Chemistry, Third edition

(Wang) #1
MEANING OF EQUILIBRIUM CONSTANTS

The experimental rate expression for the back reaction(equation (15.2)) turns out


to be equally straightforward:


rate of reaction= kb[HI(g)]^2

As the concentration of HI builds up, the rate of the back reaction increases. At equi-


librium, the rates of the forward and back reactions are equal:


kb[HI(g)]^2 =kf[H 2 (g)] [I 2 (g)]

Comparing this with equation (15.4) shows that


Kc(T)

kf
kb

●This equation applies to all equilibrium reactions (and may be used whether or


not the reaction occurs in a single stage).


●Remember that a rate constant is the rate of reaction when the substances in the


rate expression are at unit concentration, i.e. 1 mol dm^3 (see page 251). The
above equation shows that the size of the equilibrium constant depends simply
upon how much faster the forward reaction is (at unit concentrations) than the
back reaction (at unit concentrations) at that temperature. We need only to know
the equilibrium constant and one of the rate constants in order to calculate the
remaining rate constant. For example, for the reaction

H 2 (g)I 2 (g)\===\2HI(g)

●kf= 1.66 mol^1 dm^3 s^1 andKc(T)= 45.6 at 764 K. Therefore,


kb(764 K) 

1.66 mol^1 dm^3 s^1
0.0364 mol^1 dm^3 s^1
45.6

The importance of stating the chemical equation to which


the equilibrium constant applies


An equilibrium constant applies to a particular chemical equation. For example,


whereas


[SO 3 (g)]^2
Kc(T)
[SO 2 (g)]^2 [O 2 (g)]

for 2SO 2 (g)O 2 (g)\===\2SO 3 (g)

we also write


[SO 3 (g)]
Kc(T)
[SO 2 (g)][O 2 (g)]^1 ⁄^2

for SO 2 (g)^1 ⁄ 2 O 2 (g)\===\SO 3 (g)

The equilibrium constants for these expressions will have different numerical values.


The moral is simple – never quote equilibrium constants unless you write down the


chemical equation (or equilibrium expression) to which that value of Kc(T)applies.


Meaning of equilibrium constants


Look back at Fig. 15.3(b). At equilibrium, the reaction mixture consisted of 12


molecules of A and 4 of B. Since, for a fixed volume, the molar concentration is pro-


portional to the number of molecules, we can write


Kc(T)

[B]


4
0.33
[A] 12

15.3


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