Consider the following equilibrium reaction:

$$\ce{2 NOCl (g) <=> 2 NO (g) + Cl2 (g)} \qquad \Delta H > 0$$

My textbook has a statement that:

If you decrease the temperature when keeping volume constant, then reaction shifts to the reactants.

I know that since the enthalpy change $\Delta H$ is positive, we can think of the reaction as:

$$\ce{2 NOCl (g) + \text{heat} <=> 2 NO (g) + Cl2 (g)}$$

And according to Le Chatelier's principle we can say that if we decrease the temperature, then reaction favor the side have heat in it. This comes me so reasonable but there is one more thing we have to consider in my opinion: pressure.

While keeping volume constant, decreasing temperature will decrease also pressure. So reaction will favor the side that have more molecules in it. In this case, it is the opposite side of the previous one.

Is my reasoning true? If it is, how can I conclude the side determining job?

  • 1
    $\begingroup$ You need to look up the Van't Hoff Isochore; $d\ln(K_p)/dT = \Delta H^0/(RT^2)$. In an endothermic reaction ( $\Delta H^0 \gt 0$) , $K_p$ increases with increasing temperature, so more dissociation at higher temperatures and less at lower temperatures as you initially state. $\endgroup$ – porphyrin May 11 '17 at 20:44
  • $\begingroup$ @porphyrin As far as i understand, even pressure decreases (means that $Q_p$ decreases), $K_p$ always decreases much more than $Q_p$, so, it ends up with always favoring reactants.(for $\Delta{H}>0$ of course.) $\endgroup$ – user43236 May 12 '17 at 7:55
  • $\begingroup$ $K_p$ is constant with change of pressure but does change with temperature. $\endgroup$ – porphyrin May 12 '17 at 8:52
  • $\begingroup$ The change in $K_p$ i mentiond is due to temperature change. So is it what is going on? $\endgroup$ – user43236 May 12 '17 at 8:57
  • $\begingroup$ the answers here might help. chemistry.stackexchange.com/questions/73644/… $\endgroup$ – porphyrin May 12 '17 at 9:23

Since it's given that ΔH > 0 (endothermic reaction) ,the reaction will proceed backward (towards the reactant). However as you know , Partial Pressure Of Gas= Total Pressure x Mole Fraction; hence , even if the pressure is affected , the mole fraction increases/decreases in a way so that partial pressure remains constant. If you calculate Kp ( Pressure Equilibrium constant) you will know that it depends on the partial pressure of the gases involved . Since partial pressure remains same no effect takes place. Hence , pressure will have no effect on the equilibrium.

Also note that equilibrium constant is only affected by the change in temperature. Hope this helps.

EDIT:- In the light of the comments I would like to clarify ,

Partial pressure of a particular gas is the fraction of the gas multiplied by the total pressure which means it exerts a fraction of the total pressure. When I increase the total net pressure exerted by all gases , the fraction of moles of each gas decreases (according to the mathematical equation) thereby contributing lesser to the total pressure .

You can think of it as a ratio:-

If 1 mole X Gas exerts 10 bar pressure when total pressure is 20 bar , 0.5 moles of X gas will exert the same 10 bar pressure when total pressure is increased to 40 from 20 , thereby keeping the ratio same.

  • $\begingroup$ You don't even need to do all that ... Just looking at the fact that what is happening to the temperature/pressure/volume is enough to determine where the reaction will proceed. And as for calculation could you send me the Kp and moles intially or at equilibrium so that I can try as well? $\endgroup$ – Momo Senpai May 12 '17 at 3:07
  • $\begingroup$ @Abdullah I edited the answer $\endgroup$ – Momo Senpai May 12 '17 at 9:26

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