Chemical Equilibrium

A B+ A B+ C D+ A B+ C D+

• Reaction begins. • No products yet

formed.• High rate of

collisions between A & B.

• Rate of forward reaction HIGH.

• Products formed• Collisions

between reactants decrease.

• Rate of forward reaction DECREASES

• Reverse reaction begins.

• Rate of forward reaction EQUAL to rate of reverse reaction.

• Dynamic equilibrium established.

• Concentrations constant.

1. 2. 3. 4.

1. 2 & 3 4.

EQUILIBRIM REACTIONS

Most reactions DO NOT go to completion. Reactions that do not go to completion are

REVERSIBLE. Reversible reactions exist in a state of

EQUILIBRIUM. Equilibrium is reached when the rate of the

forward reaction equals the rate of the reverse reaction.

The reaction proceeding from L R as the equation is written is known as the forward reaction.

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

A B+ A B+ C D+ A B+ C D+

Chemical Equilibrium

……………………………

………………………..

……………………………

……………….

……………………………………………

A B+ A B+ C D+ A B+ C D+

Chemical Equilibrium

ConcentrationReactants

Productstime

Reaction RateForward

Reverse

DYNAMIC CHEMICAL EQUILIBRIUM

A B+ C D+

Con

cent

rati

on

time

Reactants

Products

• The equilibrium reaction does not mean the amounts of products and reactants are equal.

• If the products react more easily than the reactants but BOTH RATES are very slow.

• We say the Equilibrium is shifted to the LEFT.

• There will be higher concentration of reactants.

Ratio of: ProductsReactants

Would be LESS than one.

Equilibrium Position - Left

< 1

A B+ C D+

Con

cent

rati

on

time

Reactants

Products

ProductsReactants = 1

Equilibrium Position Middle

• If the reactants and the products both have similar reaction rates.

• The Equilibrium position will lie in the middle.

• There will be the same concentration of reactants & Products.

Ratio of:

ProductsReactants

Would be EQUAL TO one.

= 1

A B+ C D+

Conc

time

…………….

……………

• If the reactants react ……… ………… than the products.

• The Equilibrium is ………….. to the ……………….

• There will be higher concentration of ………….

Ratio of:

ProductsReactants

Would be ………………. than ONE.

Equilibrium Position - Right

1

A B+ C D+

Conc

time

Reactants

Products

• If the reactants react more easily than the products.

• The Equilibrium is shifted to the RIGHT.

• There will be higher concentration of Products.

Ratio of:

ProductsReactants

Would be GREATER than one.

Equilibrium Position - Right

> 1

A B+ C D+

Con

cent

rati

on

time

……………….

……………….

ProductsReactants < 1

Equilibrium Position SHIFT

Con

cent

rati

on

time

……………

……………

ProductsReactants > 1

Con

cent

rati

on

time

Reactants

Products

ProductsReactants = 1

To ………To …………

A B+ C D+

Con

cent

rati

on

time

Reactants

Products

ProductsReactants < 1

Equilibrium Position SHIFT

Con

cent

rati

on

time

Reactants

Products

ProductsReactants > 1

Con

cent

rati

on

time

Reactants

Products

ProductsReactants = 1

To rightSHIFTED To Left

A B+ C D+

Conc

time

reactants

products

1. More …………………….added

2. Reactant concentration ……………………

3. Forward reaction rate …………………..

4. Reactant concentration ………………/Product concentration ………….

5. Reverse reaction rate ……………………….

6. New equilibrium established - ………….. ratio of PRODUCTS/REACTANTS - equilibrium shifted to …………………

Adding Reactants

A B+ C D+

Conc.

time

reactants

products

1. More Reactant added

2. Reactant concentration increases & Forward reaction rate increases

3. Product concentration decreases & Reverse reaction rate decreases

4. Reactant concentration decreases as reactants used up.

5. Product concentration increases as new products formed.

6. New equilibrium established - higher ratio of PRODUCTS/REACTANTS - equilibrium shifted to RIGHT

Adding ReactantsReactant added

Equilibrium Shifts

New equilibrium

established

A B+ C D+

Conc

time

1. More PRODUCT added2. PRODUCT concentration increases (Instantly - Reactant

decreases )

Adding Products

Products

ReactantsNew

Equilibrium

Eqm. Shifts

3. REVERSE reaction rate increases (fast then slower)4. PRODUCT concentration decreases, REACTANT concentration increases (fast then slower) 5. Forward reaction rate increases6. New equilibrium established - lower ratio of PRODUCTS/REACTANTS - equilibrium shifted to LEFT.

Product added

Conditions of an equilibrium

At equilibrium both reactions continue to occur - the system is DYNAMIC.

The system is CLOSED – nothing added or removed.

The concentrations of reactants and of products remain constant.

Rate of the forward reaction equals rate of the back reaction.

Equilibrium can be obtained from either side.

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

(H = -13kJ/mol)

The H value always refers to the forward reaction. 13 kJ of energy is liberated for every mole of HI formed.

H2(g) + I2(g) 2HI(g) (H = -26kJ)

For the whole reaction:

Equilibrium is reached when the rate of theforward reaction equals the rate of the reverse reaction.

React

ion R

ate

Time

H2 + I2 2 HI

2 HI H2 + I2

H2 + I2 2 HI

React

ion R

ate

Time

H2 + I2 2 HI

2 HI H2 + I2

H2 + I2 2 HI

At equilibrium the concentration of all sub-stances are constant at a fixed temperature. Each equilibrium has a constant e.g. for

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

Kc = ]][ 22

2

I[H[HI]

Kc is only determined by the concentrations of solutions and gases. Pure liquids & solids are not included in the equation – their concentrations are constant.

N.B

[H2] means conc. of ...

In General:aA + bB cC + dD

When Kc has a high value, there will be more PRODUCTS – (on the RIGHT) - we say the equilibrium lies to The RIGHT (vice versa for a low value).

Kc = [C]c [D]d

[A]a [B]b

Calculations1. The balanced equation must be known.

2. The concentration of a solid or liquid remains constant - these are not included in the equation.

3. The concentration of the solvent is constant and not included in the equation.

4. The value of Kc is given without units.

Kc Calculation ExamplesWrite expressions for Kc for each of the following reactions:

A. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

B. Ca(s) + 2H2O(ℓ) Ca(OH)2(aq) + H2(g)

C. AgNO3(aq) + NaCℓ(s) AgCℓ(s)+ NaNO3(aq)

D. Na2CO3(s) + 2HCℓ (aq) 2NaCℓ(aq) + H2O(ℓ) + CO2(g)

Kc Calculation ExamplesWrite expressions for Kc for each of the following reactions:

A. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

B. Ca(s) + 2H2O(ℓ) Ca(OH)2(aq) + H2(g)

C. AgNO3(aq) + NaCℓ(s) AgCℓ(s)+ NaNO3(aq)

D. Na2CO3(s) + 2HCℓ (aq) 2NaCℓ(aq) + H2O(ℓ) + CO2(g)

Kc = [CO2]3[H2O]4

[C3H8] [O2 ]5

Kc = [Ca(OH)2][H2]

Kc = [NaNO3][AgNO3]

Kc = [NaCl]2[CO2]

[HCl]2

Kc Calculation Examples

Kc = [NH3]2

[H2]3 [N2]

3H2 + N2 2NH3

Initial: 7mol 8mol 0mol

Reacts: 6mol 2molEqm: 1mol 6mol 4mol[ ]: 1/2 6/2 4/2

7

=(4/2)2

(1/2)3 (6/2) = 10.66

Calculating K

Methane gas reacts with water vapour to produce carbon monoxide gas and hydrogen gas according to this equation:

CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)

At equilibrium: [CO] = 0.300 M [H2] = 0.800 M [CH4] = 0.400 M. If K is 5.67, calculate the concentration of water vapour

Homework: complete the worksheet: Equilibrium Exercise 1

Calculating Concentrations at Equilibrium

Carbon monoxide is a primary starting material in the synthesis of many organic compounds, including methanol, CH3OH(l). At 2000 0C, K is 6.4 x 10 −7 for the decomposition of carbon dioxide into carbon monoxide and oxygen. Calculate the concentrations of all entities at equilibrium if 0.250 mol of CO2(g) is placed in a closed container heated at 2000 0C.

If 0.50 mol N2O4(g) is placed in a 1.0 L closed container at 150 0C, what will be the concentrations of N2O4(g) and NO2(g) at equilibrium? (K=4.50)When hydrogen and iodine are placed in a closed container at 440 0C, they react to form hydrogen iodide. At this temperature, the equilibrium constant, K is 49.7. Determine the concentrations of all entities at equilibrium if 4.00 mol of hydrogen and 2.00 mol of iodine are placed in a 2.00 –L reaction vessel.

H2(g) + I2(g) ↔ 2HI(g)

The following reaction has an equilibrium constant of 25.0 1100 K.

H2(g) + I2(g) ↔ 2HI(g)

If 2.00 mol of hydrogen gas, H2(g) and 3.00 mol of iodine gas, I2(g) are placed in a 1.00 L reaction vessel at 1100K, what is the equilibrium concentration of each gas?

Predicting the effect

Le Chatelier’s Principle

If the conditions of an equilibrium system are changed, a process takes place which tends to oppose or cancel the effect of the change.

2. The equilibrium constant

Kc =

N2(g) + 3H2(g) 2NH3(g) (H < 0)

• When the concentration of the N2 is increased, the ratio (Kc) will be smaller.

• To restore the value of Kc and the equilibrium,more N2 will have to react with H2

• This will diminish the concentrations of reactants.

3

2

]][[

][

22

3

HNNHKc =

N2(g) + 3H2(g) 2NH3(g) (H < 0)

When the concentration of the N2 is increased:

• the concentration of reactants increases instantaneously

• The ratio Kc (momentarily) decreases

• Forward reaction rate increases and the concentrations of REACTANTS is LOWERED whilst the concentration of NH3 INCREASES.

• The reverse reaction rate INCREASES as more NH3 forms.

• A new equilibrium is established which has shifted to the RIGHT.

• The value of the ratio Kc is restored to its original value

Kc

time

Overall Kc UNCHANGED!

N2(g) + 3H2(g) 2NH3(g) (H < 0)

Increasing the temperature • will favour the

………………………. reaction. • In an EXOTHERMIC REACTION

The ………………………reaction will be favoured

• More ……………………..produced

• the reaction shift to the …………..

3

2

]][[

][

22

3

HNNH

Kc =

The ratio (Kc) will therefore …………………………..

Kc

time

N2(g) + 3H2(g) 2NH3(g) (H < 0)

Increasing the temperature • will favour the ENDOTHERMIC

reaction. • The REVERSE reaction will be

favoured • More REACTANTS produced• the reaction shift to the LEFT.

3

2

]][[

][

22

3

HNNH

Kc =

The ratio (Kc) will therefore DECREASE.

Kc

time

N2(g) + 3H2(g) 2NH3(g) (H < 0)

Increasing the PRESSURE • Equilibrium will shift to

…………………..pressure.• Reaction will favour the side

with the ……………………….number of MOLES of GAS – ……………

• More ………….. produced.

• Kc goes go ……………….(this eg)!

3

2

]][[

][

22

3

HNNH

Kc =

The ratio (Kc) will therefore INCREASE.

Kc

time

N2(g) + 3H2(g) 2NH3(g) (H < 0)

Increasing the PRESSURE • Equilibrium will shift to

REDUCE pressure.• Reaction will favour the side

with the LEAST number of MOLES of GAS – RHS NH3.

• More NH3 produced.• Kc goes go up (this eg)!

3

2

]][[

][

22

3

HNNH

Kc =

The ratio (Kc) will therefore INCREASE.

Kc

time

Effect of a CatalystP

ote

nti

al E

ner

gy

Po

ten

tial

En

erg

y

E prod

E reactE prod

E react

• A CATALYST lowers the ACTIVATION ENERGY of the reaction by providing a different reaction pathway.

• Activation Energy is lowered for ………………… FORWARD AND REVERSE REACTIONS.

• BOTH rates are therefore increased by the ……………………and so the EQUILIBRIUM DOES EQUILIBRIUM DOES …………………..………………….. SHIFT!!!! SHIFT!!!!

Ea Ea

Effect of a CatalystP

ote

nti

al E

ner

gy

Po

ten

tial

En

erg

y

E prod

E reactE prod

E react

• A CATALYST lowers the ACTIVATION ENERGY of the reaction by providing a different reaction pathway.

• Activation Energy is lowered for BOTH FORWARD AND REVERSE REACTIONS.

• BOTH rates are therefore increased by the same amount and so the EQUILIBRIUM DOES EQUILIBRIUM DOES NOTNOT SHIFT!!!! SHIFT!!!!

Ea Ea

N2(g) + 3H2(g) 2NH3(g) (H < 0)

Favorable conditions:• High• High• LOW• LOW

In practice a ……………………. Temperature is used. Too ………………… will slow the reaction down.

3

2

]][[

][

22

3

HNNH

Kc =

Kc

time

N2(g) + 3H2(g) 2NH3(g) (H < 0)

Favourable conditions:• High [N2] & [H2] • High Pressure• LOW Temperature• LOW [NH3]

In practice a compromise temperature is used. Too low will slow the reaction down.

3

2

]][[

][

22

3

HNNH

Kc =

Kc

time

Change ShiftIncrease in reagent concentration

In direction of using up reagent -> RHS

Decrease in reagent concentration

In direction forming more of reagent ->LHS

Increase in pressure In direction decreasing total number of moles (g)

Decrease in pressure In direction increasing total number of moles

Increase in temperature In direction of endothermic

Decrease in temperature In direction of exothermic

Catalyst is added No change

Equilibrium Disturbance

a) Conc of O2 is increased

b) Temperature is increased

c) Pressure is increased

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

Depending on circumstances, a solution may be:

Under saturated – forward rate greater than reverse (salt dissolving)

Saturated – forward & reverse the same rate (equilibrium)

Over saturated – reverse greater than forward – salt precipitates

Heterogeneous equilibrium

Depending on circumstances, a solution may be:

Under saturated – forward rate greater than reverse (salt dissolving)

Saturated – forward & reverse the same rate (equilibrium)

Over saturated – reverse greater than forward – salt precipitates

Heterogeneous equilibrium

1. All nitrates are soluble

2. All alkali metal & ammonium salts are soluble

3. Chlorides, bromides & iodides are soluble - except Ag, Hg, Cu & Pb

4. Sulphates are soluble – except Pb, Ca, Ag &Hg

5. Carbonates, phosphates & sulphates of alkali metals & ammonium are soluble

6. Hydroxides of alkali metals, ammonium & barium are soluble

7. Sulphides of alkali metals, alkaline earth metals and ammonium are soluble

The equation for the equilibrium reaction of a saturated salt solution can be represented as follows:

AB (s) A+ (aq) + B- (aq)

The equilibrium constant is called the solubility product and is calculated as follows:

Ca(OH)2(s) Ca2+(aq) + 2 OH-(aq)

Ksp = [Ca2+][OH-] 2

Temperature Change:Solubility curves show us that solubility's of most salts increase with increase in temperature.

Change in concentration :

NaCl (s) Na+(aq) + Cl-(aq)

Adding HCl to the above equilibrium, causesthe equilibrium to shift to the left. NaCl istherefore precipitated until the equilibrium is restored.

Disturbance of the equilibrium by increas-ing the concentration of one kind of ion is called the common ion effect.

Take note that the common ion effect isnot restricted to solubility equilibria only.

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

Adding a few drops of indicator and NH4Clwill show that the equilibrium will shift to the left - Explain.