CHAPTER 1CHEMICAL EQUILIBRIUM

CHE 195PROCESS CHEMISTRY

Prepared by: Mohd Shahrul Nizam Bin Salleh

1.1 REVERSIBLE AND NON-REVERSIBLE REACTIONS

•Reversible: -The reaction that takes place in both the forward and

backward directions.

Ex: Production of gaseous hydrogen iodide

-The products can react with one another under suitable conditions to give back the reactants.

-It may be noted that for reversible reactions the symbol is used between the reactants and products

•Non-revesible:-The chemical reaction in which the products formed

do not combine to give the reactants.

Ex: Potassium chlorate decomposes on heating to form potassium chloride and oxygen.

-The products cannot combine to form potassium chlorate.

-The change occurs only in one direction and the processes go to completion.

Cont’d

1.2 HOMOGENEOUS AND HETEROGENEOUS EQUILIBRIUM

• Homogeneous equilibrium : The equilibrium reactions in which all the reactants and the products are in the same phase are called homogeneous equilibrium reactions.

C2H5OH(l) + CH3COOH(l) ↔ CH3COOC2H5(l) + H2O(l)

N2(g) + 3H2(g) ↔ 2NH3(g)

2SO2(g) + O2(g) ↔ 2SO3(g)

• Heterogeneous equilibrium : The equilibrium reactions in which the reactants and the products are present in different phases are called heterogeneous equilibrium reactions.

2NaHCO3(s) ↔ Na2CO3(s) + CO2(g) + H2O(g)

1.3 Kc AND Kp EXPRESSIONS• Kc - Concentration equilibrium expression.

• The concentrations at the equilibrium point are related by a simple mathematical equation known as an equilibrium expression which is governed by an equilibrium constant K, at constant temperature.

• For any reaction in solution or a gaseous mixture:

aA + bB + cC ↔tT + uU + wW

Kc = [T]t [U]u [W]w

[A]a [B]b [C]c

• The forward reaction (RHS) are on the top line and the arithmetical product of the reactant concentrations from the backward reaction (LHS) are on the bottom line.

• In all cases the product concentrations are raised to the appropriate power (a, b, c, .. t, u, w, ..) given by the stoichiometric mole ratios of the balanced equation.

Cont’d• For heterogeneous equilibrium, K expressions do not normally include

values for solid phases, since their chemical potential cannot change since the concentration of a solid cannot change.

• The equilibrium constant, Kc is governed by temperature, which is the only factor that can alter the internal potential energy of the reactants or products. The 'rule' for the trend in K value change is provided by Le Chatelier's Principle.

• If the forward reaction is exothermic, Kc (or Kp later) will decrease in value with increase in temperature.

• If the forward reaction is endothermic, Kc (or Kp) will increase in value with increase in temperature.

• Changes in pressure or concentration have no effect on a K value for ideal mixtures of gases/liquids or solutions.

• Application of a catalyst to a reaction also has no effect on a K value.

Example Given the esterification reaction: ethanoic acid + ethanol  ethyl ethanoate +

waterCH3COOH(l) + CH3CH2OH(l) ↔ CH3COOCH2CH3(l) + H2O(l)

A mixture of 1.0 mol of ethanoic acid and 1.0 mol of ethanol was left to reach equilibrium at 25oC.On analysis of the equilibrium mixture it was found that by titration with standard sodium hydroxide solution 0.333 mol of the ethanoic acid was left unreacted.Calculate the value of the equilibrium constant Kc and give its units.

The reactant-product mole ratios are 1:1 ==> 1:1 If 0.333 mol ethanoic acid was left unreacted, then 0.667 mol of the acid

had reacted. Therefore 0.667 mol of ethanol must also have reacted, leaving 0.333

unreacted.

Cont’d

If 0.667 of the acid/ethanol reacted, 0.667 mol of  ethyl ethanoate and 0.667 mol of water must be formed.

The concentrations = mol / volume, but the volume terms cancel each other out, so substitution in the equilibrium expression can be done in terms of final numbers of moles reactants and products

Kc = [CH3COOCH2CH3(l)] [H2O(l)] \

[CH3COOH(l)] [CH3CH2OH(l)]

Kc = 0.667 x 0.667

0.333 x 0.333

= 4.01 (no units)

Example 12.0g of pure ethanoic acid was mixed with 11.5g of pure ethanol and left to stand for a

week at room temperature (298K/25oC). The mixture was then mixed with deionised water and made up to 250 cm3 in a calibrated volumetric flask. When a 25.00 cm3 aliquot of the mixture was titrated with 0.50 mol dm-3 sodium hydroxide solution using phenolphthalein indicator (colourless ==> 1st permanent pink end-point), 10.60 cm3 of the alkali was needed for complete neutralisation.

(i) CH3COOH(l) + CH3CH2OH(l) ↔ CH3COOCH2CH3(l) + H2O(l) (esterification reaction)

(ii) CH3COOH(l) + NaOH(aq) ↔ CH3COO-Na+(aq) + H2O(l) (neutralisation

titration reaction)

(a)Calculate the moles of ethanoic acid unreacted in the original mixture.

(b)Calculate the moles of ethanoic acid and ethanol in the starting mixture.

(c)Calculate the moles of ethanol left unreacted and the moles of ethyl ethanoate ester and water formed.

(d)Calculate the equilibrium constant Kc for this

esterification.

Cont’da) Moles of ethanoic acid unreacted in the original mixture.

1 mole CH3COOH : 1 mole NaOH from equation (i) above.

moles = molarity x vol(dm3), so moles acid = 0.50 x 10.6/1000 = 0.0053 mol since the 25.00 cm3 aliquot titrated is equal to 1/10th of the total mixture, the total moles of unreacted acid = 10 x 0.0053 = 0.053 mol CH3COOH left

b) Moles of ethanoic acid and ethanol in the starting mixture. Mr(CH3COOH) = 60, mol ethanoic acid = 12/60 = 0.20

Mr(CH3CH2OH) = 46, mol ethanol = 11.5/46 = 0.25

c) Moles of ethanol left unreacted and the moles of ethyl ethanoate ester and water formed.

equation CH3COOH(l) + CH3CH2OH(l) ↔ CH3COOCH2CH3(l) + H2O(l)

moles at start 0.20                 0.25                    0                             0

mol at equilibrium 0.20 - x            0.25 - x               x                             x

mol at equilibrium 0.053         0.103            0.147              0.147

Cont’d x = moles of ester or water formed, so x mol of acid or alcohol must be used to reach equilibrium, since mol ratios are 1:1 ==> 1:1 in the reaction equation. Therefore 0.20 - x = 0.053 from calculation (a), therefore x = 0.20 - 0.053 = 0.147 so all the molar quantities can be logically deduced and are shown in the final line

of the table.

c) Equilibrium constant Kc for this esterification.

Kc = [CH3COOCH2CH3(l)] [H2O(l)]

[CH3COOH(l)] [CH3CH2OH(l)]

Kc = (0.147/V) x (0.147/V)

(0.053/V) x (0.103/V)= 3.96 (no units)

Note that all the volume terms cancel out, so you can work purely in moles to calculate Kc.

Kp Expression

• Kp - Partial equilibrium expression.• For any gaseous reaction:

aA (g)+ bB(g) + cC(g) ↔tT(g) + uU(g) + wW(g)

Kp = pTt pU

u pww.

pAa pB

b pCc

• px indicates the partial pressure of x, usually in atm (atmospheres) or Pa (pascals).

• The partial pressure of a gas is defined as the pressure a gaseous component in a mixture would exert, if it alone occupied the space/volume in question.

• e.g. in air, 21% by volume is oxygen and 79% is nitrogen etc.• Therefore the fraction of molecules which are oxygen and nitrogen is 0.21

and 0.79 respectively.

Cont’d• Since air has a total pressure of 1 atm. or 101325 Pa, the partial pressures of

oxygen and nitrogen in air are:

• pO2 = 0.21 x 1 = 0.21 atm or pO2 = 0.21 x 101325 = 21278 Pa,

• pN2 = 0.79 x 1 = 0.79 atm or pN2 = 0.79 x 101325 = 80047 Pa,

• In other words, the partial pressure of a gas in a mixture pgas = its % x ptot / 100

• In a gaseous mixture the total pressure equals the sum of all the partial pressures:

• ptot = p1 + p2 + p3 etc.

• so for air : ptot-air = pO2 + pN2 + pAr + pCO2. = 1 atm or 101325 Pa

• The % by volume ratio is also the mole ratio of the gases in a mixture.• If we call x the mole fraction of gas in a mixture then:

• The mole fraction of a gas A in a mixture = xA = mol of A / total moles of all gases

• Therefore the partial pressure of gas A, pA = xA x ptot

• 2SO2(g) + O2(g) ↔ 2SO3(g) (sulphur trioxide)

Kp = pSO32 (units atm-1 or Pa-1)

pSO22 pO2

• Ammonia is synthesised by combing nitrogen and hydrogen, but the reaction is reversible.

N2(g) + 3H2(g) ↔ 2NH3(g) (ΔH = -92 kJ mol-1)

• In an experiment starting with a 1:3 ratio H2:N2 mixture, at 400oC and total pressure of 200 atm, the equilibrium mixture was found to contain 36.3% by volume of ammonia.

(a) Write out the equilibrium expression for Kp.

(b) Calculate the partial pressures of nitrogen, hydrogen and ammonia.

(c) Calculate the value of the equilibrium constant at 400oC and give its units.

Example

a) Kp = pNH32 (units atm-2 or Pa-2)

pN2 pH23

b) Partial pressures of nitrogen, hydrogen and ammonia. 36% of ammonia means its mole fraction is 0.36, so pNH3 = 0.36 x 200 = 72

atm The other 64% of gases must be split on a 1:3 ratio between nitrogen and

hydrogen. If you think of the 1:3 ratio as 1 and 3 parts out of 4 the logic becomes easy. So mole fraction of nitrogen x ptot = pN2 = 64/100 x 1/4 x 200 = 32 atm,

and mole fraction of hydrogen x ptot = pH2 = 64/100 x 3/4 x 200 = 96 atm

check: ptot = pN2 + pH2 + pNH3 = 32 + 96 + 72 = 200 atm!

Cont’d

Cont’d

c) Value of the equilibrium constant at 400oC.Kp = pNH3

2 = pN2 pH2

3 Kp = 722

32 x 963 = 1.83 x 10-4 atm-2

Converting Kc to Kp.

2NOCl(g) ↔ 2NO(g) + Cl2(g) Kc = 3.75 x 10-6 at 796oC.

Write the expression to convert

Kc to Kp: KP = Kc(RT)Δn Extract the relevant data from the question:

R = 0.0821 (Ideal Gas Constant) T = 796oC = 796 + 273 = 1069K Δn = (2 + 1) - 2 = 1 (moles of gaseous product –moles of gaseous reactant

KP = Kc(RT)Δn KP = 3.75 x 10-6(0.0821 x 1069)1 KP = 3.29 x 10-4

REACTION QUOTIENT (QC)EQUILIBRIUM CONSTANT (KC)

The concentration in the Qc can be any concentration (usually initial concentration)

What is the difference ?

The concentration in the Kc expression

must be at equilibrium concentration

but

To determine the direction in which the net reaction will proceed to achieve equilibrium

Comparison the value of Reaction quotient (Qc)

and Equilibrium constant (Kc)

The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.

IF

• Qc > Kc system proceeds from right to left to reach equilibrium

• Qc = Kc the system is at equilibrium

• Qc < Kc system proceeds from left to right to reach equilibrium

1.4 Le Chatelier’s Principle

• Le Chateliers's Principle states that if an 'instantaneous' change is imposed on an equilibrium, the position of the equilibrium will further change to minimise the 'enforced' change.

• So the system reaches a new equilibrium position• In other words, if a change is 'instantaneously' imposed, the equilibrium

attempts to restore the original situation, but it cannot do this completely BUT the change 'trend' can be predicted.

• Le Chatelier’s Principles Temperature and energy changes (ΔH)

Raising the temperature favours the endothermic direction (ΔH +ve). The system absorbs the heat energy from the surroundings to try to minimise the temperature increase

Decreasing the temperature favours the exothermic direction (ΔH -ve). The system releases heat energy to the surroundings to try to minimise the temperature decrease.

Cont’d

Gas pressure changes at constant temperature (ΔV) Increasing the pressure favours the side of the equilibrium with the least

number of gaseous molecules as indicated by the balanced symbol equation. The system attempts to reduce the number of gas molecules present to reduce the pressure increase.

Decreasing the pressure favours the side of the equilibrium with the most number of gaseous molecules as indicated by the balanced symbol equation. The system attempts to increase the number of gas molecules to minimise the pressure decrease

Concentration changes at constant temperature If the concentration of a reactant (on the left) is increased, then some of it

must change to the products (on the right) to maintain a balanced equilibrium position.

If the concentration of a reactant (on the left) is decreased, then some of  the products (on the right) must change back to reactants to maintain a balanced equilibrium position.

Cont’d

Using a catalystA catalyst does NOT affect the position of an equilibrium.BUT it does is enable you get to the point of equilibrium faster !A catalyst speeds up both the forward and reverse reactions by

providing a mechanistic pathway with a lower activation energy, but there is no way it can influence the final 'balanced' concentration ratios.

The importance of a catalyst lies with economics of chemical production

e.g. bringing about reactions with high activation energies at lower temperatures and so reducing energy requirements and time, and both reductions save money!

ExampleChanges in Concentration.

Consider the following system at equilibrium:

Fe3+(aq) + SCN-(aq) ↔ FeSCN2+(aq)(colourless) (red)

- Increasing the concentration of either Fe3+(aq) or SCN-(aq) will result in the equilibrium position moving to the right, using up the some of the additional reactants and producing more FeSCN2+(aq). The solution will become a darker red colour.

- Removing some of the Fe3+(aq) or SCN-(aq) will result in the equilibrium position moving to the left to produce more Fe3+(aq) and SCN-(aq). The solution will become less red as FeSCN2+(aq) is consumed.

Predict the direction of equilibrium position when the pressure is increase. Why?

2NO2(g) ↔ N2O4(g)(red-brown) (colourless)

What happens to the Products if:a. More Reactanct are added to the system? explain whyb. temperature is decreased? explain whyc. pressure is decreased? explain why

PRACTICE EXERCISE

PRACTICE EXERCISE

A closed system initially containing 1.000 x 10−3 M H2 and 2.000 x 10−3 M I2 at 448°C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at 448°C for the reaction

H2 (g) + I2 (g) ↔ 2 HI (g)