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Why does temperature affect Kc only and not concentration or pressure?

Can someone please explain why only temperature affects the position of Kc because I'm a bit confused.
Temperature affects the equilibrium constant (Kc) because it changes the rates of forward and reverse reactions, which are the basis of Kc. Temperature can change reaction rates by altering the frequency of collisions between atoms and molecules that are necessary for a successful reaction. For example, if a temperature change increases the amount of product formed, the numerator of the Kc expression increases, which increases the value of Kc.
In contrast, changes in pressure and concentration don't alter the value of Kc, but they can affect the rate at which equilibrium is reached.
let me know if you have any other questions
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Reply 2
I'm afraid that the answer above is confusing kinetic (rate) and thermodynamic (equilibrium) effects.

While temperature does affect both reaction rates and Kc, these are quite separate phenomena. The fundamental reason why Kc depends on temperature is because it is linked to the standard free energy change of reaction through the equation ΔG∘=−RTlnKc​. Since ΔG∘ also depends on T, then it follows Kc will depend on T. However, ΔG∘ is mostly independent of pressure and concentration, and so these do not affect the value of Kc.

If pressure is increased, the position of equilibrium may shift according to Le Chatelier's principle to counteract the change, but the ratio of concentrations at equilibrium (i.e., Kc) will remain constant at a given temperature. Similarly, changing the concentration of reactants or products will shift the equilibrium position to maintain the equilibrium constant, but the value of Kc​ itself does not change since​ it is a ratio of concentrations at equilibrium and is determined by the inherent properties of the reaction at a specific temperature. While it's true that temperature affects the rates of both the forward and reverse reactions, this is not the reason why Kc changes with temperature.
Reply 3
Original post by lordaxil
I'm afraid that the answer above is confusing kinetic (rate) and thermodynamic (equilibrium) effects.
While temperature does affect both reaction rates and Kc, these are quite separate phenomena. The fundamental reason why Kc depends on temperature is because it is linked to the standard free energy change of reaction through the equation ΔG∘=−RTlnKc​. Since ΔG∘ also depends on T, then it follows Kc will depend on T. However, ΔG∘ is mostly independent of pressure and concentration, and so these do not affect the value of Kc.
If pressure is increased, the position of equilibrium may shift according to Le Chatelier's principle to counteract the change, but the ratio of concentrations at equilibrium (i.e., Kc) will remain constant at a given temperature. Similarly, changing the concentration of reactants or products will shift the equilibrium position to maintain the equilibrium constant, but the value of Kc​ itself does not change since​ it is a ratio of concentrations at equilibrium and is determined by the inherent properties of the reaction at a specific temperature. While it's true that temperature affects the rates of both the forward and reverse reactions, this is not the reason why Kc changes with temperature.

I'm still a bit confused, I've only covered year 12 content so far
Does only temperature affect Kc as it doesn't initially increase or decrease reactants/products so the ratio of concentrations is changed and the reaction is no longer at equilibrium?
(edited 2 months ago)
Reply 4
No problem - sorry if my original explanation was not pitched at an appropriate level.

Think of Kc as something which is related to the relative stability of reactants and products at a particular temperature. As you change the temperature, their relative stability will change too, which means the position of the equilibrium will shift. The reaction still remains in equilibrium, but now at different ratios of reactants/products than at the previous temperature. The rate at which this equilibrium is reached may also change, but that is independent of the final position of the equilibrium as it depends on the kinetics of reaction.

Does that help?
Reply 5
Original post by lordaxil
No problem - sorry if my original explanation was not pitched at an appropriate level.
Think of Kc as something which is related to the relative stability of reactants and products at a particular temperature. As you change the temperature, their relative stability will change too, which means the position of the equilibrium will shift. The reaction still remains in equilibrium, but now at different ratios of reactants/products than at the previous temperature. The rate at which this equilibrium is reached may also change, but that is independent of the final position of the equilibrium as it depends on the kinetics of reaction.
Does that help?

Sorry, I am still a bit confused, would Kc be at the point the reaction is the most stable at so if the temperature is increased/decreased Kc point changes to a new equilibrium with a different ratio of reactants/products?
Reply 6
What you said is correct, but it's just a restatement of the definition of Kc. To explain WHY Kc depends on temperature, you need to link it to the relative chemical stabililty of the reactants and products at that temperature, as I did above.
Reply 7
Original post by lordaxil
What you said is correct, but it's just a restatement of the definition of Kc. To explain WHY Kc depends on temperature, you need to link it to the relative chemical stabililty of the reactants and products at that temperature, as I did above.

Do you think this would become more clear once I cover the year 2 content then ?
Reply 8
Yes, most probably. Until you learn about Gibbs free energy and connection to phase equilibrium, then it's hard to understand what Kc means apart from just being the ratio of amounts of reactants and products (which is true by definition).
Original post by Slugzie(:
Can someone please explain why only temperature affects the position of Kc because I'm a bit confused.

The explanation used at A level is that immediately after a change is made, the system is no longer at equilibrium (i.e if you were to plug all the concentrations - or rather more accurately the activities into the Kc expression, it would not give you the value of Kc, but rather some other quantity sometimes represented with the symbol Q).

The system then acts to counteract the change and restore the equilibrium, which just so happens to make it so the ratio of all participating substances in the system is the same as before if only a concentration or the pressure is changed.

The temperature, however, is a different story - all reversible reactions have an exothermic and an endothermic route. A drop in temperature will favour the exothermic route and an increase in temperature will favour the endothermic route. These changes drive the equilibrium too far in one direction for Kc to be maintained and so Kc changes value with temperature.

Lordaxil’s explanation is much better, but much of it won’t be used at A level (ΔG = -RTlnK is a minor spec point on the Edexcel syllabus and makes no appearance for other boards and is only ever used for the sake of calculating equilibrium constants - never for explaining their temperature dependence).

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