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I gather that the equilibrium constant K = 1 implies that, at equilibrium, neither the forward nor the backward reactions are thermodynamically favoured. But if pressure of the system increases and this shifts the position of equilibrium, the value of K remains to be 1.

How then can K = 1 imply that, at equilibrium, neither the forward nor the backward reactions are thermodynamically favoured?

Please can I have a simple and comprehensive explanation of what the value of K actually tells us?

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closed as unclear what you're asking by Mithoron, Mathew Mahindaratne, Tyberius, Todd Minehardt, Jon Custer Oct 1 at 17:05

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    $\begingroup$ If the pressure increases, the system will be out of equilibrium. What's unclear about that? $\endgroup$ – Ivan Neretin Sep 28 at 15:29
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I gather that the equilibrium constant K = 1 implies that, at equilibrium, neither the forward nor the backward reactions are thermodynamically favoured.

There is nothing special about K = 1. At equilibrium, no matter the value of K, there is no net reaction. Forward and backward reactions proceed at the same speed.

But if pressure of the system increases and this shifts the position of equilibrium, the value of K remains to be 1.

Yes, even a reaction with K = 1 can react. Just imagine you start with some reactant and no product. This situation is not at equilibrium. The reaction will proceed in the forward direction until Q = K. Then, you can disturb the system, changing Q, and there will be a net reaction again, changing concentrations until Q reaches K again.

Please can I have a simple and comprehensive explanation of what the value of K actually tells us?

You said "please", so I will try for a short, simple and comprehensive answer. The value of K allows you to calculate equilibrium concentrations, no matter which concentrations you started with. For a lower K, the product equilibrium concentrations will be lower than for a higher K (same initial concentrations). The opposite is true for reactant equilibrium concentrations.

It should be clear that the equilibrium concentrations for a given K are not always the same. The simplest example is doubling initial concentrations (while keeping the volume constant). If the equilibrium concentrations were equal for both cases (initial concentrations doubled or not), it would go against the conservation of matter. (We doubled the amount of species present initially, so we can't have the "old" amount of equilibrium species).

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The equilibrium constant normally written as $K_p$ for a gas phase reaction is independent of pressure but does depend on temperature. This means that if the pressure is changed then the amount of dissociation of reactants and products changes (and so does their partial pressure) until equilibrium is restored. $K_p$ is defined as the ratio of the partial pressures at equilibrium, and the exact expression depends on the type of reaction, for example $A_2\to 2A$ produces $K_p=p_{a}^2/p_{A_2}$. In your example if $K_p=1$ then $p_{a}^2=p_{A_2}$ at all pressures and this can only be true if the amount of dissociation changes.

Any equilibrium constant $K_p$ is also related to the standard free energy of the reactants and products at $1$ atm because $\Delta G^\text{o}=-RT\ln(K_p)$. If $\Delta G^\text{o}$ is positive then there will be more reactants than products and the reaction will not proceed to such an extent than if $\Delta G^\text{o}$ is negative i.e $K_p \gt 1$. (Note that just because an equilibrium constant may be large this does not imply that the reaction is fast: thermodynamics does not inform us about reaction kinetics only the final equilibrium state.)

If we know $\Delta G^\text{o}$ for reactants and products then it is possible by calculating the change (products minus reactants) to find the equilibrium constant for a reaction without having to try it experimentally.

Notes. Calculation of change in degree of dissociation for the reaction $A_2\to 2A$. Assume the degree of dissociation is $\alpha$ then $A_2$ changes by $(1-\alpha)$ and $2A$ by $2\alpha$ and the total change is $1+\alpha$. The partial pressure of $A_2$ is $\displaystyle P_{A_2}= \frac{1-\alpha}{1+\alpha}p$ with $p$ as total pressure, and for $A$ the partial pressure is $\displaystyle P_{A}= \frac{2\alpha}{1+\alpha}p$. As $\displaystyle K_p = \frac{P_A^2}{P_{A_2}}=\frac{4\alpha^2 }{1-\alpha^2}p$. As the pressure $p$ changes then so must $\alpha$ because $K_p$ is constant.

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  • $\begingroup$ “If delta G is positive then there will be more reactants than products” is not always true. “If all species initially are at standard state and delta G is positive, the reaction goes in reverse to attain equilibrium and equilibrium concentrations of reactants will be higher than those of products” is a general statement. $\endgroup$ – Karsten Theis Sep 29 at 11:50
  • $\begingroup$ I though that is what my statement said anyway, 'more reactants than products' vs 'concentrations of reactants will be higher than those of products' ? $\endgroup$ – porphyrin Sep 29 at 12:09
  • $\begingroup$ Extremely small concentrations favor dissociation while extremely high concentrations favor association. Even in a situation where at standard state, association is favored ( or dimerization like in your example), at extremely low concentrations it won’t be. So the key difference in our statements is that I specify the initial conditions as standard state. $\endgroup$ – Karsten Theis Sep 29 at 12:20
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    $\begingroup$ I think what you are saying is that (in my example) at low pressures (concentration) to keep $K_p$ constant $4\alpha^2/(1-\alpha^2)$ is large as $p$ is small and vice versa. $\endgroup$ – porphyrin Sep 29 at 14:49
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Actually K tells u nothing but it implies the ratio of the concentrations of produts and reactants when state of equilibrium has been attained which as the name suggests is always constant for a given temperature...Whenever there is increse or decrease in pressure( or Volume) the reaction shifts as the concentration gets altered and the shift is to re-establish the equilibrium....thus there should be and there is no effect in the value of K.... Hope it helps!

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    $\begingroup$ Please avoid colloquialisms, salutations and excessive punctuation, and mind proper capitalization (you also might want to fix a couple of typos along the way). In general, this somewhat addresses the OP's question, but I'm not sure this is exactly the answer OP is looking for. $\endgroup$ – andselisk Sep 28 at 11:23

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