How do I know a dynamic equilibrium hasn't just stopped? And does a catalyst change Kc?

I read the NCERT equilibrium chapter twice and one word keeps bothering me: "dynamic". The book says at equilibrium the rate of forward reaction equals the rate of reverse reaction. And that concentrations stay constant. But if nothing seems to change, how do we know both reactions are still going on and it is not simply stopped?

I have the example $N_2O_4 \rightleftharpoons 2NO_2$ , where the colour stops changing in the closed tube. Looking at the tube, I can't see anything happening.

Another thing. Questions in my test paper asked: "A catalyst is added to a reaction at equilibrium. What happens to $K_c$ ?" I wrote it increases because reaction is faster. I was marked wrong, but the teacher just said "think about both directions".

Last, I am not clear on $Q$ and $K$ . I understand that $Q$ is the same expression as $K$ but with non-equilibrium concentrations. How do I use it to say in which direction the reaction goes? And is a closed system really required?

7 answers & replies

[+]Siddharth RaoDropperSample· 19d ago

Evidence for "dynamic": isotope labelling. In NCERT-type experiments, if you add a bit of a labelled species (like deuterium) to an equilibrium mixture, you soon find the label turning up in other species too, even though the overall concentrations don't change. That can only happen if the forward and backward reactions are still going on. Same with a saturated sugar solution: add radioactive sugar crystals and the radioactivity soon appears in the solution while the amount of solid stays the same.

The net change is zero because rate forward = rate backward. Nothing has stopped, it's just balanced. Also you get the same equilibrium whether you start from reactants or from products.

[+]Rohan Sharma11thSample· 19d ago

equilibrium = concentrations of reactants and products become equal, right? that's why nothing changes. meme: both sides tied 50-50

[+]Nisha Verma12thSample· 19d ago

Rohan, no. Equilibrium means concentrations are constant, not equal. The ratio is fixed by $K$ . If $K$ is very large the mixture is mostly products, if it is very small it is mostly reactants. They're equal only if it happens that $K = 1$ in a simple 1:1 reaction. In a tube of $N_2O_4 \rightleftharpoons 2NO_2$ you can have much more of one than the other and the colour will still stay constant.

[+][–]Rohan Sharma11thSample· 19d ago
oh lol, constant not equal. 😅 saved

[+]Tanvi Deshmukh12thSample· 19d ago

Catalyst!! It speeds up forward and reverse reactions by the same factor (it lowers the activation energy for both). So it helps you reach equilibrium faster, but the equilibrium composition and $K_c$ stay the same. $K$ depends only on temperature. Catalyst doesn't change it.

And Q vs K:

$Q < K$ : reaction moves forward (products form)
$Q > K$ : reaction moves backward
$Q = K$ : at equilibrium

[+][–]Lakshmi TripathiDropperSample· 19d agoAuthor
So the catalyst helps both sides equally. The word "both directions" finally makes sense. Thank you.

[+]Arjun Nair12thSample· 19d ago

Closed system, yes. Decomposition $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ reaches equilibrium only in a closed vessel. In an open vessel the $CO_2$ escapes and reaction goes to completion. Also temperature must be constant. Physical examples: ice ⇌ water at 273 K and 1 atm, and liquid ⇌ vapour in a closed container.

One more trap: $K_c$ has units unless $\Delta n = 0$ . For $N_2O_4 \rightleftharpoons 2NO_2$ , $\Delta n = 1$ , so $K_c$ is in mol/L.

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