 (g)
■ Kp = 2.2 x 10^12 at 25° C
Given: Kp = 2.2 x 10^12
Equation → Kp = Kc (RT) ^∆n
Find: Kc
Solution:
1) Solve the equation for Kc
Kc=__Kp___
(RT) ^∆n
2) Calculate ∆n
∆n
2 - =3 = -1
...Continued
3) Substitute to calculate Kc. **The temperature must be in kelvins and the units are dropped when
writing Kc**
Kc= _____2.2 x 10 ^12_____________
(0.08206 L atm/mol K x 298K) ^-1
4) CHECK→ substitute answer back into Kp = Kc (RT) ^∆n → confirm that you get the original value of
Kc
Kp = Kc (RT) ^∆n
= 5.4 x 10^13 (0.08206 L atm/mol K
x 298 K) ^-1
= 2.2 x 10^12 ✓
Back to material… Units of K
●
In book, concentrations + partial pressures within equilibrium constant expression in units of
M and atm
○ BUT when expressing value of equilibrium constant→ units not included
■ BC formally, values of concentration/partial pressure substituted into equilibrium
constant expression are RATIOS of concentration/pressure to a reference
concentration (1 M) or reference pressure (1 atm)
● Ex: within equilibrium constant expression, pressure of 1.5 atm →
○ 1.5 atm = 1.5
1 atm
***As long as concentration units expressed in M for Kc and in atm for Kp → we can skip formality of
writing units and enter quantities directly into equilibrium expression ***
14.5: Heterogeneous Equilibria: Reactions Involving Solids and
Liquids
●
Consider : 2CO (g) ⇌ CO2 (g) + C (s)
○ Is this the equilibrium constant expression?
■ Kc = [CO2] [C]
[CO]^2
●
●
●
NO!
○ Because carbon is a SOLID, its concentration is constant
■ -If you double the amount of carbon its concentration remains same
■ Solids do not expand to fill container
● concentration only depends on density (which is constant as long as
some solid is present)
■ → pure solids (s) are NOT included in equilibrium expression
● (bc there constant value is incorporated into K)
Correct equilibrium constant expression:
Kc = [CO2]
[CO]^2
...Continued
●
Concentrations of pure liquids do NOT change either
○ pure liquids (reactants or products w (l)) also EXCLUDED from equilibrium expression
A Heterogeneous
Equilibrium
14.6: Calculating the Equilibrium Constant from
Measured Equilibrium Concentrations
●
most direct way to obtain experimental value for equilibrium constant of a reaction = measure concentrations
of reactants and products in a reaction mixture at equilibrium
○
Consider: H2 (g) + I2 (g) ⇌ 2HI (g)
■ w/ a mixture of H2 and I2 allowed to come to equilibrium at 445°C
■ **Since equilibrium constants depend on temperature, many problems state even though not
part of calculation**
■ measured equilibrium concentrations:
● [H2] = 0.11 M
● [I2] = 0.11 M
● [HI ]= 0.88 M
■ What is value of equilibrium constant? → We can write expression for Kc from balanced equation
Kc = [HI]^2
[H2] [I2]
= (0.78)^2
(0.11)(0.11)
= 5.0 x 10^1
●
for any reaction, equilibrium concentrations of the reactants and products depend on initial concentrations
(and in general vary from one set of initial concentrations to another)
○ However, equilibrium constant is always same at a given temperature (regardless of initial
concentrations)
Initial Concentrations
Equilibrium Concentrations
Equilibrium Constant
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
Kc = [HI^2]
[H2][I2]
0.50
0.50
0.0
0.11
0.11
0.78
Kc = (0.78)^2
(0.11)(0.11)
0.0
0.0
0.50
0.055
0.055
0.39
Kc = (0.39)^2
= 50
(0.055)(0.055)
0.50
0.50
0.50
0.165
0.165
1.17
Kc = (1.17)^2
= 50
(0.165)(0.165)
1.0
0.5
0.0
0.53
0.033
0.934
Kc = (0.934)^2
(0.53)(0.033)
0.50
1.0
0.0
0.033
0.53
0.934
Kc =(0.934)^2
= 50
(0.033)(0.53)
= 50
= 50
Ex: Table below shows different concentrations of
reactants/product from a different set of initial
concentrations
**No matter what initial concentrations are,
reaction always goes in direction that ensures
equilibrium concentrations, when substituted into
equilibrium expression, give same constant, K **
Previously, we calculated K from values of equilibrium
concentrations of all reactants and products→ in most
cases, only know initial concentrations of reactant(s)
and equilibrium concentrations of any one reactant or
product
→ can deduce other equilibrium concentrations
from stoichiometry of reaction!
Ex: Consider A(g) ⇌ 2B (g)
[A] initial = 1.00 M
[B] initial = 0.00 M
equilibrium reached → [A] = 0.75M
Since [A] changed by -0.25 M → can
deduce based on stoichiometry that
[B] changed by 2 x (+0.25 M) → 0.50
M
Summarized in “ICE table”
[A]
[B]
Initial
1.00
0.00
Change
-0.25
+0.50
Equilibrium
0.75
0.50
To calculate K, we use balanced equation to write
expression for equilibrium constant + then substitute
equilibrium concentrations from ICE table
K = [B]^2 = (0.50)^2 = 0.33
[A]
(0.75)
Let’s go through a guided example to find Equilibrium Constants from
Experimental Concentration Measurements!
Q: Consider the following reaction: CO (g) + 2H2 (g) ⇌
CH3OH (g)
A reaction mixture at 780°C initially w/ [CO] = 0.500 M +
[H2] = 1.00 M
At equilibrium, [CO]= 0.15 M
Find: value of equilibrium constant, K
1)
2)
Using balanced equation as a guide, create ICE table
showing known initial concentrations and
equilibrium concentrations of reactants and
products
For reactant/product whose concentration is known
both initially and at equilibrium, calculate the change
in concentration that occurs
[CO]
[H2]
[CH3OH]
Initial
0.500
1.00
0.00
Change
-0.35
-0.70
+0.35
Equil
0.15
0.30
0.35
3) Use change calculated and stoichiometric relationships
from balanced chemical equation to determine changes in
concentration of all other reactants and products (think
about +/- significance!)
4) Sum each column for each reactant and product to
determine equilibrium concentrations
5) Use balanced equation to write an expression for
equilibrium constant + substitute equilibrium
concentrations to calculate K
Kc = [CH3OH]
[CO][H2]^2
= 0.35___
(0.15)(0.30)^2
= 26
14.7: The Reaction Quotient: Predicting the Direction of
Change
●
When reactants of a chemical reaction mix→ generally react to form products→ reaction
proceeds to the right (toward products)
○ Can we predict direction change for a mixture not at equilibrium that contains both
reactants and products?
■ → reaction quotient (Qc) = ratio -- at any point in the reaction-- of the concentrations
of products raised to their stoichiometric coefficients divided by the concentrations
of the reactants raised to their stoichiometric coefficients
● aA + bB ⇌ cC + dD
Qc = [C]^c [D]^d
[A]^a [B]^b
Qp = P^c C P^d D
P^a A P^b B
●
different from equilibrium constant → at a given temperature, K w/ only one value and
specifies relative amounts of reactants and products at equilibrium
○ reaction quotient depends on current state of reaction + has many different values as
reaction proceeds
■ Ex: in a reaction mixture w/ only reactants, Qc = 0
Qc = [0]^c [0]^d
[A]^a [B]^b
● In reaction mixture w/ only products, Qc = ∞
Qc = [C]^c [D]^d
[0]^a [0]^b
● In a reaction w/ both reactants and products, each at concentration of 1 M, Qc = 1
Qc = (1)^c (1)^d
(1)^a (1)^b
● Qc → useful because Q relative to K is measure of progress of reaction toward
equilibrium
● At equilibrium, Qc = K
3 possible conditions:
Figure above shows Q as a function of concentrations
of A and B for A (g) ⇌ B (g) w/ K= 1.45
Q
K
Predicted Direction of
Reaction
0.55
1.45
To the right (toward
products)
Q<K → Q must get larger as reaction proceeds toward
equilibrium, while [reactants] decreases and [products]
increases
2.55
1.45
To the left (toward
reactants)
Q> K →Q must get smaller as reaction proceeds toward
equilibrium, while [reactants] increases and [products]
decreases
1.45
1.45
No change (at
equilibrium)
Q=K
In general…
● Q<K → Reaction goes to right (toward
products)
● Q>K → Reaction goes to left (toward reactants)
● Q = K→ Reaction is at equilibrium
Let’s try a practice question!
Consider:
I2 (g) + Cl2 (g) ⇌ 2ICl (g)
Kp = 81.9
Reaction mixture contains PI2 = 0.114 atm, PCl2 = 0.102 atm, and PICl = 0.355 atm
Is reaction mixture at equilibrium? If not, in which direction will reaction proceed?
1)
Calculate Q
Qp = ____P^2 ICl______
PI2PCl2
= (0.355)^2
(0.114)(0.102)
=10.8
1)
Compare Q to K
Qp = 10.8
Kp = 81.9
Since Qp < Kp → reaction is not at equilibrium and will proceed to the right
From the
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