# Opposite effects of Le Chatelier's Principle

How does Le Chatelier's Principle deal with a change in reaction conditions which can simultaneously effect the equilibrium in opposite ways.

For example, here's the equation for the conversion of $$\ce{N2}$$ and $$\ce{H2}$$ into $$\ce{NH3}$$. It's an exothermic reaction, so I've added heat as a product on the right side.

$$\ce{N2(g) + 3H2(g) <=> 2NH3(g) +\text{heat}}$$

If I increase the temperature (add heat) that would tend to push the reaction to the left because the reaction to the right is an exothermic reaction. But if I add heat (and the volume remains constant), then I'm also increasing the pressure, which should tend to push the reaction to the right, towards fewer moles of gaseous products.

Is there a simple explanation for how Le Chatelier's Principle works in such a situation?

It is important to know the externally caused change the system is dealing with.

If the primary change is increasing the pressure, then the equilibrium shifts to the lower volume, in case of gases the lower molar amount.

If the primary change is increasing the temperature, than the equilibrium shifts away from the direction that evolves thermal energy.

If external causes are combined, like pressurizing the system AND externally initiated heat transfer, the resulting direction of equilibrium shift would depend on what effect is stronger. As the Le Chatelier's rule is qualitative, it cannot provide answer here, but it must be quantitatively calculated via the equilibrium constant at particular temperature.

• I removed "if both is changed at the same time" because that phrase suggests it matters whether the changes are done at the same time or in sequence, when in fact it doesn't matter whether you do #1 then #2, or #2 then #1, or both together. I also changed, e.g., "If the primary change is increasing the pressure by lowering the volume, than the equilibrium shifts to the lower pressure.", because it doesn't matter how the pressure is changed, and because the new pressure could be a constant pressure, in which case the shift in equilibrium isn't changing the pressure. – theorist Sep 8 '20 at 6:25
• I guess it may be little overstepping the editor licence, by changing the meaning of the answer.. Better would-be to comment it and let the author to rephrase it. – Poutnik Sep 8 '20 at 6:25
• Yeah, in retrospect I can see I did overstep. Sorry. I thought I was making it clearer, but in fact I was actually changing the meaning. – theorist Sep 8 '20 at 6:30
• Anyway, thanks for the attention, clarity attempts always count :-) – Poutnik Sep 8 '20 at 6:33
• About the final constant pressure case, it would be like a continuous external cause, reacting on the system change. – Poutnik Sep 8 '20 at 6:48