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Given an exothermic reaction $\ce{N2 + 3H2 -> 2NH3}$ (which was initially in equilibrium). The temperature is then increasing with time. I am supposed to predict the yield of ammonia with time.

According to an exothermic reaction, the equilibrium constant K is supposed to decrease with increase in temperature. But it is found that the yield increases initially before decreasing continuously as expected. What causes the initial increase in the yield?

I suppose that the heat overcoming the activation energy could provide an explanation, but is there any other concept which is not fully reliant on the kinetics of the reaction?

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closed as unclear what you're asking by Mithoron, airhuff, A.K., aventurin, Tyberius Oct 22 '18 at 15:04

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  • $\begingroup$ Can you provide the reaction? $\endgroup$ – A.K. Dec 8 '15 at 16:46
  • $\begingroup$ Reaction: N2 + H2 -> NH3 $\endgroup$ – user232216 Dec 9 '15 at 7:59
  • $\begingroup$ @A.K Can you answer now? I'm still not clear with this question. $\endgroup$ – user232216 Dec 10 '15 at 14:48
  • $\begingroup$ No your question is on hold you should rephrase it and elaborate to show more effort to understand the material as the hold states. $\endgroup$ – A.K. Dec 10 '15 at 14:54
  • $\begingroup$ @A.K. Is this fine now? $\endgroup$ – user232216 Dec 10 '15 at 15:02
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Look, you say: "which was initially in equilibrium". Then the temperature are stable. In the reaction you have four mols in the reactants and 2 mols in the products. When you increase the temperature the system pressure increase too.

So to increase the stability the system "walk" for the reactants, because this decrease the pressure (2 mols in products and 4 mols in reactants).

If you remember the gas ideal law, the pressure is directly proportional to the mol numbers. So this is the reason because the temperature favor the products.

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  • $\begingroup$ Thank you for commenting on this post. How can you conclude that the pressure increases? No constraint is given to the volume of the system. And I was asking why the yield increases initially, which I suppose Le Chatelier's principle cannot explain. $\endgroup$ – user232216 Dec 11 '15 at 13:07
  • $\begingroup$ The reactants and products are gases, if the system stay open they will travel across the Universe. :) So, the system must be closed. If the system is closed a increase in the temperature increase the pressure. So the system will choose a condition where the gas mol number be smaller. $\endgroup$ – Koba Dec 11 '15 at 14:28
  • $\begingroup$ What about variance of K with temperature? How can you say that yield increases? The yield is supposed to decrease for a reaction where heat is released. I am confused. Thanks again. $\endgroup$ – user232216 Dec 11 '15 at 14:42
  • $\begingroup$ For example, let's think that K = 1 (1 atm) when you increase the temperature the reaction goes to a new equilibrium point. Using PV = nRT (R and V are constant). If the temperature increase, P increase and n decrease. If the n decrease that means the reaction will go to products (because in the products the mol number of gas is smaller in relationship to the reactants). Now, at the new temperature we have a new K, the equilibium was displaced. The must be K' > 1. $\endgroup$ – Koba Dec 11 '15 at 20:55
  • $\begingroup$ Oh, and the Le Chatelier's principle can explain this situation. The "system response" to temperature increase is: displace the equilibrium to te products for decrease the pressure. $\endgroup$ – Koba Dec 11 '15 at 21:09
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To answer your question in short, according to Le Chatelier's Principle the endothermic(reverse) reaction will be favored. So from the equation, it is clear that the reverse reaction produces a greater number of moles as compared to the forward one. Therefore because the amount of moles of a compound is directly proportional to the mass of that compound then the reverse reaction will produce more yield that the forward reaction which will produce less yield.Thus less ammonia will be produced. Hope this helps.

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