# Is Kc only affected by temperature?

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Thread starter 7 years ago
#1
if you increases the concentration of a substance that is present on the reactant side, will the equilibrium constant of the product be the same as Kc is only affected by temperature? Also, is Kc constant , at any temperature?
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7 years ago
#2
(Original post by otrivine)
if you increases the concentration of a substance that is present on the reactant side, will the equilibrium constant of the product be the same as Kc is only affected by temperature? Also, is Kc constant , at any temperature?
Kc is only affected by temperature ...

It is different at different temperatures
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7 years ago
#3
Kc is temperature dependent, so changing the temperature will change its value. The type of change depends on whether it's an exothermic or endothermic reaction.

If the reaction is exothermic, increasing the temperature will reduce Kc and vice-verca. This is because if you increase the temp., you drive the equilibrium backwards (in the endothermic direction), and therefore increase the concentration of reactants and decrease the concentration of products.

If endothermic, increasing the temperature will increase Kc, and vice-verca.
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Thread starter 7 years ago
#4
(Original post by charco)
Kc is only affected by temperature ...

It is different at different temperatures
is it correct to say that, if you add more reactant the equilibrium constant does not change because the equilibrium constant is constant at any range of temeperatures?
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Thread starter 7 years ago
#5
(Original post by AlexanderCLea)
Kc is temperature dependent, so changing the temperature will change its value. The type of change depends on whether it's an exothermic or endothermic reaction.

If the reaction is exothermic, increasing the temperature will reduce Kc and vice-verca. This is because if you increase the temp., you drive the equilibrium backwards (in the endothermic direction), and therefore increase the concentration of reactants and decrease the concentration of products.

If endothermic, increasing the temperature will increase Kc, and vice-verca.
is it correct to say that, if you add more reactant the equilibrium constant does not change because the equilibrium constant is constant at any range of temperatures?
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7 years ago
#6
(Original post by otrivine)
is it correct to say that, if you add more reactant the equilibrium constant does not change because the equilibrium constant is constant at any range of temperatures?
Nothing affects the value of kc apart from temperature ...

and stop calling me 'dear', I am neither expensive nor one of your lovers!
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7 years ago
#7
(Original post by otrivine)
is it correct to say that, if you add more reactant the equilibrium constant does not change because the equilibrium constant is constant at any range of temperatures?
To me that makes no sense. Changing concentrations or partial pressures makes no difference to Kc or Kp (so long as the overall pressure stays the same).

What does make a difference? Temperature, enthalpy change, entropy change, pressure - you can link these with equations though I think the pressure effect is negligible for any systems except gaseous ones. Enthalpy change and entropy change are often taken as constant but the reality is that they are also temperature dependent so I thought I'd throw them in the mix. In any case, you need both to work out the Kc at a certain temperature.

I don't think I missed anything out ... (note that lowering the activation energy, e.g. by using a catalyst, quickens the rate at which equilibrium is reached but does not affect the equilibrium constant)

Edit: I answered at this level because your OP talked about Kc values rather than "equilibrium" or something. You seem to be above the level of talking in terms of Le Chatelier's Principle. However if you wish we could give purely qualitative explanations.
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Thread starter 7 years ago
#8
(Original post by charco)
Nothing affects the value of kc apart from temperature ...

and stop calling me 'dear', I am neither expensive nor one of your lovers!
But you are precious because you are helping me in Chemistry

OK
but the statement that I wrote , Kc is constant at any given temperature? Is this statement wrong or right
2
7 years ago
#9
(Original post by otrivine)
But you are precious because you are helping me in Chemistry

OK
but the statement that I wrote , Kc is constant at any given temperature? Is this statement wrong or right
Just butting in here but it doesn't seem to me you understand the principles at all The answer to your question is very simple but won't help you as much as understanding it properly.

Read up this website: http://www.chemguide.co.uk/physical/...ia/change.html We can go from there.
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Thread starter 7 years ago
#10
(Original post by Big-Daddy)
Just butting in here but it doesn't seem to me you understand the principles at all The answer to your question is very simple but won't help you as much as understanding it properly.

Read up this website: http://www.chemguide.co.uk/physical/...ia/change.html We can go from there.
I understand Kc very well but in the exam it says that is this statement true,Kc is constant at any given temperature? yes or no
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7 years ago
#11
(Original post by otrivine)
I understand Kc very well but in the exam it says that is this statement true,Kc is constant at any given temperature? yes or no
Are you a troll? I'm asking you to read a webpage and evidently you haven't looked at it yet and are still coming back to ask questions.

I'm not responding until you answer your own question, which is trivially simple. You will NOT get an answer from me until you read the website in detail.
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Thread starter 7 years ago
#12
(Original post by Big-Daddy)
Are you a troll? I'm asking you to read a webpage and evidently you haven't looked at it yet and are still coming back to ask questions.

I'm not responding until you answer your own question, which is trivially simple. You will NOT get an answer from me until you read the website in detail.
No, I am not!

Look, Kc is only affected by temperature, Kc increases if forward reaction is endothermic and Kc decreases if forward reaction is exothermic.

If Kc is less than 1 then is reactant favoured but if Kc is more than product favoured
1
7 years ago
#13
I'm currently revising the areas of undergrad chem that covers the equilibrium constant function, so here goes!!!!

Kc/Kp or more correctly K (the true theoretical function, not the experimental function) is derived by assuming chemical equilibrium for the system. You are perturbing this when you add more of a component so that the equilibrium is disrupted. It will take time (how long is a kinetics problem, not determined by any thermodynamics) to return to the equilibrium position.
K is a function of T, a complex one as the simple version you use at A-level is a simplification allowed by making reasonable assumptions. Across a small temperature range it is reasonable to write deltaGr = -RTlnK, which describes how the equilibrium position shifts with changing temperature.

Now, otravine..... Why not try actually doing some work for a change. Wonder why you get all this negative rep? Well it's because you ask questions/try to get answers that require no effort on your part. This system relies on the goodwill of others and you will only have that if you are capable of putting in at least the effort we are putting in trying to write a helpful response. You cannot expect somebody to paraphrase a whole textbook/website so that you can just copy and past the answer into your homework.
0
7 years ago
#14
(Original post by otrivine)
Kc is only affected by temperature,
So what's the answer to your own question "is Kc constant at a given temperature"? If, as you say, the only thing that will change Kc is changing the temperature?

The reality is that it's a bit more complicated than that as JMaydom is beginning to go into but I would need more sign that you are listening carefully to try and explain myself. Kc is not only affected by temperature, but at A-level you might say it is.
0
7 years ago
#15
At one particular temperature, Kc will remain the same no matter what changes elsewhere, e.g. in concentration, pressure, or addition of a catalyst. If the temperature changes, so will Kc.
0
7 years ago
#16
(Original post by JMaydom)
I'm currently revising the areas of undergrad chem that covers the equilibrium constant function, so here goes!!!!
Not gonna lie, I was also hoping to do the same here

(Original post by JMaydom)
Kc/Kp or more correctly K (the true theoretical function, not the experimental function) is derived by assuming chemical equilibrium for the system.
Can't you show the temperature dependence for Kc and Kp the same way?

(Original post by JMaydom)
Across a small temperature range it is reasonable to write deltaGr = -RTlnK, which describes how the equilibrium position shifts with changing temperature.
And the lovely way ΔSO cancels out and you only need 1 constant and ΔHO to get the constant at any other temperature (only works for temperature-independent entropy/enthalpy, I know - I'm guessing you still have to work with ΔGO if the entropy/enthalpy are temperature dependent?)
0
7 years ago
#17
(Original post by Big-Daddy)
Not gonna lie, I was also hoping to do the same here

Can't you show the temperature dependence for Kc and Kp the same way?

And the lovely way ΔSO cancels out and you only need 1 constant and ΔHO to get the constant at any other temperature (only works for temperature-independent entropy/enthalpy, I know - I'm guessing you still have to work with ΔGO if the entropy/enthalpy are temperature dependent?)
I'm thinking in terms of the statistical mechanics derivation of K. This works off the molecular partition function, not any thermodynamic function in particular.
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2 years ago
#18
I think that changing concentration or pressure at equilibrium is just physical pressure on system which doesn't change heat content of system so kc remain constant but temperature change at equilibrium disturb heat content of system so one species may be more favoured than other hence equilibrium constant disturb ........if I am wrong kindly correct me
0
2 years ago
#19
Ok bro you are greAt
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