Consider the following reaction as an example: $$\ce{NaNO3(s) <=> NaNO2(s) + 1/2O2(g)}$$ Once this reaction has achieved equilibrium and then if we added pure $\ce{NaNO3}$ to this system, then is it correct to say that it will favour forward reaction?

My Thoughts

I have gone through a lot of questions related to this topic in Chemistry Stack Exchange and all of them have explained in detail that why it will not affect the value of equilibrium constant (as their "concentration" would not change and I fully agree with that), but no one answers the core question. That if we add or remove a pure solid/liquid then why wouldn't "moles" of other compounds in the reaction change which are included in equilibrium expression?

  • $\begingroup$ I meant that let's say we add some moles of nano3 (s) to the above example then some moles of both nano2 and o2 would form by standard reaction stoichiometry. $\endgroup$ Sep 6 '20 at 6:48
  • $\begingroup$ Then- 1) Wouldn't change in moles of o2 affect reaction quotient? (Keeping Temperature constant, I know 'Kp' wouldn't change but Reaction Quotient 'Q' should change and hence forward reaction should be more favoured (have higher rate than backward reaction) $\endgroup$ Sep 6 '20 at 6:56
  • $\begingroup$ 2) If what I said above is true, then wouldn't that mean that changing amount of solid also affects equilibrium but not Kp $\endgroup$ Sep 6 '20 at 6:59
  • $\begingroup$ It isn't concentration but activity $\endgroup$ Sep 6 '20 at 7:11
  • $\begingroup$ Where is the carbon on the right side of the chemical equation coming from? $\endgroup$
    – MaxW
    Sep 6 '20 at 7:22

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