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
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.
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