CHEMICAL EQUILIBRIUM - RATES OF REACTION kF

Reactants products kB

Chemical reactions are a dynamic process, that is, reactions involve both forward and reverse processes.

Chemical Equilibrium is reached by as reaction mixturewhen the rates of forward and reverse reactions becomes equal.

kF = kB

NO net change appears obvious although the system is still in constant motion.

LAW OF MASS ACTION

k equilibrium constantaA + bB cC + dD

k = Products Reactants

k = [C]c [D]d = Q reaction quotient [B]b [A]a

Write the equilibrium equation for:a. HC2H3O2 H+ + C2H3O2

-

b. H2O + H2O H3O+ + OH-

c. 4NH3(g) + 302(g) 2N2(g) + 6H2O(g)

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

Predicting the direction of reaction:

Q > K forms more reactants

Q = K equilibrium

Q < K forms more products

Note: 1. kf = 1 kr

2. k = kn If the balanced equation is multipliedby a factor then the K(& Q) is multiplied by the exponent.

K as either Kc or Kp

Kc = the equilibrium constant using concentrations.

Kp = the equilibrium constant using pressureP = n P x n

RT v •• v

Kp = Kc(RT) n gas

Write Kp the Kc for:1. N2(g) + 3H2(g) 2NH3(g)

2. N2O4(g) 2NO2(g)

3. Calculate kp if kc = 0.105 for #14. 2SO3(g) 2SO2(g) + O2(g)

if Kc = 4.07 x 10-3, what is kp?

N2 + O2 2NO

a. Since Q is very small Q<<<1, very little, NO will form @25°C. The equilibrium lies to the left favoring reactants.

N2 + O2 = 2NO

b. Kp = (PNO)2 n = 2 - 2 = 0 PN2PO2 kp = Kc

Kc = [NO]2

[N2][O2]

c. Kp = PN2 PO2

(PNO)2

d. n = 2 - 2 = 0 kp = kckc = [N2][O2]

[NO]2

DIRECTION OF REACTIONS AND Keg

1. The following reaction is a means of “fixing” nitrogen:N2(g) + O2(g) 2 NO(g)

A. If the value for Q at 25°C is 1 x 10-30, describe the feasibility of this reaction for Nitrogen fixation.

B. Write the equilibrium expression, Kc

C. Write the equilibrium expression for 2NO(g) N2(g) + O2(g)

D. Determine the Kc for “C”

HOMOGENEOUS EQUILIBRIAH2(g) + I2(g) 2HI(g)

Kp = (PHI)2

(PI2)(PH2) Kc = ?

2O3(g) 3O2(g)

Kp = (PO2)3

(PO3)2 Kc = ?HETEROGENEOUS EQUILIBRIA

2H2O(l) H3O+(aq)_ + OH-(aq)

Kc = [H3O+][OH-]

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)

Kc = ?

HETEROGENEOUS EQUILIBRIA

Substance in more then 1 phase

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

Kc = [CaO][CO2] [CaCO3]

How do the [ ] of solid express?A. D = g/cm3 = mol MW g/mol cm3

Pure solids & liquid have constant [ ]Kc = constant [CO2]

constantKc = Kc con = [CO2]

con

Q. Each of the mixtures listed below was placed in a closed container and allowed to stand. Which of these mixtures is capable of attaining the equil, expressed by 1

a) pure CaCO3

b) CaO & PCO2 > Kp

c) solid CaCO3 & PCO2 > Kp

d) CaCO3 & CaO

CALCULATING THE Keq

1. In one experiment, Haber introduced a mixture of H2

& N2 into a reaction vessel and allowed the systemto attain chemical equilibrium at 472°C. The equilibrium mixture of gases were analyzed and found to contain 0.1207M H 2, 0.0402M N2, and0.00272M NH3. Calculate Keq.

2. Nitryl Chloride, NO2Cl, is in equilibrium in a closed container with NO2 and Cl2.

2 NO2Cl(g) 2 NO2(g) + Cl2(g)

Calculate Keq if [NO2Cl] = 0.00106M[NO2] = 0.0108M & [Cl2] = 0.00538M

3. For the Haber processN2(g) + 3H2(g) 2NH3(g)

Kp = 1.45 x 10-5 at 500°C

If an equilibrium mixture of the three gas started with partial pressures of 0.928 atm for H2 and 0.432 atm for N2, what is the partial pressure of NH3?

4. A 1.00 L flask is filled with 1.00 mol of H2 and 2.00 mol I2 at 448°C Kc is 50.5.

What are the equilibrium concentrations of H2, I2 & HI?

Workshop CALCULATING Keq

1. A mixture of 5.0 x 10-3 mol of H2 and 1.0 x 10-2 mol of I2 is placed in a 5.0L container at 448°C and allowed to come to equilibrium. Analysis of this equilibrium mixture shows that the [HI] is 1.87 x 10-3 M.

Calculate Kc: H2(g) + I2(g) 2HI(g)

2. At 448°C the equilibrium constant Kc for the reaction below is 50.5.

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

Predict how the reaction will proceed to reach equilibrium if the initial amount of HI is 2.0 x 10-2 mol,H2 is 1.0 x 10-2 mol, and I2 is 3.0 x 10-2 mol in a 2.00 L container.

APPLICATION OF Keq

1. Predicting the direction of reaction Q = reaction quotient

at equil Q = K Q > K species on Rt (prod)(no net Rx) react to form left

K = [Equil] Q = [Non Equil]Q < K forms more products

Goal: Calculate Q to determine state of Rx, equil, more product or more reaction

1. At 448°C the equilibrium constant Kc for the reaction is 50.5.

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

Predict how the Rx will proceed to reach equil at 448°C if the initial amount of HI is 2.0 x 10-2 mol, 1.0 x 10-2 mol, H2 and 3.0 x 10-2 mol I2 in at 2.0 L container.

[HI]° = 2 x 10-2 mol/2.0L = 1.0 x 10-2M[H2]° = 1.0 x 10-2mol/2.0L = 5.0 x 10-3M[I2]° = 3.0 x 10-2mol/2.0L = 1.5 x 10-2M

Q = Prod = [HI]2 = (1.0 x 10-2)2 = 1.3 React [H2][I2] (5 x 10-3)(1.5 x 10-2)

since K = 50.5, Q = 1.3, Q < K [HI] will need to increase and [H2][I2] will decrease to reach equilibrium.

2. At 1000K the value of Kc for the reaction

2S03(g) 2SO2(g) + O2(g)

is 4.07 x 10-3. Calculate the value for Qand predict the direction in which the reactionwill proceed towards equil if the initial concentration of reactants are:

[SO3] = 2 x 10-3 M[SO2] = 5 x 10-3 M[O2] = 3 x 10-2 M

Q = 0.2 reaction will proceed from Rt to left forming SO3.

CALCULATING Keq

1. Sulfur Trioxide decomposes at High temperature in a sealed container.

2 SO3(g) 2 SO2(g) + O2 (g)

Initially the vessel is filled at 1000K with SO3(g) at a concentration of 6.09 x 10-3M. At equilibrium, the[SO3] is 2.44 x 10-3M. Calculate Kc.

2. Calculate the value for Q and predict the direction in which the reaction will proceed towards equilibrium if the initial concentrations are:

[SO3] = 2.0 x 10-3M[SO2] = 5.0 x 10-3M[O2] = 3.0 x 10-2M

LE CHATELIER’S PRINICIPLE

“If a system at equilibrium is disturbed by a change in temperature, pressure, or concentration of one of its components, that system will shift it’s equilibrium positionas to ‘counteract’ the effect of the disturbance.”

Equilibrium can be disturbed by:

- adding or removing components- a change in pressure- a change in volume- a change in temperature

PREDICTING THE DIRECTION OF THE SHIFT

I. CATALYSTA catalyst increases the rate at which equilibriumis achieved but not the composition of the equilibrium mixture.

II. THE REACTION QUOTIENTQ < K: the reaction shifts to the productsQ > K: the reaction shifts to the reactants

III. CHANGES IN VOLUMEReducing the volume of a gas at equilibrium causesthe system to shift in the direction that reduces the number of moles of gas.

IV. CHANGES IN TEMPERATUREWhen heat is added to a system, the equilibrium

shifts in the direction that absorbs heat. Cooling has the opposite effect and shifts the equilibrium towards the side which produces heat.

Endothermic Reactions:heat + Reactants Products

An increase in temperature leads to a shift towards the products, a decrease leads to a shift towards the reactants. (Keq increases)

Exothermic Reactions:Reactants Products + heat

An increase in temperature leads to a shift towards reactants. (Keq decreases)

Example 1:

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

In which direction will the equilibrium shift when each of the following changes are made to a system at equilibrium?

A) add N2O4

B) remove NO2

C) increase the total pressure by adding N2

D) increase the volume of the container

E) decrease the temperature

Answer: N2O4 2NO2

A) The system will adjust so to decrease [N2O4] shiftsto products

B) Shifts to reactants (NO2 removed)

C) N2 will increase total pressure but since it is not involved in Rx the partial pressures of N2O4 and

NO2 are unchanged, no shift.

D) volume shifts to more moles of gas

E) Temp - Rx is endo heat + N2O4 2 NO2

Temp shifts so more heat produced K is affected

Example 2:

PCl5(g) PCl3(g) + Cl2(g) H = 88 kJ

In which direction will the equilibrium shift when each of the following changes are made to a system at equilibrium?

A) add Cl2

B) temperature is increased

C) the volume of the reaction system is decreased

D) PCl5 is added

E) a catalyst is added

1. The equilibrium constant for the Haber processat 472°C is Kc = 0.105. A 2.00 L flask is filled with 0.500 mol of ammonia and is then allowed to reach equilibrium at 472°C. What are the equilibriumconcentrations?

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

2. For the reaction

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

at a certain temperature Kc equals 450. What willhappen when 0.10 mol of PCl5, 1.0 mol of PCl3, andl.5 mol of Cl2 are added to a 2.0-L container and the system is brought to the temperature at which Kc=450. What are the equilibrium concentrations?

EFFECT OF VARIOUS DISTURBANCES ON AN EQUILIBRIUM SYSTEM

DISTURBANCE NET DIRECTION OF REACTION EFFECT ONVALUE OF K

Concentration Increase (reactant) Toward formation of product None Decrease (reactant) Toward formation of reactant None

Pressure (volume) Increase P Toward formation of lower

amount (mol) of gas1 None Decrease P Toward formation of higher

amount (mol) of gas None

Temperature Increase T Toward absorption of heat Increases H°rxn>0

Decreases if H°rxn<0 Decrease T Toward release of heat Increases H°rxn<0

Decreases H°rxn>0Catalyst added None; rates of forward and

reverse reactions increase equally None

VAN’T HOFF EQUATION

Changes in K due to T

In K2 = - H°rxn ( 1 - 1) K1 R T1 T2

R = 8.314 J/mol KT = Kelvin

The formation of methanol is an important industrialreaction in the processing of new fuels. At 298K, Kp = 2.25 x 104 for the reaction

CO(g) + 2 H2(g) CH3OH(l)

If H°rxn = -128 kJ/mol CH3OH, calculate kp at 0°C.

Q 2. CH4(g) + CO2(g) 2CO(g) + 2 H2(g)

A. What is the percent yield of H2 when equimolar mixture of CH4 and CO2 with a total pressure of 20.0 atm reaches equilibrium at 1200K at which Kp = 3.548 x 106?

B. What is the percent yield of H2 for this system at 1300K, at which Kp = 2.626 x 107?

C. Use the Van’t Hoff equation to find H°rxn.