Chemistry Form 6 Sem 1 06

CHEMISTRY LOWER 6 SEMESTER 1 CHAPTER 6 CHEMICAL EQUILIBRIA

6.0 Introduction
Among the reactions discussed so far, most of them are irreversible reactions. For example, the dissociation of hydrogen peroxide, H2O2 is an irreversible process. Equation : 2 H2O2 (l)  2 H2O (l) + O2 (g)
So, if the graph of concentration of hydrogen peroxide and hydrogen gas against time is plotted

6.1 Reversible reaction and dynamic equilibria
Actually, the example cited above, is one of a few chemical reactions proceed in only one direction, Most of the chemical reaction occur around us are usually reversible.
At the start of a reversible process, the reaction proceeds toward the formation of products. As soon as some product molecules are formed, the reverse process begins to take place and reactant molecules are formed from product molecules.
Chemical equilibrium is achieved when the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant.
An example of a reversible reaction, is the dissociation of dinitrogen tetraoxide, N2O4, into nitrogen dioxide, NO2, where the chemical equation can be written as N2O4 (g) 2 NO2 (g)
Diagram below shows 3 different graphs of changes in the concentration of N2O4 and NO2. In each case, equilibrium is established to the right of the vertical line

a) Initially only NO2 present Observation : Since there are only NO2, reaction will move to backward to form some N2O4. at the same time concentration of NO2 decrrease

b) Initially only N2O4 present Observation : Since there are only N2O4, reaction will move to forward to dissociate some N2O4 and form NO2. At the same time concentration of N2O4 decrrease

c) Initially a mixture of NO2 and N2O4 is present. Observation : Depending on the situation, concentration of N2O4 and NO2 may varies. From the graph, the concentration of NO2 decrease as it form more N2O4.

6.2 Homogeneous Equilibria and Equilibrium Constant
The term homogeneous equilibrium applies to reactions in which all reacting species are in the same phase. Example of the dissociation of N2O4 above is one of homogeneous gasphase equilibrium since both reactant and product are gases. N2O4 (g) ↔ 2 NO2 (g)
As mentioned earlier, a reversible reaction is to say achieved a chemical equilibrium, when the rate of forward reaction is the same as rate of backward reaction.
Consider the following equilibrium reaction. aA + bB cC + dD
Since a reversible reaction only take one single steps, hence the stoichiometry coefficient shall be the order of reaction in which it can be expressed as the following
rate of forward reaction ;
rate of backward reaction ;




When reaction achieved equilibrium rate of forward reaction = rate of backward reaction

So, k [A]a[B]b = k–1 [C]c[D]d c

d

k [C ] [ D] = a b k −1 [ A] [ B ]

k = cons tan t k −1

Since Or

k Kc = k −1

6.2.1 Equilibrium constant of concentration, Kc
According to equilibrium law (mass law action) states that, at a given temperature, when the amount of concentration of reactants in a chemical reaction, is divided by the amount of concentration of products is a constant where the concentration of each substance is raised to the power equal to coefficient in the balanced reversible chemical equation
For example, in a chemical reaction : pA (aq) + qB (aq) rC (aq) + sD (aq) where A and B are reactants ; C and D are products ; p , q , r , s are stoichiometry coefficient of reaction
Kc of the reaction above can be written as

Note that the [X] is used to express concentration for both products and reactants. Hence, in terms of equilibrium constant, KC, is known as the equilibrium constant of concentration. The ‘c’ stands for concentration. Since the unit of concentration for products / reactants is mol dm-3. So, the unit of KC depend on the total concentration of products over the reactants.
Kc can also be used to express substances in gaseous phase


Chemical equation a)

CH3COOH (aq) + C3H7OH (aq) ↔ CH3COOC3H7 (aq) + H2O

Equilibrium constant, KC

[CH3COOC3H 7 ][H 2O] Kc = [CH3COOH][C3H 7 OH] 2+ 2

b) 2 Fe3+ (aq) + 2 I- (aq) ↔ 2 Fe2+ (aq) + I2 (aq)

[Fe ] [I 2 ] Kc = [Fe3+ ]2 [I − ]2

c) 2 N2O5 (g) ↔ 4 NO2 (g) + O2 (g)

[ NO2 ]4 [O 2 ] Kc = [ N 2O5 ]2

d) AgNO3 (aq) ↔ Ag+ (aq) + NO3- (aq)

[Ag+ ][ NO3− ] Kc = [AgNO3 ]

e)

H2SO4 (aq) + 8 HI (aq) ↔ 4 I2 (g) + H2S (g) + 4 H2O (l)

[H 2S][I 2 ]4 [H 2O]4 Kc = [H 2SO4 ][HI]8

Example 1 : Consider the following equation 3A + 2 B ↔ 2 C + D Given the concentration of A, B, C and D are 1.20 mol dm-3 ; 0.920 mol dm-3 ; 1.10 mol dm3 ; 1.30 mol dm-3 respectively. Calculate the KC for the reaction above KC = [C]2[D] / [A]3[B]2 Kc = (1.10)2(1.30) / (1.20)3(0.920)2 KC = 1.08 mol-2 dm6 Example 3 :When 60.0 g of sulphur dioxide, SO2 react with 46.0 g of oxygen to form 58.7 g of sulphur trioxide in 1 dm3 vessel. Calculate the KC of system.[RAM S = 32 , O = 16] 2 SO2 + O2 ↔ 2 SO3 mol = 60.0 / 64 46.0 / 32 58.7 / 80 = 0.9375 = 1.438 = 0.7338 KC = [SO3]2 / [SO 2]2[O2] Kc = (0.7338/1)2 / (0.9375/1)2(1.438/1) KC = 0.426 mol-1 dm3

Example 2 : Given the KC for following equation 2E + F ↔ G +2H is 4.30. Given the concentration of E, F and G are 0.35 mol dm-3 ; 0.46 mol dm-3 ; 0.30 mol dm-3 respectively. Calculate the concentration of H in the reaction above. KC = [H]2[G] / [F] [E]2 4.30 = (H)2(0.30) / (0.46)(0.35)2 [H] = 0.90 mol dm-3 Example 4 : An equilibrium mixture contains 1.25 mol of hydrogen, 2.00 mol of bromine and 0.50 hydrogen bromide in a 4.0 dm3 of vessel. Calculate the KC of the system H2 + Br2 ↔ 2 HBr KC = [HBr]2 / [H2] [Br2] Kc = (0.50/4.0)2 / (1.25/4.0) (2.00 /4.0) KC = 0.100

6.2.2 Equilibrium constant of partial pressure, Kp
Different from Kc, which can be use to express the equilibrium constant of both aqueous and gaseous phase, Kp (equilibrium constant of partial pressure) is specifically an equilibrium constant used to expressed for gaseous substance.
For example, in the reaction of production of ammonia, N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)
We can expressed equilibrium constant as both Kc and Kp as followIn terms of KC, it is written as Equilibrium constant of concentration,

Equilibrium constant of partial pressure,

Kc

Kp

2

[ NH3 ] Kc = [ N 2 ][H 2 ]3

KP =

(PNH3 )

2

(PN 2 )(PH 2 )

3

a)

H2 (g) + I2 (g)

b)

2 SO2 (g) + O2 (g)

c)

d)

PCl5 (g)

2 N2O5 (g)

2 2 HI (g) K = (PHI ) P (PH 2 )(PI2 )

2 SO3 (g)

PCl3 (g) + Cl2 (g)

KP =

KP =

4 NO2 (g) + O2 (g) KP =

e)

N2O4 (g)

2 NO2 (g) KP =

(PNO2 ) 2 (PN 2O4 )

(PSO3 ) 2 (PSO2 ) 2 (PO2 )

(PPCl3 )(PCl2 ) (PPCl5 )

(PNO2 ) 4 (PO2 ) (PN 2O5 ) 2

Example 5 : The conversion of oxygen to ozone at high altitude can be represented by the following equation. 2 O3 (g). 3 O2 (g) In an equilibrium mixture, the pressure of oxygen gas and ozone are 0.120 atm and 0.200 atm respectively. Calculate the KP of reaction.

KP =

(PO3 ) 2 (PO2 )

3

KP = (0.200)2/ (0.120)3 KP = 23.1 atm-1

Example 6 : A mixture of hydrogen and nitrogen gas is made up in the ratio of 3 : 1 by volume, and is left to attain equilibrium at 1.01 kPa. The temperature is kept at 650 K. At constant temperature and pressure, the equilibrium mixture contains 20% of ammonia. Calculate the KP at equilibrium. N2 + 3 H2 ↔ 2 NH3 Initial : V 1 : 3 (ratio) 0 At eq : 1 : 3 = 80% 20% 20% 60% Since Pa = xa . Ptot (0.20)(1.01) (0.60)(1.01)

(0.20)(1.01)

Pgas : 0.202 kPa 0.606 kPa 0.202 kPa Kp = (PNH3)2 / (PN2) (PH2)3 KP = (0.202)2/ (0.202) (0.606)3 KP = 0.908 kPa-2

Example 7 : Chlorine and carbon monoxide react to form phosgene, COCl2, according to the equation. Cl2 (g) + CO (g) COCl2 (g). Chlorine gas was placed in a reaction vessel connected to a manometer and its pressure was found to be 0.69 atm. Carbon monoxide, with an initial pressure of 0.48 atm was then allowed to react with Cl2 in the vessel. At equilibrium, the total pressure was found to be 0.82 atm. Calculate KP. Initial : At equilibrium :

At equilibrium : KP KP KP

CO (g) + Cl2 (g) COCl2 (g) 0.48 atm 0.69 atm 0 atm (0.48 – x) (0.69 – x) x Ptotal = 0.82atm So, (0.48 – x) + (0.69 – x) + x = 0.82atm x = 0.35 atm 0.13 atm 0.34 atm 0.35 atm = (PCOCl2) / (PCl2) (PCO) = (0.35) / (0.13)(0.34) = 7.9 atm-1

6.2.3 Relationship between KC and KP.
In general, KC is not equal to KP, because the partial pressures of reactants and products are not equal to their concentrations expressed in moles per liter. A simple relationship between KP and Kc can be derived as follows. Let us consider the following equilibrium in the gas phase w A (g) + x B (g) y C (g) + z D (g) Equilibrium constant of partial pressure, Kp Expressing Equilibrium constant

Kp

Equilibrium constant of concentration, Kc




Assuming ideal gas behaviour, where PV = nRT, where V is the volume of container. Rearrange the equation @ P = [conc]RT




For each partial pressure of the gas PA = [A] RT

PB = [B] RT

PC = [C] RT

PD = [D] RT




Substituting the partial pressure of each gas into the equilibrium constant of partial pressure




Factorised RT from equation will make the equation become :




The general equation relating KC and KP can be expressed

KP = KC (RT)∆n where ∆n = (total mol of gaseous products) - (total mol of gaseous reactants)
From the equation, there may be a possibility where KP = KC is when the total amount of mol of product is equal to the total amount of mol of reactant in a balanced chemical equation.

6.3 Heterogeneous equilibria
Heterogeneous equilibrium results from a reversible reaction involving reactants and products that are in different phases
Consider a heterogeneous equilibrium system of calcium carbonate, when it is heated in a vessel CaCO3 (s) ↔ CaO (s) + CO2 (g)
In the heterogeneous system, the equilibirum constant can be expressed as




*KC' is used in this case to distinguish it from the final form of equilibrium constant derived shortly




The “concentration” of a solid, like its density, is an intensive property and does not depend on how much of the substance is present. For example, the “molar concentration” of calcium carbonate (density: 2.83 g/cm3) at 20°C is the same, whether we have 1 gram or 1 ton of the metal [CaCO3] =




For this reason, the terms [CaCO3] and [CaO] are themselves constants and can be combined with the equilibrium constant. We can rearrange and simplify equation above by writing




where KC, the “new” equilibrium constant, is conveniently expressed in terms of a single concentration, that of CO2. Note that the value of KC does not depend on how much CaCO3 and CaO are present, as long as some of each is present at equilibrium

Another example of heterogeneous equilibria can be explained using hydrolysis of ester. CH3COOCH3 (aq) + H2O (l) ↔ CH3COOH (aq) + CH3OH (aq)
In the heterogeneous system, the equilibirum constant can be expressed as





Concentration of a liquid, similar to solid, is another intensive property and does not depend too, on ow much the substance presence. In 1 liter (1 dm3) of pure water, the molarity can be calculated accordingly

[H2O] = *Similar to the example above, Kc does not depend on how much the concentration of water used during hydrolysis as long as water is present in equilibrium




From both of example above, in expressing equilibrium constant of a heterogeneous equilibria, pure solid and pure liquid can be ignored from the expression of the equilibrium constant

Reversible reaction

1. CO2 (g) + C (s)

2.

2 CO (g)

3 Fe (s) + 4 H2O (g) Fe3O4 (s) + 4 H2 (g)

3.

Ca(OH)2 (s) + 2 NH4Cl (aq) CaCl2 (aq) + 2 NH3 (g) + H2O (l)

KC

[CO]2 Kc = [CO2 ]

[H 2 ]4 Kc = [H 2O]4

KP

(PCO ) 2 KP = (PCO2 )

KP =

(PH 2 ) 4 (PH2O ) 4

[CaCl2 ][ NH3 ]2 Kc = [ NH4Cl]2 K P = (PNH3 ) 2

9.3.1 Dissociation reactions where large molecule decomposed
Dissociation reaction ~ Reaction ……………………………………………….. to form smaller molecules …………………………………………………………………………….
For example, 2 NO2 (g) N2O4 (g) 2 N2O5 (g) 4 NO2 (g) + O2 (g) 2 HI (g) H2 (g) + I2 (g) PCl5 (g) PCl3 (g) + Cl2 (g)
Most dissociation reactions are reversible reaction, meaning that the dissociation process proceed until a certain extent before reaching equilibrium.
The extent of dissociation of the compound is measured by degree of dissociation …………………………… completely dissociated. If
If α = 1 it indicates molecules are …………..……. partially dissociated. α < 1, molecules are ……………


20 Example, if α = 0.20, …………% of molecules are dissociated.




The following example shows the method how is usually Kc is calculated when a system with single reactant is dissociated into its component until it achieved equilibrium. PCl5 (g)

Initial mol

↔

PCl3 (g) +

Cl2 (g)

1 mol

0

0

degree of dissociation, α

- α

+ α

+α

At equilibrium

1-α

α

α

Example 8 : At a temperature of 398oC, hydrogen iodide (HI)has a degree of dissociation of 25.0% a) Calculate the KC of the above reaction at 398oC. a) 2 HI (g) ↔ H2 (g) + I2 (g) Initial : 1.00 0.00 0.00 When 25% dissociate : – 0.25 + 0.25 / 2 + 0.25 / 2 At equilibrium 0.75 0.125 0.125

KC =

[H 2 ][I 2 ] [HI]2

( 0.V125 )( 0.V125 ) KC = = 0.028 0.75 2 (V )

Example 9 : At 300 K and 1.0 atm, dinitrogen tetraoxide, N2O4, is 20% dissociated. a) Calculate the equilibrium constant of pressure for the reaction b) Calculate the degree of dissociation at 300 K and the total pressure of 0.20 atm. c) Calculate the pressure that must be applied so that the degree of dissociation is lowered to 15% a) N2O4 (g) ↔ 2 NO2 (g) Initial 1.00 mol 0 mol When 20% dissociated – 0.20 + 2 (0.20) At equilibrium 0.80 mol 0.40 mol Total mol = 1.2 mol

PN 2O4

0.80 2 = ×1.00 atm = atm 1.20 3 KP =

(PNO2 ) 2 (PN 2O4 )

PNO2

0.40 1 = ×1.00 atm = atm 1.20 3

( 13 ) 2 K P = 2 = 0.17 atm (3)

Calculate the degree of dissociation at 300 K and the total pressure of 0.20 atm. b) N2O4 (g) ↔ 2 NO2 (g) Initial 1.00 mol 0 mol When α dissociated –α +2α At equilibrium 1.00 – α +2α Total mol = 1.00 + α


PN 2O4

1− α = × 0.200 atm 1+ α

PNO2

2α = × 0.20 atm 1+ α

2α 2 α ( × 0.20 atm) 2 (PNO2 ) KP = = 0.167 atm K P = 1 + α (PN 2O4 ) 1− α ( × 0.200 atm) 1+ α 2

α = 0.416

Calculate the pressure that must be applied so that the degree of dissociation is lowered to 15% c) N2O4 (g) ↔ 2 NO2 (g) Initial 1.00 mol 0 mol When α dissociated – 0.15 + 2 (0.15) At equilibrium 0.85 mol 0.300 mol Total mol = 1.15


PN 2O4

0.85 = × P atm 1.15

PNO2

0.30 = × P atm 1.15

0 . 30 ( PNO2 ) ( × P atm) 2 KP = = 0.17 atm K = 1.15 = 0.17 atm P ( PN 2O4 ) 0.85 ( × P atm) 1.15 2

P = 1.8 atm

Example 10 : At 500oC, the KP for the reaction N2 (g) + 3 H2 (g) 2 NH3 (g) is 1.5 x 10-5 atm-2. Calculate the pressure that has to be applied to the system containing a mixture of N2 and H2 in the molar ratio 1 : 3 so that half of the reactants are converted to ammonia at equilibrium N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) Initial 1.00 mol 3.00 mol 0 mol When ½ dissociated – 0.50 – 3 (0.50) + 2 (0.50) At equilibrium 0.50 mol 1.50 mol 1.00 mol Total mol = 0.50 mol + 1.50 mol + 1.00 mol = 3.00 mol

0.50 1.00 1.50 PN 2 = × P atm PH 2 = × P atm × P atm PNH3 = 3.00 3.00 3.00 2 1 (PNH3 ) 2 ( P ) 5 −2 5 −2 3 KP = = 1 . 5 x 10 atm K = = 1 . 5 x 10 atm P (PN 2 )(PH 2 )3 ( 16 P)( 12 P)3 P = 5.96 x 10-3 atm

Example 11 : Consider the heterogeneous equilibrium MO (s) + H2 (g) M (g) + H2O (g) At ToC and pressure of 120 kPa, 1.5 mol of metal oxide, MO and 1.0 mol of H2 reacted to produce 0.012 mol of M and 0.012 mol H2O at equilibrium. Calculate a value for the equilibrium constant, KP at ToC. MO (s) + H2 (g) ↔ M (g) + H2O (g) Initial : 1.5 mol 1.0 mol 0 mol 0 mol Mol dissociate : - 0.012 0.012 + 0.012 mol + 0.012 mol At equilibrium : 1.488 mol 0.988 mol 0.012 mol 0.012 mol total mol = 0.988 mol + 0.012 mol + 0.012 mol = 1.012 mol

0.988 0.012 PH 2 = × 120 kPa =117 kPa PM = × 120 kPa =1.42 kPa 1.012 1.012 0.012 PH 2O = × 120 kPa =1.42 kPa 1.012 KP =

(PH 2O )(PM ) (PH 2 )

=

1.42 kPa x 1.42 kPa 117 kPa

KP = 0.017 kPa

9.4 Position of Equilibrium
In general, the equilibrium constant helps us to predict the direction in which a reaction mixture will proceed to achieve equilibrium and to calculate the concentrations of reactants and products once equilibrium has been reached. Since [products] KC ∝ [reactatns]





For example, The equilibrium constant Kc for the formation of hydrogen iodide from molecular hydrogen and molecular iodine in the gas phase H2 (g) + I2 (g) ↔ 2HI(g) KC = 54.3 at 430°C. Suppose that in a certain experiment we place 0.243 mole of H2, 0.146 mole of I2, and 1.98 moles of HI all in a 1.00 L container at 430°C. Inserting the starting concentrations in the equilibrium constant expression, we write

QC

[ HI ]02 = [ H 2 ]0 [ I 2 ]0

⇒

(1.98) 2 QC = (0.243)(0.146)

⇒

Qc = 111

Reaction quotient (Qc) is a value obtained by substituting the initial concentrations into the equilibrium constant expression in order to determine position of equilibrium compare to equilibrium constant (KC) .
When value of Qc is large, we shall have more products over reactants while, when value of Qc is small, we shall have more reactants over products.
To determine the direction in which the net reaction will proceed to achieve equilibrium, we compare the values of QC and KC.


QC < KC

The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants must be converted to products. The system proceeds from left to right (consuming reactants, forming products) to reach equilibrium. At this moment, we described it as equilibrium shift to right

QC = KC

• The initial concentrations are equilibrium concentrations. The system is at equilibrium. At this moment, we described it as equilibrium remain unchanged

QC > KC

• The ratio of initial concentrations of products to reactants is too large. To reach equilibrium, products must be converted to reactants. The system proceeds from right to left (consuming products, forming reactants) to reach equilibrium. At this moment, we described it as equilibrium shift to left

6.6













Factors that Affect the Position of Equilibrium

Once a system has achieved equilibrium, it is possible to change the chemical composition of equilibrium mixture by changing the condition of the reaction. In other words, position of equilibrium can be altered. 2 important factors which must be considered in chemical equilibria are the position of equilibrium the rate at which equilibrium is achieved. The factors which affect the position of equilibrium can be deduced using Le Chatelier's principle. Le Chatelier’s Principle stated that if an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position




The factors that affected the position of equilibrium are
the concentration of reactants and products
the temperature of the experiment
the pressure in reactions involving gases

9.4.1 The Effect of Concentration on Equilibrium.
If the concentration of the reactants are increased in a reaction at shift its position to right forming equilibrium, the equilibrium will ……………………………….., decreased products By this way, some of the reactants will be more ……………… …………....., the same and vice versa. The equilibrium constant will remain …………………..
Changing the concentration of one or more of the reactants or products of a reaction at equilibrium disturbed the original equilibrium, but the equilibrium constant remain the same and the system value of ……………………………….. changes until equilibrium is re-established.
Example, in the reaction of the formation of nitrogen dioxide 2 NO (g) + O2 (g) ↔ 2 NO2 (g)
Equilibrium is established at one point.

[NO]

[O2]

[NO2]

 Then, the equilibrium is disturbed by adding nitrogen monoxide gas into the gas vessel. When this happened, concentration of NOincrease ……….  According to Le Chatelier’s Principle, equilibrium will shift to right ……… NO to decrease the concentration of …………… increase  At the same time, concentration of NO2 : …………………. decrease concentration of O2 : …………………

Substance altered

Equilibrium position

[H2] increased

Shift to right

[I2]

[Fe2+] decreased

Shift to right

[Fe3+]decrease [I2]

[H2] decrease

Shift to left

[N2]

4. 2 SO2 + O2 ↔ 2 SO3

[SO3] increased

Shift to left

[SO2] increase [O2] increase

5. N2O4 ↔ 2 NO2

[NO2] decreased

Shift to right

[N2O4]decrease

Equation

1. H2 + I2 ↔ 2 HI

2. 2 Fe3+ + 2 I- ↔ 2 Fe2+ + I2

3. N2 + 3 H2 ↔ 2 NH3

Concentration of other substances

decrease

increase

[HI] increase

increase

[NH3]decrease

9.4.2 The effect of Pressure on a system in equilibrium
Change in pressure of system will only affect the position gas of equilibrium of reaction involving ……….. pressure
Result in a change in the ……………of both forward and backward reactions. position of equilibrium mixture
Result in a change in the ………… rate, k Equilibrium, KC at constant
Will not change …………. or …………………. temperature.
Le Chatelier’s principle stated that, When the pressure of a gaseous system increased, equilibrium will less shift to the position with ……..…… total amount of gaseous mole
When the pressure of a gaseous system decreased, equilibrium will more total amount of gaseous mole shift to the position with ……..….. constant
Equilibrium constant, KC and KP will remain …………………










Consider the reaction ; N2O4 (g) ↔ 2 NO2 (g). The pressure in the vessel increased According to Le Chatelier’s principle, the system will react to (i) decrease the pressure of the system (ii) decrease the concentration of all species involved. The results are :
Position will shift to the position with the least total number of left moles. From the equation above, equilibrium shift to ……………… increase where are NO : ……………… decrease
Concentration of N2O4 : …………… 2 constant
KC and KP remain ………………………

Equation

2 SO2 (g) + O2 (g) ↔ 2 SO3 (g)

Factor Equilibriu Changes of concentration of altered m position substances

P↑

Shift to right

[SO2] decrease

[SO3] increase

4 NH3 (g) + 5 O2 (g) ↔ 4 NO (g) + 6 H2O (g)

P↓

Shift to right

[NH3] decrease

[NO] increase

CO (g) + SO3 (g) ↔ CO2 (g) + SO2 (g)

P↓

No change

[CO] No change

[SO3] No change

CaCO3 (s) ↔ CaO (s) + CO2(g)

P↑

Shift to left

[CO2] decrease

2 N2O5 (g) ↔ 4 NO2 (g) + O2(g)

V↑

Shift to right

[N2O5] decrease

[NO2] increase

Adding a noble gas (inert gas) on an equilibrium. (i) Adding an inert gas into the system at a constant pressure and the volume of vessel remain unchanged lowered ………………..… but ……………….. the partial pressure of the gases involved in the system. (ii) This will cause the equilibrium to shift to the direction more with the ……………….. total number of moles.


Equation

Factor Equilibriu Changes of concentration of altered m position substances

PCl5 (g) ↔ PCl3 (g) + Cl2 (g)

+ Ne

Shift to right

[PCl5] decrease

[PCl3] increase

2 H2 (g) + O2 (g) ↔ 2 H2O (g)

+ Ar

Shift to left

[H2] increase

[H2O] decrease

+ Kr

No change

[I2] remain

[HI] remain

2 HI (g) ↔ H2 (g) + I2 (g)

9.4.3 Effect of Temperature on an Equilibrium
In a reversible reaction, the changes in temperature will affect Rate of forward reaction rate constant • ………………………. • k, …………………………. Rate of backward reaction Equilibrium constant • ………………………. • K, …………………………. Arrhenius equation to
In kinetic chemistry earlier, we used ……………. explain the effect on temperature toward the rate of reaction E − A RT

k = Ae




In equilibrium, the effect of temperature toward an equilibrium system Van’t Hoff equation can be explained by using …………………

∆H ln K = − +c RT


At different temperature, Van’t Hoff equation can be derived to become

K2 ∆H  1 1   −  ln = K1 R  T1 T2 

K = equilibrium constant, KC ∆H = enthalpy change R = gas constant, 8.31 J mol-1 K-1 T = temperature in Kelvin, K

Exothermic reaction

Endothermic reaction KC

KC

T • From the graph, the equilibrium decrease with constant …………...... temperature.

T • From the graph, the equilibrium increase constant …………......with temperature.

According to Le Chatelier’s principle, the changes of temperature will affect the position of equilibrium by o When temperature increased, equilibrium will shift to the position where decrease the temperature, which is toward the system is countered by …………….. endothermic the position of …………………………. process o When temperature decreased, equilibrium will shift to the position where increase the temperature, which is toward the system is countered by …………….. exothermic the position of …………………………. Process

Equation

Process

2 HI (g) ↔ H2 (g) + I2 (g)

Endothermic

Factor Equilibriu altered m position

T↑

Shift to right

Rate constant

Equilibriu m constant

increase increase

2 H2O2 (l) ↔ 2 H2O (l) + O2 (g)

Exothermic

T↑

Shift to left

CH4(g) + 2O2(g) ↔ CO2 (g) + 2H2O

Exothermic

T↓

Shift to decrease right

increase

T↓

Shift to decrease left

decrease

2 NO2 (g) ↔ N2O4 (g)

Endothermic

4 NO2 (g) + O2 (g) ↔ 2 N2O5 (g)

Exothermic

T↑

Shift to left

increase decrease

increase decrease

Example 12 At a specific temperature T, a 1.00 dm3 flask was found to contain an equilibrium mixture of 0.500 mol of sulphur dioxide, 0.100 mol of oxygen and 4.60 mol of sulphur trioxide. a) Calculate the equilibrium constant, KC of the reaction. b) Calculate the number of moles of oxygen that must be forced into the mixture to increase the number of moles of sulphur trioxide at equilibrium to 4.70 mol. 2 SO2 (g) At equilibrium

[SO3 ]2 Kc = [SO2 ]2 [O 2 ] b) At equilibrium Amount of mol insert At new equilibrium

[SO3 ]2 Kc = [SO2 ]2 [O2 ]

0.500 mol

+

O2 (g)

↔

2 SO3 (g)

0.100 mol

4.60 mol

60 2 [ 14..00 ] K c = 0.500 2 0.100 ⇒ K C = 846 mol−1 dm3 [ 1.00 ] [ 1.00 ]

2 SO2 (g) 0.500 mol

+

O2 (g) 0.100 mol

– 0.100 mol

– 0.0500 + x

0.400 mol

0.0500 + x

↔

2 SO3 (g) 4.60 mol + 0.100 4.70 mol

70 2 [ 14..00 ] K c = 0.400 2 0.0500+ X = 846 mol−1 dm3 [ 1.00 ] [ 1.00 ]

X = 0.113 mol

Example 13

Consider the equilibrium N2O4 (g) 2 NO2 (g). At 30oC and 1.0 atm., 30% of N2O4 is dissociated a) calculate the equilibrium constant, KP, for the reaction b) calculate the pressure needed to increase the dissociation to 40% at fixed temperature. a) N2O4 (g) ↔ 2 NO2 (g) Initial 1.00 mol 0 mol When α dissociated – 0.30 + 2 (0.30) At equilibrium 0.70 mol 0.60 mol Total mol = 1.30

PN 2O4

0.70 0.60 = ×1.00 atm = 0.538 atm PNO2 = ×1.00 atm = 0.462 atm 1.30 1.30 2 (PNO2 ) 2 ( 0 . 462 ) KP = KP = = 0.40 atm (PN 2O4 )

(0.538)

calculate the pressure needed to increase the dissociation to 40% at fixed temperature N2O4 (g) ↔ 2 NO2 (g) Initial 1.00 mol 0 mol When α dissociated – 0.40 + 2 (0.40) At equilibrium 0.60 mol 0.80 mol Total mol = 1.40

b)

PN 2O4

0.60 = × P atm = 0.4286 P atm 1.40

KP =

(PNO2 ) 2 (PN 2O4 )

PNO2

0.80 = × P atm = 0.5714 P atm 1.40

(0.5712P) 2 KP = = 0.396 atm (0.4286P)

P = 0.53 atm

Example 14 Consider the reaction N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)
The KP values at different temperatures are given below.




T (oC)

25

127

227

327

427

Kp (atm-2)

6.76 x 105

4.07 x 101

3.55 x 10-2

1.66 x 10-3

7.76 x 10-5

Plot a graph of ln KP against 1/T. From your graph, determine the value of ∆H of the reaction 1/T (K-1)

3.36 x 10-3

2.50 x 10-3

2.00 x 10-3

1.67 x 10-3

1.43 x 10-3

ln Kp (atm-2)

13.4

3.71

– 3.34

– 6.40

– 9.46

Gradient = 3.71 – (–6.40) / 2.50 x 10-3 – 1.67 x 10-3 Gradient = – ∆H / R = 12180 ∆H = – 101 kJ / mol

6.5 The Industrial Application of Chemical Equilibria
3 main aspects considered by Chemical Industries in producing chemical substance Short time High yield Low cost i) ………………… ii) …………………… iii) ………………………
In this Chapter, there are 2 industrial processes which are discussed.

ammonia

1. Haber process – process of synthesizing ……………. in industry. N2 (g) + 3 H2 (g)












2 NH3 (g)

∆H = – 92 kJ / mol

Consider the Haber reaction using Le Chatelier’s Principle In order to produce large amount (high yield) – equilibrium must shift right to ………………. Pressure : to make equilibrium shift to right, the pressure must be high lesser ………as right side of the equation has …………total amount of moles Temperature : to make equilibrium shift to right, the temperature must low exothermic reaction. be …………as the forward reaction is an ………………. of product but it will Adding catalyst (Iron) will not change the yield ……………………. rate of reaction only increase the ………………………

acid in industries 2. Contact Process – process of synthesizing sulphuric …………………. 2 SO2 (g) + O2 (g)

2 SO3 (g) ∆H = -90 kJ/mol




Consider the Contact process reaction using Le Chatelier’s Principle




In order to produce large amount (high yield) – equilibrium must shift

right to ……………


Pressure : to make equilibrium shift to right, the pressure must be

high as right side of the equation has …………total lesser ………. amount of moles


Temperature : to make equilibrium shift to right, the temperature

exothermic low as the forward reaction is an ……………………reaction. must be ………


yield of product but it will only Adding catalyst will not change the ……………………. of reaction increase the rate …………………