Get two 10 mL graduated cylindersPlace 5 mL in one, and 10 mL in the otherPlace a straw in each cylinder and transfer water from one tube to the other

Make a graph of Volume vs # of transfers

# of transfers

Volume mL

Volume mL

0 10.0 5.0123456789

10111213141516171819202122232425

VolumemL

# of transfers

If a volatile liquid is placed in an open container underneath an enclosed space the liquid will evaporate until the air in the closed container is saturated with particles of the vapour.

When the rate of evaporation is equal to the rate of condensation, no net change is occurring. The system is in a state of dynamic equilibrium since both events occur simultaneously.

Most elementary reactions are, in theory, reversible so they are capable of establishing a dynamic equilibrium. The gas, NO2 is composed of small rapidly moving particles which collide randomly. Sometimes these collisions result in the formation of N2O4 molecules which are capable of spontaneously reforming NO2 molecules after breaking apart. This gas is really a mixture of 2 gases where the rates of formation of each from the other is equal when in dynamic equilibrium.

NO2

N2O4

NO2 -> N2O4 Movie

This dynamic equilibrium is represented by the following equation:

NO2(g) + NO2(g) N2O4(g)

2NO2(g) N2O4(g)

or

A system at equilibrium can be disturbed in a number of ways. When disturbed it tends to respond in a way which offsets the disturbance. This is called LeChatelier’s Principle. To illustrate this consider the following system.

If water is added to the left water flows from

left to right

If water is removed from the left water flows from

right to left

This principle can be applied to an equilibrium. Consider this system

A(g) + B (g) C(g) + D(g)

If some A is added to the system the forward reaction will go faster, using up A and B faster than it can be replaced by C and D reacting in the reverse reaction. The net result can be shown by the following graphs.

A(g) + B (g) C(g) + D(g)

[A]

[B]

time

time

[C]

[D]

time

time

Some A is added

New Equilibrium is established

A(g) + B (g) C(g) + D(g)

[A]

[B]

time

time

[C]

[D]

time

time

Some B is addedNew Equilibrium is established

A(g) + B (g) C(g) + D(g)

[A]

[B]

time

time

[C]

[D]

time

time

New Equilibrium is established

Some C is added

A(g) + B (g) C(g) + D(g)

[A]

[B]

time

time

[C]

[D]

time

time

New Equilibrium is established

Some D is added

A(g) + B (g) C(g) + D(g)

[A]

[B]

time

time

[C]

[D]

time

time

Some A is removed

New Equilibrium is established

A(g) + B (g) C(g) + D(g)

[A]

[B]

time

time

[C]

[D]

time

time

Some B is removedNew Equilibrium is established

A(g) + B (g) C(g) + D(g)

[A]

[B]

time

time

[C]

[D]

time

time

New Equilibrium is established

Some C is removed

A(g) + B (g) C(g) + D(g)

[A]

[B]

time

time

[C]

[D]

time

time

New Equilibrium is established

Some D is removed

2A(g) + B (g) 4C(g) + 3D(g)

[A]

[B]

time

time

[C]

[D]

time

time

Some A is added

New Equilibrium is established

2x

x

4x

3x

4A(g) + 2B (g) 3C(g) + D(g)

[A]

[B]

time

time

[C]

[D]

time

time

Some B is removedNew Equilibrium is established

2xx

4x

3x

4A(g) + 2B (g) 3C(g) + D(g)

[B]

time

[C][D]

Some B is removed

2x

x3x

[A]

4x

or shown all on 1 graph

How do changes in Temperature affect a system at equilibrium? To consider this let’s look at this equilibrium.

2NO2(g) N2O4(g)

When this equilibrium is upset by heating, it becomes darker brown. Does this suggest this reaction is endothermic or exothermic?Hint: Increasing temperature is just like adding heat.

BrownColourless

2NO2(g) N2O4(g) + heat

If an equilibrium is cooled, the equilibrium shifts to the side with the heat as the equilibrium responds to this disturbance by replacing the heat which has been removed. Graphs for the system above are:

[NO2]

time

2x

2NO2(g) N2O4(g) + heat

The system is plunged into ice water(removal of heat)

[N2O4]

time

x

2A(g)+B (g) + heat 4C(g) + 3D(g)

[A]

[B]

time

time

[C]

[D]

time

time

The temperature is increased

New Equilibrium is established

2x

x

4x

3x

How do changes in Volume affect a system at equilibrium? To consider this let’s look at this equilibrium.

2NO2(g) N2O4(g)

2NO2(g) N2O4(g)

When the volume is decreased what happens?

When the volume is decreased what happens?

2NO2(g) N2O4(g)

When the volume is decreased what happens?

2NO2(g) N2O4(g)

The gas appears browner since it is more concentrated

When the volume is decreased what happens?

2NO2(g) N2O4(g)

Then it gets lighter, due to the NO2 changing into N2O4

When the volume is decreased what happens?

2NO2(g) N2O4(g)

Then it gets lighter, due to the NO2 changing into N2O4

When the volume is decreased what happens?

2NO2(g) N2O4(g)

Then it gets lighter, due to the NO2 changing into N2O4

When the volume is decreased what happens?

2NO2(g) N2O4(g)

This happens because one N2O4 molecule takes up less space than 2 NO2 molecules

2NO2(g) N2O4(g)

When this equilibrium is upset by a decrease in volume, it shifts in a direction which decreases the total number of molecules. Since there is less space it makes sense to reduce the number of molecules present. If the volume is doubled ….

2NO2(g) N2O4(g)

When the volume is doubled ……

2NO2(g) N2O4(g)

When the volume is doubled ……

2NO2(g) N2O4(g)

When the volume is doubled ……

2NO2(g) N2O4(g)

It first gets lighter since the NO2 molecules are less concentrated

2NO2(g) N2O4(g)

Then it gets darker because with more space available the N2O4 molecules can form twice as many NO2 molecules

2NO2(g) N2O4(g)

Then it gets darker because with more space available the N2O4 molecules can form twice as many NO2 molecules

2NO2(g) N2O4(g)

Then it gets darker because with more space available the N2O4 molecules can form twice as many NO2 molecules

2NO2(g) N2O4(g)

Then it gets darker because with more space available the N2O4 molecules can form twice as many NO2 molecules

[NO2]

time

2x

2NO2(g) N2O4(g) + heat

The volume is doubled.

[N2O4]

time

x

2NO2(g) N2O4(g) + heat

This shift in equilibrium can be demonstrated mathematically using rate equationsWhat are the rate equations for forward and reverse reactions?

rf = k[NO2]2, rr = k[N2O4]if concentrations of NO2 and N2O4 are 1 mol/L

rf = k[1]2 = 1k rr = k[1] = 1k

2NO2(g) N2O4(g) + heat

This shift in equilibrium can be demonstrated mathematically using rate equationsWhat are the rate equations for forward and reverse reactions?

rf = k[NO2]2, rr = k[N2O4]if concentrations of NO2 and N2O4 are 1 mol/0.5L

rf = k[2]2 = 4k rr = k[2] = 2kHow does this impact on Equilibrium?

2NO2(g) N2O4(g) + heat

What happens when the volume is changed from 1 L to 2 L? Assume 1 mol of each gas.

rf = k[NO2]2, rr = k[N2O4]

[NO2]and [N2O4]are 1 mol/2L = 0.5 mol/L

rf = k[0.5]2 = 0.25k, rr = k[0.5] = 0.5kHow does this impact on Equilibrium?

2NO2(g) N2O4(g) + heat

rf = k[0.5]2 = 0.25k, rr = k[0.5] = 0.5k

Graph Rate change vs time

rf=rr

rf=rr

2NO2(g) N2O4(g) + heat

Describe the graph above

rf=rr

rf=rr

2NO2(g) N2O4(g) + heat

The increase in volume leads to decreased probability of collisions so both rates drop, rf

drops twice as much, since rf = 0.25k, rr = 0.5k.More N2O4 is used than is made so rr slows down. More NO2 is made than is used so rf increases until both rates are equal and a new equilibrium is established.

rf=rr

rf=rr

Under a specific set of conditions of temperature and pressure all participants in a dynamic equilibrium are related mathematically. To determine this mathematical relationship the following data will be considered for this equilibrium:

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

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

[PCl3] [Cl2] [PCl5]

1.40 4.60 5.15

3.40 5.60 15.23

1.40 3.50 3.92

0.93 0.37 0.28

4.53 0.29 1.05

Hint: Try to find a mathematical constant

[PCl5][PCl3][Cl2]

= 0.800

This means for any equilibrium aA + bB cC + dD

Ke =[C]c[D]d

[A]a[ B]b

Find the equilibrium constant for this reaction:

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

If [H2](g) = 0.25 molL-1, [I2](g)= 0.45 molL-1, [HI](g)= 0.0035 molL-1

Ke = [HI]2

[H2]1[ I2]1

Ke = [0.0035]2

[0.25]1[ 0.45]1

Ke = 1.1 x 10-4

For this equilibrium 2A + B C + D

If Ke is very large there is a lot ofC and D and very little ofA and B (product favoured reaction)If Ke is very small there is a lot ofA and B and very little of C and D (reactant favoured reaction)

If an equilibrium involves more than one phase (solids and aqueous ions for instance) the equilibrium expression only considers freely mobile particles since they alone obey the Laws of Mass Action. For example the Ke expression for this equilibriumPbI2(s) <=====>Pb2+(aq) + 2I1-(aq) isKe = [Pb2+][I1-]2 since the solid PbI2 is not composed of freely mobile particles

1. For the equilibrium PCl3(g) + Cl2(g) <=====> PCl5(g)0.50 mol of PCl5(g) is placed in a 5.0 L container. At equilibrium 0.14 mol of Cl2(g) is present. Calculate the equilibrium concentrations of each component and find Ke.

PCl3(g) + Cl2(g) <=====> PCl5(g)

(molL-1)

Initialshift@E

O.5 mol

5.0 L= 0.10 molL-1

0.028

Where did the 0.028 molL-1 of the Cl2 come from?It came from the reaction of 0.028 molL-1 of the PCl5. When the Cl2 was created an equal amount of PCl3 was madeThe amount of PCl5 at equilibrium can now be determined by subtraction.

0.028

+0.028 +0.028 -0.028

0.072

Ke can now be calculated. Ke = 0.072 / (0.028)(0.028) =

92

2.0 For the equilibrium CO(g) + 2H2(g) <=====> CH3OH(g)1.60 mol of CH3OH(g) is placed in a 4.0 L container. At equilibrium 0.24 mol of CO(g) is present. Calculate the equilibrium concentrations of each component and find Ke.

CO(g) + 2H2(g) <==> CH3OH(g)

(molL-1)

Initialshift@E

1.60 mol

4.0 L= 0.40 molL-1

0.12

Where did the 0.060 molL-1 of the CO come from?It came from the reaction of 0.060 molL-1 of the CH3OH. When the CO was created, double the amount of H2 was made. The amount of CH3OH at equilibrium can now be determined by subtraction.

0.060

2x 0.060 +0.060 -0.060

0.34

Ke can now be calculated. Ke = 0.34 / (0.060)(0.12)2 =

3.9 x 102 mol-2L2

3.0 For the equilibrium I2(g) + H2(g) <=====> 2HI(g)H2(g) and I2(g) were placed in a container at 728 K and allowed to reach equilibrium (eq.)At eq. the [H2] = 0.23 M, [I2] = 0.28 M, and

[HI] = 3.4 M. If in another 2.0 L container at the same temperature 0.64 mol of H2 and 0.64 mol of I2

are injected. Find the equilibrium concentrations of all 3 gases.

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

Since all the concentrations are given for a specific temperature find the Ke at this temperature.

Ke = [HI]2

[I2] [H2] =

[3.4]2

[0.28] [0.23]

Ke = 179.5, this value can now be used to answer the rest of the question

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

(molL-1)

Initialshift@E

0.64 mol

2.0 L= 0.32 molL-1

Since there is no HI in the container initially, the system, in order to reach equilibrium, must produce some. To do this it must use up equal quantities of H2 and I2. Since this quantity is unknown let it be x.

-x - x +2x

0.32 0.32

The balanced equation shows twice as much HI is produced.

Equilibrium values can now be calculated.

0.32 - x 0.32 - x 2x

Since Ke has already been calculated (179.5) the value of x can be determined using the equilibrium expression.179.5 = (2x)2 / (0.32- x)2

To avoid a quadratic equation take the square root of each side. 13.40 = 2x / 0.32 - x

4.287 - 13.40 x = 2xx = 0.278 so [I2] = [H2] = 0.04 M, [HI] = 0.56 M

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

(molL-1)

Initialshift@E

0.64 mol

2.0 L= 0.32 molL-1

-x - x +2x

0.32 0.32

0.32 - x 0.32 - x 2x