- The value of the equilibrium constant of a reaction is the same, at constant 'T' and 'P'. However, if either temperature 'T' or pressure 'P' or both are changed, the value of the equilibrium constant may also change. For example, the value of Kp (equilibrium constant in terms of partial pressures) for the reaction,
at 700 K and 800 K are 1.5 x 10-4 and 1.4 x 10-5 respectively.
- The value of equilibrium constant of a reaction does not depend upon the concentrations of the various species involved in the reaction. For example, the value of the equilibrium constant for the reaction
is 48 at 720 K. Whatever may be the concentrations of H2(g), I2(g) and HI(g), the value of the equilibrium constant remains the same provided temperature 'T' and pressure 'P' are held constant.
- For a reversible reaction, the equilibrium constant for the backward reaction is inverse (reciprocal) of the equilibrium constant for the forward reaction. For example, if the equilibrium constant for the forward reaction,
is K, then the equilibrium constant for the backward reaction,
is 1/K. From these two relationships, we can show,
- The value of the equilibrium constant of a reversible reaction gives an idea about the more favoured direction of the reaction. For example, when, (i) K>> 1, the forward direction is more favoured. As a result, at equilibrium, the concentration of the products will be much higher than the concentration of the reactants. Very large value of K, e.g.,
K = 1010 means that the reaction proceeds in the forward direction to almost completion.(ii) When, K = 1, both directions of a reversible reaction are equally favoured and the concentration of the reactants and products at equilibrium will be comparable (nearly equal).
Effect of catalyst
A catalyst has no effect on the value of the equilibrium constant of a reaction. This is because, a catalyst affects the rates of forward and backward reactions to the same extent. A catalyst however, helps in attaining the state of equilibrium faster.
Stoichiometry of a chemical equation
The numerical value of equilibrium constant depends upon the stoichiometry of the chemical equation used in representing the reaction. For example, the reaction
can also be expressed as
Then, the corresponding equilibrium constants are
When the coefficients are doubled, then the K for the new reaction is square of the original 'K'. In general when we multiply the terms in an equation by a certain value, we must change the 'K' for the equation to a power equal to that value.
Thus, if n= ½, Knew = (K)1/2 or
The value of the equilibrium constant also depends upon the units in which concentrations is expressed.
8. The equilibrium constant for the reaction
at 298 K is 138. What is the value of equilibrium constant for the reaction,
at the same temperature. Comment upon the values of the two equilibrium constants.
Equilibrium constant for the two reactions may be written as follows
From above, we have
The values of K1 and K2 show that the value of equilibrium constant for a reaction depends upon the stoichiometry of the chemical equation for the reaction.9. At 700 K, the equilibrium constant for the reaction
is 54.8. If 0.5 mole/litre of HI(g) is present at equilibrium at 700 K, what are the concentrations of H2(g)and I2(g), assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?
Let the concentrations of H2(g) and I2(g) at equilibrium be 'x' mol L-1. Then for the reaction,
The equilibrium constant, is,
But, K = 54.8.So,
Thus, at equilibrium, H2(g) = 0.067 mol/L and I2(g) = 0.067 mol/L.10. 1 mole of H2O and 1 mole of CO are taken in a 10 litre vessel and heated to 725 K. 40 per cent of water (by mass) reacts with carbon monoxide at equilibrium according to the equation:
Calculate the equilibrium constant for the reaction.
The reaction is,
Initial amount: 1 mol 1 mol
= 0.4 mol = 0.4 molAmounts at equilibrium:(1-0.4)mol (1-0.4)mol 0.4 mol 0.4 mol
= 0.6 mol = 0.6 mol 0.4 mol 0.4 molVolume of the reaction vessel is 10 L.
The equilibrium constant of this reaction is given by,
11. One mole of oxygen and 2 moles of SO2 are heated in a closed vessel of 1 dm3 capacity at 1098 K. At equilibrium 1.6 moles of sulphur trioxide are formed. Calculate the equilibrium constant for the reaction.
According to the equation, if 2 moles of SO3 are formed, 2 moles of SO2 and 1 mole of O2 are used up. Hence 1.6 moles of SO3 are formed from 1.6 moles of SO2 and 0.8 mole of O2. Therefore,
SO2 left over at equilibrium = 2 - 1.6 = 0.4 moleO2 left over at equilibrium = 1 - 0.8 = 0.2 mole
At equilibrium [SO2] = 0.4 mole dm-3[O2] = 0.2 mole dm-3
[SO3] = 1.6 mole dm-3