7.3 The Equilibrium Constant Opposing Rates Law of Chemical Equilibrium Equilibrium Constant Equilibrium Constant

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7.3 The Equilibrium ConstantOpposing RatesLaw
of Chemical EquilibriumEquilibrium
ConstantEquilibrium Constant
Temperature(pp.334-338)

• SCH4U Grade 12 Chemistry, University Preparation

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References

• McGraw-Hill Ryerson Chemistry 12
• Nelson Chemistry 12

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Past discussions

• Reactions occur at a specific rate and can be
  influenced by different factors
• Reactions are reversible
• Some reactions are spontaneous/favourable
• This depends on enthalpy, temperature and entropy
• We can predict if a reaction is spontaneous using
  Gibbs Free Energy (?G ?H - T?S)
• ?G gt 0 reverse reaction is favourable
• ?G lt 0 forward reaction is favourable
• ?G 0 reaction is at equilibrium
• Equilibrium occurs when opposing changes occur
  simultaneously at the same rate

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Lets evaluate the reversible reaction of N2O4(g)
2 NO2(g)
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Lets evaluate the reversible reaction of N2O4(g)
2 NO2(g)
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Equilibrium is reached from the forward and
reverse direction N2O4(g) 2
NO2(g)
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Law of Chemical Equilibrium

• At equilibrium, there is a constant ratio between
  the concentration of the products and reactants
  in any change.

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The Equilibrium Constant

• Forward Reaction
• N2O4(g) ? 2 NO2(g)
• Forward Rate
• kfN2O4

• Reverse Reaction
• 2 NO2(g) ? N2O4(g)
• Reverse Rate
• krNO22

• At equilibrium
• Forward Rate Reverse Rate
• kfN2O4 krNO22

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Keq the equilibrium constant

• At equilibrium, there is a constant ratio between
  the concentration of the products and reactants
  in any change.
• You calculate the equilibrium constant by
  dividing the forward rate constant by the reverse
  rate constant
• kf Keq
• kr
• So kf NO22 Keq
• kr N2O4

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Equilibrium Constant for any equilibrium equation

• The equilibrium equation is usually expressed in
  terms of molar concentrations.
• So the equilibrium constant is both Keq and Kc
• For the chemical equation
• aP bQ cR dS
• so Kc RcSd
• PaQb
• The equilibrium expression depends on the
  stoichiometry of the reaction and the
  concentration of the products reactants.
• Usually, units are not included when using or
  calculating the value of Kc.

Products(s)
Reactant(s)
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Example

• QUESTION
• Write the equilibrium expression for the
  following reaction
• N2(g) O2(g) 2NO(g)
• ANSWER
• Kc NO2
• N2O2

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Effect of temperature on the equilibrium constant

• The value of the equilibrium constant depends
  only on temperature.
• Changing the temperature changes the rates of the
  forward reverse reactions differently because
  the forward reverse reactions have different
  activations energies.
• The starting concentrations have no effect on the
  equilibrium constant.

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Example 2

• QUESTION
• A solution of phosphorus pentachloride gas was
  kept at 250 K in a 2.5L reaction flask.
• PCl5(g) PCl3(g) Cl2(g)
• When the equilibrium mixture was analyzed, it was
  found to contain 0.025 mol of PCl5(g),
• 0.05 mol of PCl3(g) and 0.01 mol of Cl2(g).
• Calculate the equilibrium constant for this
  reaction.

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Sample Problem - Analysis

• ANALYSIS OF QUESTION
• You want to calculate the value of Kc
• ? Write out the equilibrium expression
• You were given the volume of the container and
  the amount of the chemicals
• ? Calculate concentration by taking the amounts
  (mol) and divide by the volume (L) to get the
  molar concentrations (mol/L)

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Sample Problem - Solution

• PCl5(g) PCl3(g) Cl2(g)
• PCl5(g) 0.025 mol / 2.5 L 0.01 mol/L
• PCl3(g) 0.05 mol / 2.5 L 0.02 mol/L
• Cl2(g) 0.01 mol / 2.5 L 0.004 mol/L
• Kc PCl31Cl21 0.020.004 0.008
• PCl51 0.01
• SoThe equilibrium constant for this reaction at
  250 K is 0.008.