UNIT 3 CHEMICAL EQUILIBRIUM. Introduction to Chemical Equilibrium Focus has always been placed upon chemical reactions which are proceeding in one direction.

UNIT 3

CHEMICAL EQUILIBRIUM

Introduction to Chemical Equilibrium

Focus has always been placed upon chemical reactions which are proceeding in one direction (forward)

We assume that chem. rxn. proceed until one of the reactants is entirely consumed

However many rxns are reversible, i.e. they can occur in either direction

Products regenerate the original reactants

For a given forward reaction aA + bB ---> cC + dD

One can conceive of the reverse process cC + dD ---> aA + bB

product molecules (C & D) are not present at the beginning of a reaction, thus reverse reaction is non-existent

as [products] the reverse rxn. is initiated

For a given forward reaction aA + bB ---> cC + dD

the rate of the forward rxn. b/c [reactant] is reduced by conversion to products & the rate of the reverse reaction (as product concentrations build up)

the rate of of the reverse rxn. will be proportional to some product of the concentrations of C and D

For a given forward reaction aA + bB ---> cC + dD

eventually a point in time is reached when the Rate Forward Rxn. = Rate Reverse Rxn.

This is the point of CHEMICAL EQUILIBRIUM unless the system is disturbed by a temp.

change or by adding excess reactant or product, the equality of RATES is maintained

CONCENTRATIONS of all chemical species reach "equilibrium" (i.e. []’s remain constant, but NOT the same)

The variation of reaction rates with time is illustrated graphically as follows:

Chemical Equilibrium = Dynamic Equil.

once equilibrium is reached, the rxn. seems to be stopped

we cannot visibly see any changes in reality the reactions are still taking

place; chem. equil. is a dynamic or changing process

the observable properties and [] of reactants and products are constant

A chemical system is said to be in a state of equilibrium if it meets the following criteria:

1. The system is closed. 2. The observable macroscopic properties are

constant. 3. The reaction is sufficiently reversible so

that observable properties change and then return to the original rate when a factor that affects the rate of the reaction is varied and then restored to its original value.

The Equilibrium Constant

Consider the forward and reverse processes for some general "atom exchange" reaction

aA + bB ---> cC + dDa, b, c & d = coefficients for substances A, B,

C & D respectively

Assuming the forward rxn. occurs by a one-step mechanism, then…

Rate Forward Rxn. = Rf = kf[A][B]Similarly, assuming the reverse rxn. is also a one-step

mechanism, then

Rate Reverse Rxn. = Rr = kr[C][D]"Principle of Microscopic reversibility”

activated complex in the reverse rxn. is identical to the activated complex for the forward rxn.

Thus, the energy profile for the forward and reverse rxns can be illustrated by the same plot:

Ef = activation energy of forward rxn.Er = activation energy of reverse rxn.

The activated complex is identical for both directions, thus ∆H = Er - Ef

NOTE: If ∆H = 0, then

Er = Ef

At equilibrium Rf = Rr

Thus, kf[A][B] = kr[C][D]

By rearrangement we can obtain the expression [C]c[D]d = kf = Keq (equil. constant)

[A]a[B]b kr

Keq describes the relative []’s of reactants & products @ equilibrium

Keq ratio is always a constant value for a given rxn. regardless of []’s of reactants and products @ equil.

Every reversible rxn. obeys this relationship & has a specific Keq

“Law of chemical equilibrium” every reversible rxn. proceeds to a state of equil. & has a specific ratio of the [reactant] : [products] expressed by Keq

Examples: Write the equil. expressions for the following:1. 2CO(g) + O2(g) 2CO2(g)

Keq = [CO2]2 .

[CO]2[O2]

2. CO(g) + 3H2(g) CH4(g) + H2O(g)

Keq = [CH4][H2O] .

[CO][H2]3

Examples: 3. At 100°C 1.0L of gas at equilibrium contains

0.0045 mol N2O4 and 0.030 mol NO2. Calculate the Keq.

2NO2 N2O4

Keq = [N2O4] = (0.0045) = 5.0

[NO2]2 (0.030)2

NOTE: Keq has no units

What is indicated by the magnitude of the Keq value?

Keq does NOT tell you anything about the time it takes to reach equil.

Keq tells you the extent to which a rxn. proceeds to completion

Keq >> 1 (“>>” means significantly)

numerator (products) must be much larger than denominator (reactants)

@ equilibrium this rxn. system consists mainly of products and is considered to proceed to completion

“Equilibrium lies to the right” (toward product side)

Keq << 1

a very small value, as the denominator (reactants) is much larger than the numerator (products)

system mostly reactants reaction barely gets going

“Equilibrium lies to the left” (toward reactant side)

In general…

Keq > 1 products favored @ equilibrium

(spontaneous rxn.)Keq < 1 reactants favored @ equilibrium

Assignment Pg. 419 – 420, # 11 – 15,Pg. 425 – 426, # 43 – 46, # 48 –

50

Calculation of Concentrations at Equilibrium

Use “ICE” to help you find the concentrations of each substance at equilibrium.

I nitial

C hange

E quilibrium

Examples:

1. When 0.40 moles of PCl5 is heated in a 10.0 L container, an equilibrium is established in which 0.25 moles of Cl2 is present.

a) What is the number of moles of PCl5 and PCl3 at equilibrium?

b) What are the equilibrium concentration of all three components?

a)            PCl5 <=====> PCl3 + Cl2

Initial 0.40 mol ------ ------- Change 0.25 mol +0.25 mol +0.25 mol Equil. 0.15 mol 0.25 mol 0.25 mol

b) [PCl5] = 0.15 moles = 0.015 moles/L 10.0 L [PCl3] = [Cl2] = 0.25 moles = 0.025 moles/L 10.0 L

Example:

2. The equilibrium constant for the reaction represented below is 50 at 448°C

H2(g) + I2(g) <=====> 2 HI(g) a) How many moles of HI are present at equilibrium when

1.0 moles of H2 is mixed with 1.0 moles of I2 in a 0.50 L container and allowed to react at 448°C?

b) How many moles of H2 and I2 are left unreacted? c) If the conversion of H2 and I2 to HI is essentially

complete, how many moles of HI would be present? d) What is the percent yield of the equilibrium mixture?

a)

The concentrations at the start are: [H2] = 1.0 mol/0.50 L = 2.0 mol/L

[I2] = 1.0 mol/0.50 L = 2.0 mol/L

[HI] = 0 moles therefore 0 mol/L

Fill in the numbers under the reaction for the starting concentrations.

Let 'x' = number of moles of H2 consumed per litre

H2 + I2 <=====> 2 HI I 2.0 M 2.0 M 0 M C x x 2(+x)E 2.0 x 2.0 x 2x

Fill in the finish concentrations into the Keq equation. and solve for 'x‘.

7.1 = 2x . 2.0 - x

50 = (2x)2 . 14.2 - 7.1x = 2x

(2.0 - x)(2.0 - x) x = 1.56 mol/L 50 = (2x)2

.

(2.0 - x)2

You are solving for 'x' which is in moles/L (M)

therefore you can automatically fill in

the units for any value of 'x'.

Now to answer the question. What are the final reactant and product concentrations?

[H2] = 2.0 M x = 2.0 M 1.56 M = 0.44 M

[I2] = 2.0 M x = 2.0 M 1.56 M = 0.44 M

[HI] = 2x = 2 • 1.56 M = 3.12 M

b) From part a) you get the M of H2 and I2 left unreacted. This is the concentration not the answer to the question. You are asked for the number of moles left unreacted therefore moles = concentration • volume [H2] = [I2] = 0.44 mol/L • 0.5 L= 0.22 moles

Therefore 0.22 moles of H2 and 0.22 moles of I2 are left unreacted.

c) If you look at the equation again you can see that there is exactly the right amount of H2 to react with I2. Theoretically if both the H2 and I2 where all used up then we should make 2 moles of HI. This is the theoretical yield because it is what could be made in theory if everything went to completion.

d) The percentage yield is calculated by using the actual yield from part a) and the theoretical yield from part b). You may use either moles or moles/L but make sure that they are both the same.

Percentage Yield = Actual Yield • 100 Theoretical Yield

= 1.56 M • 100 2.00 M= 78% yield (note that the moles/L

units cancel)

Example: (Challenging)

3. How many moles of HI are present at equilibrium when 2.0 moles of H2 is mixed with 1.0 moles of I2 in a 0.50 L container and allowed to react at 448°C. At this temperature Keq = 50.

H2(g) + I2(g) <=====> 2 HI(g)

The concentrations at the start are: [H2] = 2.0 mol/0.50 L = 4.0 mol/L [I2] = 1.0 mol/0.50 L = 2.0 mol/L [HI] = 0 moles therefore 0 mol/L

Let 'x' = number of moles of H2 consumed per litre

H2 + I2 <=====> 2 HI I 4.0 M 2.0 M 0 M C x x 2(+x)E 4.0 x 2.0 x 2x

50 = (2x)2 . (4.0 - x)(2.0 - x)

50 = 4x2 .

8.0 - 6.0x + x2 400 - 300x + 50x2 = 4x2 46x2 - 300x + 400 = 0 (standard form) Use the quadratic equation to solve for 'x'.      

In this case a = +46, b = -300 and c = +400.

We only started with 4 mols/L of H2 to begin with, so 4.7 is to large. It is therefore the unreal root. So let x = 1.9 mol/L.

Substitute 'x =1.9 mol/L' back up into the original concentration calculations to see how much is left after reaction.

[H2] = 4.0 M x = 4.0 M 1.9 M = 2.1 M

[I2] = 2.0 M x = 2.0 M 1.9 M = 0.1 M

[HI] = 2x = 2 • 1.9 M = 3.8 M

The moles of HI present at equilibrium would be:

moles = concentration • volume = 3.8 moles/L • 0.5 L = 1.9 moles

Homogenous & Heterogeneous EquilibriaHomogenous Equilibria

equilibrium conditions for rxns in which all the reactants & products are in the same state eg. 2CO(g) + O2(g) 2CO2(g)

Heterogenous Equilibria equilibrium conditions for rxns that involve substances in more than 1 state

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

Earlier, we expressed [] as M (for gases)

[] of solids & liquids in a chem. rxn. do not change substantially and so are not included in the equilibrium constant

[solid & liquid] change during a rxn, left out of Keq

eg. C(s) + H2O(g) CO(g) + H2(g)

Keq = [CO][H2] .

[H2O]

The Reaction Quotient

It is hard to tell if a rxn. has reached equilibrium The “reaction quotient” (Q) is used to

determine if a rxn. is @ equilibrium Q is calculated like Keq except the []’s are taken

@ the time of measurement, not necessarily @ equilibrium

Thus we can compare the value of Q to the value of Keq to see how the rxn. is progressing

Rxn. is @ equilibrium when Keq = Q

Example:

At 473°C, nitrogen gas is reacted with hydrogen gas to produce ammonia. A measurement was taken as the reaction proceeded and it was found that there was 0.15 M of nitrogen gas, 0.0020 M of hydrogen gas, and 0.15 M of ammonia. Given that the Keq is 0.105, is the reaction at equilibrium?

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

temp. = 473°C [NH3] = 0.15 M

Keq = 0.105 [N2] = 0.15 M

[H2] = 0.0020 MKeq = [NH3]2 Q = (0.15)2 .

[N2][H2]3 (0.15)(0.0020)3

= 1.9 x 107

Keq Q this rxn. is NOT @ equilibrium

Which direction will rxns proceed to reach equilibrium?

Q < Keq

Q is less than Keq

denominator is too large and numerator is too small

too much of reactants, too little of products

rxn. will proceed to the right (in the direction of products)

Q > Keq

Q is greater than Keq

denominator is too small and numerator is too large

too much products, too little reactants

rxn. will proceed to the left (in the direction of reactants)

Example:

COCl2(g) CO(g) + Cl2(g) Keq = 170

At the time of measurement:[CO] = 0.15 M[Cl2] = 0.15 M

[COCl2] = 1.1 x 103 M

Is the rxn. @ equilibrium? If not, which direction will it proceed?

Q = [CO][Cl2] = (0.15)(0.15) . = 20.

[COCl2] (1.1 x 103)

Q < Keq thus rxn. is NOT @ equilibrium

there is too much reactants rxn. will shift right in the direction of products