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For the reaction:

S O 2 ( g ) + N O 2 ( g ) N O ( g ) + S O 3 ( g )

the concentration of the reagents is as follows:

[SO 3 ] = 0.2 mol.dm - 3

[NO 2 ] = 0.1 mol.dm - 3

[NO] = 0.4 mol.dm - 3

[SO 2 ] = 0.2 mol.dm - 3

Calculate the value of K c .

  1. K c = [ N O ] [ S O 3 ] [ S O 2 ] [ N O 2 ]
  2. K c = ( 0 . 4 × 0 . 2 ) ( 0 . 2 × 0 . 1 ) = 4
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For the reaction:

S ( s ) + O 2 ( g ) S O 2 ( g )

  1. Write an equation for the equilibrium constant.
  2. Calculate the equilibrium concentration of O 2 if K c =6 and [ S O 2 ] = 3 m o l . d m - 3 at equilibrium.
  1. K c = [ S O 2 ] [ O 2 ]

    (Sulfur is left out of the equation because it is a solid and its concentration stays constant during the reaction)

  2. [ O 2 ] = [ S O 2 ] K c
  3. [ O 2 ] = 3 m o l . d m - 3 6 = 0 . 5 m o l . d m - 3
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Initially 1.4 moles of NH 3 (g) is introduced into a sealed 2.0 dm - 3 reaction vessel. The ammonia decomposes when the temperature is increased to 600K and reaches equilibrium as follows:

2 N H 3 ( g ) N 2 ( g ) + 3 H 2 ( g )

When the equilibrium mixture is analysed, the concentration of NH 3 (g) is 0.3 mol · dm - 3

  1. Calculate the concentration of N 2 (g) and H 2 (g) in the equilibrium mixture.
  2. Calculate the equilibrium constant for the reaction at 900 K.
  1. c = n V

    Therefore,

    n = c × V = 0 . 3 × 2 = 0 . 6 m o l
  2. Moles used up = 1.4 - 0.6 = 0.8 moles

  3. Remember to use the mole ratio of reactants to products to do this. In this case, the ratio of NH 3 :N 2 :H 2 = 2:1:3. Therefore, if 0.8 moles of ammonia are used up in the reaction, then 0.4 moles of nitrogen are produced and 1.2 moles of hydrogen are produced.

  4. NH 3 N 2 H 2
    Start of reaction 1.4 0 0
    Used up 0.8 0 0
    Produced 0 0.4 1.2
    Equilibrium 0.6 0.4 1.2
  5. [ N 2 ] = n V = 0 . 4 2 = 0 . 2 m o l . d m - 3
    [ H 2 ] = n V = 1 . 2 2 = 0 . 6 m o l . d m - 3
  6. K c = [ H 2 ] 3 [ N 2 ] [ N H 3 ] 2 = ( 0 . 6 ) 3 ( 0 . 2 ) ( 0 . 3 ) 2 = 0 . 48
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Hydrogen and iodine gas react according to the following equation:

H 2 ( g ) + I 2 ( g ) 2 H I ( g )

When 0.496 mol H 2 and 0.181 mol I 2 are heated at 450 o C in a 1 dm 3 container, the equilibrium mixture is found to contain 0.00749 mol I 2 . Calculate the equilibrium constant for the reaction at 450 o C.

  1. Moles of iodine used = 0.181 - 0.00749 = 0.1735 mol

  2. The mole ratio of hydrogen:iodine = 1:1, therefore 0.1735 moles of hydrogen must also be used up in the reaction.

  3. The mole ratio of H 2 :I 2 :HI = 1:1:2, therefore the number of moles of HI produced is 0.1735 × 2 = 0.347 mol.

    So far, the table can be filled in as follows:

    H 2 (g) I 2 2HI
    Start of reaction 0.496 0.181 0
    Used up 0.1735 0.1735 0
    Produced 0 0 0.347
    Equilibrium 0.3225 0.0075 0.347
  4. c = n V

    Therefore the equilibrium concentrations are as follows:

    [H 2 ] = 0.3225 mol.dm - 3

    [I 2 ] = 0.0075 mol.dm - 3

    [HI] = 0.347 mol.dm - 3

  5. K c = [ H I ] [ H 2 ] [ I 2 ] = 0 . 347 0 . 3225 × 0 . 0075 = 143 . 47
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The equilibrium constant

  1. Write the equilibrium constant expression, K c for the following reactions:
    1. 2 NO ( g ) + Cl 2 ( g ) 2 NOCl
    2. H 2 ( g ) + I 2 ( g ) 2 HI ( g )
  2. The following reaction takes place: Fe 3 + ( aq ) + 4 Cl - FeCl 4 - ( aq ) K c for the reaction is 7.5 × 10 - 2 mol.dm - 3 . At equilibrium, the concentration of FeCl 4 - is 0.95 × 10 - 4 mol.dm - 3 and the concentration of free iron (Fe 3 + ) is 0.2 mol.dm - 3 . Calculate the concentration of chloride ions at equilibrium.
  3. Ethanoic acid (CH 3 COOH) reacts with ethanol (CH 3 CH 2 OH) to produce ethyl ethanoate and water. The reaction is: CH 3 COOH + CH 3 CH 2 OH CH 3 COOCH 2 CH 3 + H 2 O At the beginning of the reaction, there are 0.5 mols of ethanoic acid and 0.5 mols of ethanol. At equilibrium, 0.3 mols of ethanoic acid was left unreacted. The volume of the reaction container is 2 dm 3 . Calculate the value of K c .

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Source:  OpenStax, Siyavula textbooks: grade 12 physical science. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11244/1.2
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