Chapter 13 Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time Equilibrium is reached.

Chapter 13Chemical Equilibrium

The state where the concentrations of all reactants and products remain constant with time

Equilibrium is reached when rates of forward and reverse reactions are equal

Equilibrium Condition

Dynamic Condition- rates of forward and reverse reactions are equal

Law of Mass Action- general description of the equilibrium condition

- generic equation

jA + kB lC + mD

A,B,C,D chemical substances

J,k,l,m coefficients

Equilibrium Expression

K=[C]l [D]m

[A]j [B]k

Write the equilibrium expression for:

1) PCl5(g) PCl3(g) + Cl2(g)

2) Cl2O7(g) + 8H2(g) 2HCl(g) + 7H2O(g)

Calculating the Values of K

Calculate the equilibrium constant, K, for the following reaction at 25°C.

H2(g) + I2(g) 2HI(g)

If the equilibrium concentrations are [H2] = 0.106 M, [I2]= 0.022M, and [HI]=1.29M

Using the same example, calculate the equilibrium concentration of HI if H2 =.81M and I2 = .035 M and K= 7.1x 102

Equilibrium Expressions involving Pressures

Ideal Gas Law

PV= nRT

P= pressure(atm)

V= volume(L)

n= # of moles of gas(mol)

R= universal gas constant= .08206Latm/Kmol

T= temp. in K(273)

Rearrange the Ideal Gas Law for Pressure

P=(n/v) RT or

P= CRT where C= molar concentration

of the gas

Kp= [C]l[D]m/[A]j [B]k

What is the relationship between Kc and Kp

Kp= K(RT)n

n= sum of the gaseous products coefficients minus the sum of the gaseous reactants coefficients

Ex. 13.4Calculating Values of Kp

Calculate the value of Kp for the following reaction at 25°C if PNOCl = 1.2 atm, PNO= 5.0x10-2 atm, and PCl2= 3.0x10-1atm.

2NO(g) + Cl2(g) 2NOCl(g)

Calculating K from Kp

Calculate the value of K at 25°C for the reaction

2NO(g) + Cl2(g) 2NOCl(g)

If Kp= 1.9x103

Kp= K(RT)n

T = 273 +25= 298Kn= 2-(2+1)= -1

Heterogeneous Equilibria

Involves more than one phase

CaCO3(s) CaO(s) + CO2(g)

K= [CaO] [CO2] / [CaCO3]

K= [CO2]

Write the expressions for K and Kp

NH4NO2(s) N2(g) + 2H2O(g)

HCl(g) + NH3(g) NH4Cl(s)

Applications of the Equilibrium Constant

Reaction Quotient(Q)- using initial concentrations in the law of mass action

N2(g) + 3H2(g) 2NH3(g)

Compare Q to K 3 Possible Cases1) Q= K, equilibrium , no shift2) Q K, system shifts to left3) Q K, system shifts to right

Predicting the shift

H2(g) + I2(g) 2HI(g)

K= 7.1x102 at 25°C

a. Q= 427

Q K shift to right

b. Q=1522

QK , shift left

Con.

[H2]0= .81M [I2]0=.44M [HI]0=.58M

Q= ?

[H2]0=.078M [I2]0=.033M [HI]0=1.35M

Q=?

Example 13.8

N2O4(g) 2NO2(g)

Kp=.133atm PN2O4=2.71atm

What is PNO2?

Example 2,p. 627

1.00 L flask initially contained 0.298 mol PCl3(g) and 8.70x10-3 mol PCl5(g). After reaching equilibrium 2.00x10-3 mol of Cl2(g) was found in the flask. Calculate the equilibrium concentrations of all species and value of K.

PCl5(g) PCl3(g) + Cl2(g)

K= [PCl3][Cl2]/[PCl5]

Solving Equilibrium Steps

1. Write a balanced equation

2. Write the equilibrium expression

3. List the initial concentrations

4. Calculate Q and determine the direction of the shift

5. Use ICE box

6. Solve for unknown(s)

Ex. 13.10, p. 628

CO(g) + H2O(g) CO2(g) + H2(g)

K= 5.10

Calculate the equilibrium concentration of all species if 1.00 mol of each component is mixed in a 1.000 L flask.

Example, p. 632

H2(g) + F2(g) 2HF(g)

K= 1.15x102

Suppose 3.000 mol of H2 and 6.000 mol of F2 are mixed in 3.000 L flask. Calculate the equilibrium concentration for all species.

Le Chatelier Principle

If a system at equilibrium is subjected to a stress(change), the equilibrium will shift in an attempt to reduce the stress

1. Effect of a change in concentration

a. If a reactant/product is added, the system will shift away from the added component

b. If a reactant/product is removed, the system will shift toward the removed component

2. Effect of a Change in Pressure

a. Add or remove a gaseous reactant/product - add, shift away

-remove, shift towardsb. Add an inert gas

- there is no effect on equilibriumc. Change the volume of the container

- when the container volume is reduced, the system will shift toward the side involving the smaller # of gaseous molecules

- when the volume is increased, it will shift toward the side with the larger # of molecules

Example of Pressure

N2(g) + 3H2(g) 2NH3(g)If volume is decreased, then ?Shift right, 4 2If volume is increased, then ?Shift left, 2 4Determine the shift if the volume is reduced:

P4(s) + 6Cl2(g) 4PCl3(l)

Shift right, P4(s) and PCl3 (l), look only at Cl2

PCl3(g) + 3NH3(g) P(NH2)3(g) + 3HCl(g)No effect 4 on each side

3. Effect of a change in Temperature

a. If H is positive, it is an endothermic reaction energy is viewed as a reactant

b. If H is negative, it is an exothermic reaction energy is viewed as a product

N2(g) + O2(g) 2NO(g) H= 181kJ

181kJ + N2(g) + O2(g) 2NO(g)Shift right

2SO2(g) + O2(g) 2SO3(g) H= -198kJ

2SO2(g) + O2(g) 2SO3(g) + -198kJShift left

Summarizing Le chatelier

N2O4(g) 2NO2(g) H = 58kJ

58kJ + N2O4(g) 2NO2(g)

Ksp

The Solubility-Product Constant Deals with equilibria associated with solids

dissolving to form aqueous solutions

AgCl(s) Ag+(aq) + Cl-(aq)

K= [Ag+] [Cl-]/ [AgCl]

so

Ksp= [Ag+] [Cl-]

Ksp Examples

Determine the Ksp of calcium fluoride given that its molar solubility is 2.14x 10-4 M.

CaF2(s) Ca2+ (aq) + 2F-(aq)

Calculate the molar solubility(mol/L) of silver chloride that has a Ksp= 1.77x10-10.

AgCl(s) Ag+(aq) + Cl-(aq)

Con.

Calculate the molar solubility of Tin(II) hydroxide if Ksp=5.45x10-27.

Sn(OH)2(s) Sn2+(aq) + 2OH-(aq)