Thus, by changing the pressure of a system, we do not change the equilibrium constant but we can, in fact, cause a system that is initially at equilibrium to suddenly find itself not at equilibrium after the pressure change.

Consider an ideal gas system at equilibrium as follows A(g) 2B(g)

The equilibrium expression for this system would be

Let n be the initial number of moles of A  with no B initially.  Equilibrium will be reached after a certain fraction, a , of A is converted into B at pressure p.  Thus, we can write

	A	2 B
Initially	n	0
Change	-αn	+2αn
Equilibrium	(1-a)n	+2αn

the mole fractions at equilibrium would be

So now, our equilibrium expression (ea. 8.15) becomes

Thus, we have

hence, the amount of reaction α depends on pressure, even though K does not.  As p increases, α decreases, in other words, the equilibrium shifts towards the reactants.

The calculated shift in this example, corresponds to Le Châtelier's principle:

    When a system at equilibrium is subjected to a stress, the system will shift it's equilibrium to reduce the stress.  In this case, the stress is a change in pressure and we see that as p is increased, the shift is towards reactants, which would have the effect of reducing the pressure by a small amount since there are 2 moles of B(g) lost for every 1 mol of A(g) produced in that shift.

In the last step of equation 8.17, we saw a new kind of K constant.  This one uses mole fractions to describe the equilibrium constant.  K_c is not a true equilibrium constant.

Equation 8.17 seems to imply that the equilibrium constant does depend on pressure but what it is really saying is that  K_c depends on pressure such that for a change in p\pº, there is an equal and opposite change in  K_c so that their product remains constant.

Changes in Temperature

Le Châtelier's principle tells us about how a temperature change would affect exothermic and endothermic reactions.

• endothermic reactions would shift towards products if the temperature is increased.

• exothermic reactions would shift towards reactants if the temperature is increased.

The van't Hoff equation, which can be derived from equation 8.6, gives us the connection between a change in temperature and the change in the equilibrium constant.

The first form of this equation shows that for an exothermic reaction (negative enthalpy change) the slope of lnK versus T (and therefore the slope of K versus T) is negative.  In other words, as we increase the temperature of an exothermic system, the equilibrium constant gets smaller. 

From a molecular perspective, we can see that this shift is related to the partition functions of the reactants (A) and products (B)

Figure 7.8 (from Atkins) helps us visualize this dependence.  figure (a) shows an endothermic system, where the manifold of states for (B) is higher than the manifold of states for (A).  At low temperature, the population the lower manifold of states (A) is favored.  In figure (b), the reverse is true.  As temperature is increased from low to high, the population shifts towards the higher manifold of states.  Note that in this case, the entropy effects are very small as the energy level spacings are nearly the same in manifold A as in manifold B.

We are more familiar with the integrated form of the van't Hoff equation.

In this form, we can determine an equilibrium constant at one temperature, if we know the value at another temperature and if we also know the enthalpy for the change in question.

Example:  The syntheses of ammonia [N₂(g) + 3 H₂(g)  2 NH₃(g) ] has an equilibrium constant K = 6.1 x 10⁵ at 298K.  What is the equilibrium constant at 500K?  The standard enthalpy of reaction is -92.0 kJ/mol

After some algebra, we get K = 0.18.  This makes sense since the reaction is exothermic and we raised the temperature so the equilibrium constant should go down.