Chapter 14: Chemical

EquilibriumSections 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

● n = # moles■ → Kp = Kc (RT) ^∆n

○ IF the total # moles of gas is the same after the reaction as before → ■ ∆n = 0 ■ Kp = Kc

R= constant on reference table (.08206 L atm/ mol K)

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

2) Calculate ∆n ∆n =

Kc=__Kp___ (RT) ^∆n

2 - 3 = -1

...Continued

3) Substitute to calculate Kc. **The temperature must be in kelvins and the units are dropped when writing Kc**

4) CHECK→ substitute answer back into Kp = Kc (RT) ^∆n → confirm that you get the original value of Kc

Kc= _____2.2 x 10 ^12_____________ (0.08206 L atm/mol K x 298K) ^-1

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

● Concentrations of pure liquids do NOT change either ○ pure liquids (reactants or products w (l)) also EXCLUDED from equilibrium

expression

...Continued

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 equationKc = [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 = 50 (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 = 50 (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)

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”

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)

[A] [B]

Initial 1.00 0.00

Change -0.25 +0.50

Equilibrium 0.75 0.50

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 MFind: value of equilibrium constant, K

1) Using balanced equation as a guide, create ICE table showing known initial concentrations and equilibrium concentrations of reactants and products

2) For reactant/product whose concentration is known both initially and at equilibrium, calculate the change in concentration that occurs

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

[CO] [H2] [CH3OH]

Initial 0.500 1.00 0.00

Change -0.35 -0.70 +0.35

Equil 0.15 0.30 0.35

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

Figure above shows Q as a function of concentrations of A and B for A (g) ⇌ B (g) w/ K= 1.45

3 possible conditions:

In general…● Q<K → Reaction goes to right (toward

products) ● Q>K → Reaction goes to left (toward

reactants) ● Q = K→ Reaction is at equilibrium

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

Let’s try a practice question!Consider: I2 (g) + Cl2 (g) ⇌ 2ICl (g)Kp = 81.9Reaction 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

2) Compare Q to K

Qp = ____P^2 ICl______ PI2PCl2= (0.355)^2(0.114)(0.102)=10.8

Qp = 10.8Kp = 81.9Since Qp < Kp → reaction is not at equilibrium and will proceed to the right

From the College Board→

Questions?