Download - Dynamic equilibrium FromAS Reversible reactions In a closed system Both forward and reverse reactions occur at equal rates. Macroscopic properties remain.

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Dynamic equilibrium

• FromAS• Reversible reactions• In a closed system• Both forward and reverse reactions occur at

equal rates.• Macroscopic properties remain constant.• Macroscopic?• Concentration, pressure, temperature, mass,

volume i.e. properties we can see or measure.

• So dynamic at molecular level.

Week 17

© Pearson Education Ltd 2009This document may have been altered from the original

Graph to show how the concentrations of N2O4 and NO2 change until equilibrium is achieved

Le Chatelier’s Principle

• ‘Which way will the balance lie?’• Le Chatelier did cry.• ‘It will move to cancel out• Whichever change you bring about!’• ‘When any of the conditions affecting the

position of a dynamic equilibrium are changed, then the position of the equilibrium will shift to minimise that change.’

The Haber Process

• N2(g) + 3H2(g) 2NH3(g) ΔH -93kJmol-1

• Complete the table for the above reaction:Change Effect on yield of

ammonia

Increase pressure Increase yieldDecrease pressure Decrease yieldIncrease temperature Decrease yieldDecrease temperature

Increase yield

Catalyst No change

Week 17

© Pearson Education Ltd 2009This document may have been altered from the original

• Deduce expressions for the equilibrium constant, Kc, for homogeneous reactions.

• Determine the units for Kc.

The Equilibrium Law

• In a dynamic equilibrium the concentrations of products and reactants are determined by Kc (capital K). – the equilibrium constant in terms of concentration.

• This relationship is given by the Equilibrium Law• For a reaction:• aA + bB cC + dD• Kc = [C]c[D]d

• [A]a[B]b

• Where the equilibrium concentrations of reactants and products are raised to the power of the stoichiometric ratios in the balanced equation for the reaction.

Units of Kc

• As with rate constants the units of K must be worked out for each reaction.

• E.g. For the reaction:• N2(g) + O2(g) 2NO(g)

• Kc = [products] [reactants] = [NO]2

[N2][O2]Units = (mol.dm-3)2

(mol.dm-3)(mol.dm-3)NO UNITS

Week 17

© Pearson Education Ltd 2009This document may have been altered from the original

• Calculate the value of an equilibrium constant, Kc, including its units.

• Calculate the concentration or quantities of substances present at equilibrium.

Calculating Kc

• Propanone reacts with hydrogen cyanide in the following reaction:

• (CH3)2C=O + HCN CH3C(OH)(CN)CH3

• A mixture initially containing 0.05mol.dm-3 propanone and 0.05mol dm-3 HCN in ethanol is left to reach equilibrium at room temp.

• At equilibrium the concentration of product is 0.0233 moldm-3

• Calculate Kc

Calculating Kc

• Since the reaction is 1:1 then 1 mole of each reactant produces 1 mole of product.

• Since the product concentration at equilibrium is 0.0233 mol.dm-3

• The concentration of each reactant must have gone down by 0.0233 mol dm-3.

propanone HCN 2-hydroxy-2methylpropanitrile

Initial conc(mol.dm-3) 0.0500 0.0500 0

Equilibrium conc (mol.dm-3)

0.05-0.0233= 0.0267

0.05-0.0233=0.0267

0.0233

•Step 3

•Write the equilibrium constant for this reaction in terms of concentrations:

•Kc = [product]

• [propanone][hydrogen cyanide]

• = 0.0233

• 0.0267 x 0.0267

• = 32.7 dm3 mol-1

•(Units =mol.dm-3 =1 )

• mol.dm-3 x mol.dm-3 mol.dm-3

Try this:

• Calculate the equilibrium constant for the following reaction:

• H2(g) + CO2(g) CO(g) + H2O(g)

• The initial concentration of hydrogen is 10.00mol.dm-3 and of carbon dioxide is 90.00 mol.dm-3. At equilibrium 9.47 mol.dm-3 of carbon monoxide are formed.

H2 CO2 CO H2O

InitialMol.dm-3

10.00 90.00 0 0

Equil mol.dm-3

10.00-9.47= 0.53

90.00 – 9.47 = 80.53

9.47 9.47

•If 9.47 mol.dm-3 of carbon monoxide are formed so too must 9.47 mol.dm-3 steam.

•Since the reaction is 1:1 then that concentration of each reactant must have been used up

Calculating Kc

• Kc = [products]

[reactants] = [CO][H2O]

[H2][CO2]

= 9.47 x 9.47 0.53 x 80.53 = 2.10 no units

Try this:

• Find Kc for the reaction between ethanol and ethanoic acid.

• 0.10 mol of ethanol were mixed with 0.10 mol of a solution of ethanoic acid and allowed to reach equilibrium.

• The total volume of the system is 20.0cm3.• We find by titration that 0.033mol

ethanoic acid is present once equilibrium is reached.

• C2H5OH(l) + CH3CO2H(l) CH3CO2C2H5(l) + H2O(l)

• Tabulate the data:• (The reaction is 1:1 so every mole of acid used

uses 1 mole ethanol)• If started with 0.10 moles acid and 0.033 remain

then (0.10-0.033) have been used to make the ester.

ethanol acid ester water

InitialMoles

0.10 0.10 0 0

Equil moles

0.033 0.033 0.01-0.033 = 0.067

0.067

EquilMol.dm-3

0.033/0.02=1.65

0.033/0.02=1.65

0.067/0.02=3.35

0.067/0.02=3.35

• Kc = [products]

• [reactants]

• = [ester][water]• [acid][ethanol]• = (0.067/0.02)(0.067/0.02)• (0.033/0.02)(0.033/0.02)• = 4.1

• The volume and units cancel out so Kc has no unit.

• Ester hydrolysis:• Ethyl ethanoate + water ethanoic acid +

ethanol• Exactly 1 mole ester was mixed with exactly 1

mole water (from dilute HCl – the catalyst) and allowed to reach equilibrium. The equilibrium mixture was analysed and found to contain 0.300 moles of ethanoic acid. Calculate the value for Kc at the temperature of the reaction.

• Find the equilibrium amounts not given.• i.e. It’s a 1:1 reaction again.• If 0.300 moles of ethanoic acid formed then so

too must 0.300 moles ethanol.• If 0.300 moles ethanoic acid have been formed

0.300 moles ester must have reacted. 0.700 must remain as must 0.700 moles water.

• Fill in a table:

Ester Water Ethanoic acid

Ethanol

Initial moles

1 1 0 0

Equil moles

1-0.300= 0.700

1-0.300=0.700

0.300 0.300

Equil conc mol.dm-

3

0.700/V 0.700/V 0.300/V 0.300/V

• Kc = [products]• [reactants]• = (0.300/V)(0.300/V)• (0.700/V)(0.700/V)• = 0.300x0.300• 0.700x0.700• = 0.184 (no units – why not?)• All the V terms cancel out.

• This question refers to the gaseous equilibrium:

• A(g) 2B(g)

• 4.0 moles of A was placed in a 20.0dm3 container and heated to 320K until equilibrium had been established. The equilibrium mixture was found to contain 1.50 moles A. Calculate Kc at this temperature.

• So:• 4-1.5 moles A used up.• Each mole of A produces 2moles B 2.50 moles A produces (2x 2.5) = 5 moles B

A B

Initial amount(moles)

4 0

Equilibrium amount (moles)

1.5 2 x (4-1.5)=5

Equilibrium conc mol.dm-3

1.5/20 =0.0750 5.00/20 = 0.250

• Kc = [B]2

• [A]

• = (0.250)2

• 0.0750• =0.833mol.dm-3

The Value of Kc

• Since Kc = [products]

• [reactants]

• The size of Kc gives information about the position of the equilibrium.

• A value of 1 would indicate the position of the equilibrium is bang in the middle.

• When Kc is >1 the equilibrium lies RIGHT.

• When Kc is <1 the equilibrium lies LEFT.

• If Kc is very large e.g.1010 then the reaction is regarded as gone to completion.

• If Kc is very small e.g.10-10 then there is effectively no reaction.

The Effect of Temperature on Kc

• Consider the following:

Example Endothermic reaction2HI(g) H2(g)+I2(g)

Exothermic reaction2SO2(g) + O2(g) 2SO3(g)

Temp increase

Equilibrium position shifts RIGHT – more PRODUCTS

Equilibrium position shifts LEFT- more REACTANTS

Temp decease

Equilibrium position shifts LEFT- more REACTANTS

Equilibrium position shifts RIGHT- more PRODUCTS

• This means that Kc MUST change since the values used to calculate it have changed.

• The Effect of Temperature Changes on Kc

Endothermic reaction. ΔH positive

Exothermic reaction. ΔH negative

Temperature increase

Kc increases Kc decreases

Temperature decrease

Kc decreases Kc increases

The Effect of Concentration, Pressure or Catalyst on Kc

• Kc is ONLY affected by temperature changes.

• This can be seen in experimental data.

• Since Kc controls the relative concentrations of reactants and products in a dynamic equilibrium equilibria must shift to restore Kc.

• This means that none of the above can change Kc at constant temperature.