# How does the equilibrium evolve if the pressure of reactants or products are altered?

Nov 22, 2016

......The equilibrium should shift to a side where the macroscopic change in conditions should be relieved..........in this instance we don't really know how the equilibrium will evolve.

#### Explanation:

This is just an extension of Le Chatelier's principle. Pressure can certainly be related to concentration, and if the concentration of one reactant or product is altered, then the equilibrium evolves so as to oppose the the change ($\text{oppose "!=" counteract}$). There is, of course, a catch.

The partial pressure of a reactant/product must be altered, not just the total pressure. If we have a gaseous equilibrium such as the one shown:

$A \left(g\right) + B \left(g\right) r i g h t \le f t h a r p \infty n s C \left(g\right) + D \left(g\right)$

; a change in the total pressure INCREASES the partial pressure of ALL the reactants and products, and the equilibrium should remain static. If the total pressure is increased by pumping in an inert gas, the equilibrium will also be unaffected, because no change has been made to the partial pressure of an individual gas.

If, however, the partial pressure (and concentration) of $B \left(g\right)$ is increased, for instance by pumping in more $B \left(g\right)$, then the equilibrium should move to the right. This is precisely equivalent to the effect of an increase in concentration of one of the reactants at equilibrium. In your example, we don't really know how the equilibrium will evolve.

Anyway let's look at a practical example: the fixation of dinitrogen.

$\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s N {H}_{3} \left(g\right)$

Given all I have said, it might be surmised that if we increase the partial pressures of dihydrogen, and dinitrogen, then this important equilibrium should be moved to the right. And so in fact it is. However, we could go a step farther than this. Ammonia is condensable. Certainly it is more condensable than dinitrogen and dihydrogen (why so?). A low temperature, however, means an unacceptably poor rate of reaction, and the modern Haber-Bosch process performs a balancing act in manipulating the equilibrium, versus maintaining the rate.

A cold finger (i.e. a cold well or trap) could be introduced to the reaction, so that the newly formed ammonia condenses on the finger, and is thus removed from the equilibrium, and thus drives the equilibrium to the right. The modern industrial process achieves good yields and turnovers by manipulating these considerations.