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$$\ce{2NOCl <=> 2NO + Cl2}$$

What will be the effect on equilibrium concentration of $\ce{NOCl}$ when equal moles of $\ce{NOCl}$ and $\ce{NO}$ are introduced in the mixture at constant temperature?

Now this is what I think

Increasing the moles will result in increased concentration of both $\ce{NOCl}$ and $\ce{NO}$. But there also lies a second product $\ce{Cl2}$. Since there's an increase in $\ce{NO}$ and $\ce{NOCl}$, there should be an equal increase in the concentration of $\ce{Cl2}$ to keep the equilibrium-constant constant. But since there's no addition of $\ce{Cl2}$ molecules there should be a decrease in $\ce{NOCl}$ concentration which will decompose further to increase the molar ratio of $\ce{Cl2}$ and hence achieve equilibrium. Please enlighten me where my logic is flawed

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Given the reaction $$\ce{2NOCl <=> 2NO + Cl2}$$ we can formulate the equilibrium as $$\mathrm{K} = \dfrac{\ce{[NO]^2[Cl2]}}{\ce{[NOCl]^2}}$$

The problem then asks "What will be the effect on equilibrium concentration of NOCl when equal moles of NOCl and NO are introduced in the mixture at constant temperature?"

Now if we add equal amounts of $\ce{NOCl}$ and $\ce{NO}$ we don't have to do it at the same time to end up with the final equilibrium.

So let's add the $\ce{NOCl}$ first. We now have two possibilities (assuming that it is very unlikely that K=1 exactly).

  • If K>1 then more $\ce{NO}$ and $\ce{Cl2}$ form.

  • If K<1 then less $\ce{NO}$ and $\ce{Cl2}$ form.

Now let's add the $\ce{NO}$. This has to push the reaction to the left.

So the net result is it depends. You have too have enough data to plug into the equilibrium expression and other equations to work the net result out.

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Your logic is true. When you increase the concentration of NOCl and NO, Chlorine become limiting compound. So it will affect the equilibrium. Increase in amount of NOCl, will leads to more decomposition but it will also affected by concentration of NO in product. So in decomposition process, NO is controlling compound. And new equilibrium will achieved with more concentration of NOCl and NO and slightly more of Chlorine compare to initial concentration .

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  • $\begingroup$ Yeah but that is not the correct answer, the correct answer is the position of equilibrium remains unchanged and hence the concentration of NOCl remains constant $\endgroup$ – mathnoob123 Apr 23 '16 at 20:16

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