How does temperature affect the dissociation constant?

Remember that dissociation is a reversible process, so it will reach equilibrium even if the reaction conditions are changed.

If temperature is increased, equilibrium shifts in favour of the endothermic direction in an attempt to revert the temperature change; the inverse is true of a temperature decrease.

The dissociation constant of an object defines how inclined it is to break up into its component parts: the higher the dissociation constant, the less likely it is that the larger object will dissociate. It is mathematically determined as follows:

#"For: ""A"_x"B"_y rightleftharpoons x"A" + y"B"#

#"K"_d = (["A"]^x["B"]^y)/(["A"_x"B"_y])#

Thus, as the concentration of the components increases, so too does the dissociation constant; by the same token, the association constant, the tendency of the components to coalesce to form the larger object, falls as concentration of components increases.

Thus, since a lower value for #"K"_d# results in dissociation to a greater extent, the shifting equilibrium in favour of the reaction that forms the larger object will reduce the dissociation constant and so in fact increase the likelihood of dissociation.

Since the effect of temperature on the position of equilibrium is dependent on whether the forward reaction is endothermic or exothermic; by the nature of reversible reactions, the reverse will be inverted in nature.

And so to conclude:

• Equilibrium will shift in favour of the endothermic reaction if temperature is increased.

• Equilibrium will shift in favour of the exothermic reaction if temperature is decreased.