Eqm const (1 Viewer)

[synthesisFR]

Help i am very confused after gtting mixed responses
consider two systems
N2 + 3f2 (reverse sign) 2NF3 keq is 4.2x10^3
N2 + 3H2 (rev sign) 2NH3 key is 6.8 x 10^-2

which reaches equlilirbium fastest:
the first one because its keq is bigger so faster achievement of eqm
the second one because it lies on the reactant side so less products have to be converted to products?

im very confused my chem teacher saying the first one others saying the second one

 

[nsw..wollongong]

how can u deduce the ROR and how fast equilibrium is reached when Keq isn't a measure of like... speed. It only tells us conc. of react + prod. at equilibrium.

but maybe... the second Keq<first Keq = so forward react. is less favourable = second system might take longer to reach equilibrium compared to system one??

 

[nsw..wollongong]

omfg i just realised i basically reworded what u said

 

[Eagle Mum]

Help i am very confused after gtting mixed responses
consider two systems
N2 + 3f2 (reverse sign) 2NF3 keq is 4.2x10^3
N2 + 3H2 (rev sign) 2NH3 key is 6.8 x 10^-2

which reaches equlilirbium fastest:
the first one because its keq is bigger so faster achievement of eqm
the second one because it lies on the reactant side so less products have to be converted to products?

im very confused my chem teacher saying the first one others saying the second one

Your chem teacher is correct.
The reasoning in your first statement is correct.
Your second statement confuses cause and effect (& the wording “so less products have to be converted to products” is, itself, confusing as well). It isn’t that less reactants need to be converted to products to reach equilibrium so that direction of reaction reaches equilibrium faster (if I understand your line of reasoning correctly), but rather that this direction of reaction is slow, so produces less products by the time the system reaches equilibrium.

 

[wizzkids]

In general, there is no connection between the equilibrium constant K_eq and the rate at which a reaction converges to equilibrium.
Therefore I consider the question to be invalid. I am happy to hear alternative arguments.
However, as an interesting aside, we can state that the equilibrium constant is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction (btw, this is outside the HSC syllabus). But this only applies when we are considering the same reactants and products.
You cannot draw any conclusions about relative rate of reactions by comparing K_eq values for two different reactions.
Reaction rates (reaction mechanisms, first order, seconds order and third or reaction rates, catalysis etc) is not part of the HSC syllabus. The only place you will find reaction rate mentioned in the HSC syllabus is in relation to the Gibbs Free Energy Change as a predictor of spontaneous or non-spontaneous (Module 4), and in relation to collision theory (Module 3).
The Heats of Formation are a better indication of rate of reaction.
N₂ + 3F₂ 2NF₃
ΔH^o = -125 kJ/mol of NF₃
N₂ + 3H₂ 2NH₃
ΔH^o = +264 kJ/mol of NH₃
Notice that reaction 1 is exothermic at standard temperature and pressure (STP), and reaction 2 is endothermic at STP.
Reaction 1 is spontaneous and goes to completion, whereas reaction 2 is non-spontaneous at STP (and it requires very non-standard conditions before you get any yield at all).

 

[carrotsss]

In general, there is no connection between the equilibrium constant K_eq and the rate at which a reaction converges to equilibrium.
Therefore I consider the question to be invalid. I am happy to hear alternative arguments.

Yeah I agree, there are many reactions which take longer/shorter to reach equilibrium, and it has nothing to do with Keq, it’s a weird question

 

[synthesisFR]

Yeah I agree, there are many reactions which take longer/shorter to reach equilibrium, and it has nothing to do with Keq, it’s a weird question

IT WAS IN MY CHEMISTRY EXAM SET BY MY CHEM TEACHER WITH A PHD IN CHEMISTRY WTF
how ttf do i argue with her lmao also this task was last yr u just rememered it so its too late anyways)