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Chemical Equilibria Part I: Tackling standard A Level / Prelims examination questions
  • focuschemistry
  • August 30, 2017
  • Chemical Enquilibria

Chemical Equilibria is usually taught before Chemistry of Aqueous Solutions in junior colleges, though there are schools would do Chemical Equilibria in the 1st year, and Chemistry of Aqueous Solutions in the 2nd year.

Many A level Chemistry students usually do not have a major problem with Chemical Equil, but let us investigate a simple question from a prelim paper and see how it can be modified to make it more difficult.

MJC 2011 Paper 3

A mixture was prepared using 1.00 mol of ethanoic acid and 2.00 mol of ethanol. At a given temperature, the mixture was left to reach dynamic equilibrium according to the following equation.

     (COOH)2(l) + 2C2H5OH(l)  ⇋ COO(C2H5)2 (l) + 2H2O (l)     ΔH<0

The equilibrium mixture contained 36.8 g of ethanol.

Calculate the value of Kc.

Solutions

Step 1: Calculate number of moles of ethanol formed at equilibrium.

No. of moles of ethanol at equilibrium

=36.8
————
2(12)+6+16
= 0.8

Step 2: Mole ratio table : Fill in the data provided (calculated from) by the question

In Moles(COOH)2+2C2H5OH=(COOC2H5)2+ 2H2O
Initial    1.00    2.00      0      0
Change
Equilibrium    0.80

Step 3: Work out the amount of C2H5OH reacted from the beginning (initial) till equilibrium is reached.

Amount of C2H5OH reacted = 2.00 – 0.8 = 1.2

Step 4: Work out the amounts of other reactants /products reacted and formed by mole ratio

In Moles(COOH)2+2C2H5OH=(COOC2H5)2+ 2H2O
Initial   1.00    2.00      0     0
Change    -0.6   -1.20   +0.6   +1.2
Equilibrium    0.80

Step 5: Work out the amount of reactants and products at equilibrium by taking the data in the 1st row ADD the data in the 2nd row.

In Moles(COOH)2+2C2H5OH=(COOC2H5)2+ 2H2O
Initial   1.00    2.00      0     0
Change   -0.6    -1.20    +0.6  +1.2
Equilibrium   0.40    0.80      0.6    1.2

Step 6: Work out Kc

Kc =[(COOC2H5)2][H2O]2
—————————
[(COOH)2][C2H5OH]2
=(0.6/V)(1.2/V)2
——————–
(0.4/V)(0.8/V)2
= 3.375= 3.38(3SF)

where v = volume of the reaction

Note :

1. Volume of the reaction is needed to compute concentrations, but it is not given in the question. The reason is that the volume, v, is cancelled out in the computation of Kc. Though the volume is cancelled out, it must still be included in the working, to show the examiner that we are clear Kc is a ratio of concentrations and NOT moles.

In some cases, when the sum of the powers of the numerator and denominator are not the same, the volume will remain in the Kc expression. In those cases, the volume of the reaction will be given. The volume of the reaction is the SUM of the volumes of the aqueous reactants.

2. Water is a solvent in many reactions, and thus are not included in the computation of Kc. However, water is included in this case as water is not a solvent for the product, an ester. Ester is an oil which is not miscible in water, and thus water has to be included as well.  Thus, water is excluded when all other reactants and products are in the aqueous state.

Points of Discussion:

Note that MJC question is very straightforward where there are only reactants and NO products at the beginning and thus, it is very obvious the reaction has to proceed FROM LEFT TO RIGHT.

Another clue about the direction of the reaction is the amount of ethanol in the beginning and at the end of the reaction. There is less ethanol left after the reaction has reached equilibrium and thus, the reaction has to shift forward to give that observation.

Consider what happens if the question is changed such that the initial conditions are:

Case 1:

In Moles(COOH)2+2C2H5OH=(COOC2H5)2+ 2H2O
Initial   1.00      0    2.00    1.00
Change     ?      ?     ?    ?
Equilibrium

Case 2:

In Moles(COOH)2+2C2H5OH=(COOC2H5)2+ 2H2O
Initial   1.00    1.00   2.00   1.00
Change
Equilibrium

Does the reaction shift forward or backward for Case 1 and 2?

For Case 1, since there is NO C2H5OH in the beginning, the reaction cannot shift forward. It can only shift backwards. Thus the change in the amounts of reactants and products looks like:

In Moles(COOH)2+2C2H5OH=(COOC2H5)2+ 2H2O
Initial  1.00    0   2.00   1.00
Change  +X  +2X    -X   -2X
Equilibrium1.00+X    2X  2.00-X1.00-2X

What happens in Case 2? None of the reactants and products start with 0 moles in the beginning. Does it shift forward or backwards?

There are 2 ways to find out.

Method 1: Assume it shifts forward and calculate X given Kc value. If the value of X found is a positive number, then the reaction is indeed shifting forward. If the value of X is found to be negative, then it should be shifting backwards, i.e.

In Moles(COOH)2+2C2H5OH=(COOC2H5)2+ 2H2O
Initial   1.00   1.00   2.00   1.00
Change    -X   -2X   +X  +2X
Equilibrium1.00+X2-2X2.00+X1.00+2X

Method 2: Determine the reaction quotient of the reaction and compare with Kc.

If Kc > reaction quotient: reaction shifts forward

If Kc < reaction quotient: reaction shifts backward

If Kc = reaction quotient: reaction neither shifts forward nor backward; it has achieved equilibrium.

What is reaction quotient and how exactly does a comparison of reaction quotient with Kc tell us about the direction of the reaction? Find out more at our A Level Chemistry classes at www.FocusChemistry.com NOW!

Proceed to Part II of Chemical Enquailibria…