# In the synthesis of ammonia, what is the effect of a decrease in the reaction temperature?

Apr 29, 2016

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

#### Explanation:

The reaction is exothermic as written, and $\Delta H = - 92 \cdot k J \cdot m o {l}^{-} 1$.

Le Chatelier's principle states that a system at equilibrium that is stressed will move so as to counteract the external perturbation.

What does this mean?

It means that when the conditions at equilibrium are changed, the equilibrium will move (left or right) so as to oppose the effect. It will not wholly resist it.

So what does this mean for the Haber process? I can write it this way:

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

The $\Delta$ symbol represents the heat that is evolved by the reaction. If the system AT EQUILIBRIUM is cooled, the equilibrium will move so as to counteract the decrease. That is the equilibrium will move to the right as written. This will not really affect the yield, but it will favour the product side of the equilibrium.

Industrially dinitrogen fixation is conducted at high pressures (favouring the forward reaction!) but also at high temperatures (which tends to favour the reactant side). Unfortunately, at low temperatures the rate of reaction is unacceptably low. The 1 thing that helps this reaction is that the product is condensable. Removal of the ammonia (by condensation on a cold finger for instance) can drive the equilibrium to the right by Le Chatelier's principle.