Chemistry, Third edition

EFFECTS OF CHANGING CONCENTRATION

concentration of HI in Table 15.2 should be twice the lossin concentration of H 2 or

I 2 that takes place by the time equilibrium has been reached. Within experimental

error, this is seen to be the case. For example, in experiment 1, the equilibrium con-

centration of HI is 30.86 mol dm^3. The reduction in I 2 concentration is (20.00 

4.58) = 15.42 mol dm^3 , almost half of 30.86.

Substitution of the equilibrium concentrations into equation (15.4) gives:

For experiments 1 and 3, Kc(T)= (30.86)^2 /(4.58)^2 = 45.4

For experiment 2, Kc(T)= (14.94)^2 /(0.72)(6.84) = 45.3

This confirms that although the starting concentrations of reactant are different,

Kc(T)remains the same (within experimental error) at that temperature.

Now look at experiment 3 more closely. Experiment 3 starts with 40.00 mol of

pure HI. The HI decomposes into H 2 and I 2 , and [HI] then falls until equilibrium is

reached. 40.00 mol of HI is the amount of HI that would be produced ifwe were able

to convert all of the H 2 and I 2 in experiment 1 to HI. This is the reason that the equi-

librium compositions of experiment 1 (in which we started with pure reactants) and

experiment 3 (in which we started with pure product) are identical, confirming our

earlier statement (page 266) that the same equilibrium composition may be

achieved from both forward and reverse directions.

273

Calculations involving equilibrium concentrations

(i)Pure iodine and hydrogen gases were mixed and allowed to reach equilibrium. The

equilibrium concentrations of iodine and hydrogen gases at 490 °C were each found to be

3.00 mol dm^3. What is the equilibrium concentration of gaseous hydrogen iodide? What

were the starting concentrations of reactants?

(ii)The equilibrium concentrations (in mol dm^3 ) for a mixture of HI, I 2 , and H 2 at 527 °C were

[HI(g)] = 19.1, [I 2 (g)] = 3.50 and [H 2 (g)] = 2.50. Does Kc(T)for the reaction increase or

decrease with increasing temperature?

Exercise 15E

Case 2: Esterification of ethanol used to illustrate the

effect of concentration on equilibrium composition

The reaction of ethanol with ethanoic acid (in the presence of sulfuric acid catalyst)

produces a sweet-smelling compound known as an ester. The production of an ester

is called esterification. The name of the particular ester produced in this reaction is

ethyl ethanoate:

CH 3 COOH(l)C 2 H 5 OH(l)\===\CH 3 COOC 2 H 5 (l)H 2 O(l)

ethanoic acid ethanol ethyl ethanoate

From this we obtain

[CH 3 COOC 2 H 5 (l)][H 2 O(l)]

Kc(T)

[CH 3 COOH(l)][C 2 H 5 OH(l)]

(15.7)

Table 15.3 shows the composition of the equilibrium mixture for three initial

mixtures of differing concentration. Substituting the equilibrium concentrations

into equation (15.7) gives