# Effect of temp/conc on Kc and Kp!

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#1
It's easy to understand that values of Kc and Kp increase while temperature increases cuz the conc of products increases, and as the products are numerator in Kc/Kp's equation, so they increase. But when I was reading the explanation about effect of conc on these constants, it was written that equilibrium will be restored in this case, so it won't have any effect. My question is, the equilibrium will also be restored in the temperature increase case, so that shouldn't also effect their values.
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10 years ago
#2
Well now.. we need a bit more info here on the energetics of the reaction.

You need to distinguish between rate and equilibrium position.

Temperature always increases the rate of reactions (because it increases the collision frequency and the average energy of the particles reacting.)

But the effect of temp on the equilibrium depends on whether the reaction is exo or endothermic. In the case of an exothermic forward reaction, increasing temp will tend to push the equilib to the left (ie towards reactants). In the case of an endothermic reaction, the converse is true. Though in both cases, the equilibrium position, whatever that is, will be reached more quickly.
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#3
(Original post by Plato's Trousers)
Well now.. we need a bit more info here on the energetics of the reaction.

You need to distinguish between rate and equilibrium position.

Temperature always increases the rate of reactions (because it increases the collision frequency and the average energy of the particles reacting.)

But the effect of temp on the equilibrium depends on whether the reaction is exo or endothermic. In the case of an exothermic forward reaction, increasing temp will tend to push the equilib to the left (ie towards reactants). In the case of an endothermic reaction, the converse is true. Though in both cases, the equilibrium position, whatever that is, will be reached more quickly.

Hmm, yeah, I know that. I'm only asking why the increase in concentration doesn't effect values of Kp/Kc, while temperature does? =s
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10 years ago
#4
When you say concentration, are you now referring to gas phase reactions? If so, you need to use Kp for that (and use partial pressures in your equilibrium equation). In that case, the effect of pressure will depend on the number of molecules of reactants compared to products.

If you have the same number of reactant molecules as product molecules, then pressure will have no effect (as pressure increases/decreases the same in the numerator and denominator).

If you have more molecules of product than reactants, then increasing pressure will reduce Kp (and vice versa)

If you have fewer molecules of product than reactants, then increasing pressure will increase Kp (and vice versa)
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#5
(Original post by Plato's Trousers)
When you say concentration, are you now referring to gas phase reactions? If so, you need to use Kp for that (and use partial pressures in your equilibrium equation). In that case, the effect of pressure will depend on the number of molecules of reactants compared to products.

If you have the same number of reactant molecules as product molecules, then pressure will have no effect (as pressure increases/decreases the same in the numerator and denominator).

If you have more molecules of product than reactants, then increasing pressure will reduce Kp (and vice versa)

If you have fewer molecules of product than reactants, then increasing pressure will increase Kp (and vice versa)
Wow, now what's that? All the chem books of my A level course tell that Kp/Kc are only effected by temperature and nothing else...Is that only for my level? =(
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10 years ago
#6
you must have done Le Chatelier's principle, no? Basically, if you apply a change, the equilibrium will shift to whichever side minmises the effect of the change.

So if you squeeze (ie increase pressure) it will move to the side with the least molecules, so it takes up less space and thus reduces the effects of the pressure increase. And vice versa.

You should have been taught that at the same time as being taught about temperature.

Naughty teacher!
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#7
LOL! I know all that, maybe you ain't understanding my question. =) Anyways, I saw an old thread at TSR: http://www.thestudentroom.co.uk/showthread.php?t=702813 which answers my question. But the thing is why isn't it effected by conc changes? =(( Anyways, no problem if you can't understand my language. I'll wait for another reply! =)
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10 years ago
#8
(Original post by Zishi)
LOL! I know all that, maybe you ain't understanding my question. =) Anyways, I saw an old thread at TSR: http://www.thestudentroom.co.uk/showthread.php?t=702813 which answers my question. But the thing is why isn't it effected by conc changes? =(( Anyways, no problem if you can't understand my language. I'll wait for another reply! =)

Ok, so you are not talking about gas reactions, but liquid phase reactions. (I wish you'd be specific!). In a solution, concs don't affect the Kc because the higher the conc of reactant you put in, your just get more product. Since the Kc is [products]/[reactants], there's no change. Kc just measures the ratio of the concs.

Got it now?
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#9
(Original post by Plato's Trousers)
Ok, so you are not talking about gas reactions, but liquid phase reactions. (I wish you'd be specific!). In a solution, concs don't affect the Kc because the higher the conc of reactant you put in, your just get more product. Since the Kc is [products]/[reactants], there's no change. Kc just measures the ratio of the concs.

Got it now?
Yeah, but when in endothermic reaction temp is increased the products will increase and reactants will decrease, but the equilibrium will be maintained afterall, so why Kc will increase with temperature? =s
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10 years ago
#10
(Original post by Zishi)
Yeah, but when in endothermic reaction temp is increased the products will increase and reactants will decrease, but the equilibrium will be maintained afterall, so why Kc will increase with temperature? =s
well both Kc and Kp would increase with temperature because of the thermodynamics outcome of the gibbs equation,
At eqm,

delta G = -RTlnK
K = eqm constant = exp (-delta G/RT)

Any increase in T would increase K.
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#11
(Original post by shengoc)
well both Kc and Kp would increase with temperature because of the thermodynamics outcome of the gibbs equation,
At eqm,

delta G = -RTlnK
K = eqm constant = exp (-delta G/RT)

Any increase in T would increase K.
Well, that's outta syllabus then. I guess I'd only need to learn it as a fact. Thanks Plato's Trousers and Shengoc.
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10 years ago
#12
(Original post by Plato's Trousers)
When you say concentration, are you now referring to gas phase reactions? If so, you need to use Kp for that (and use partial pressures in your equilibrium equation). In that case, the effect of pressure will depend on the number of molecules of reactants compared to products.

If you have the same number of reactant molecules as product molecules, then pressure will have no effect (as pressure increases/decreases the same in the numerator and denominator).

If you have more molecules of product than reactants, then increasing pressure will reduce Kp (and vice versa)
If you have fewer molecules of product than reactants, then increasing pressure will increase Kp (and vice versa)
kp is unaffected by pressure changes in all situations ...
2
10 years ago
#13
(Original post by charco)
kp is unaffected by pressure changes in all situations ...

increasing pressure drives the equilib over to the right as you have 2 molecules on the right and 4 on the left
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10 years ago
#14
(Original post by Plato's Trousers)

increasing pressure drives the equilib over to the right as you have 2 molecules on the right and 4 on the left
Equilibrium proportions change but NOT kp
1
10 years ago
#15
(Original post by charco)
Equilibrium proportions change but NOT kp
how come?

isn't

in which case, if the proportions of reactants and products change, so must
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10 years ago
#16
(Original post by Plato's Trousers)
how come?

isn't

in which case, if the proportions of reactants and products change, so must

Not quite:

kp = P(products)coefficients /P(reactants)coefficients

Although the equilibrium concentrations (partial pressures) change the value of Kc and Kp do not.

perhaps this will explain it:

1
10 years ago
#17
(Original post by charco)
Not quite:

kp = P(products)coefficients /P(reactants)coefficients

Although the equilibrium concentrations (partial pressures) change the value of Kc and Kp do not.

perhaps this will explain it:

sorry, yes, you are of course (as ever) quite right. It's precisely because the doesn't change that the equlibrium moves to the right when the number of molecules decreases.

1
4 years ago
#18
please kp or kc increases whiles temperature increases for only endothermic reactions
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4 years ago
#19
(Original post by Josph)
please kp or kc increases whiles temperature increases for only endothermic reactions
I hope you aren't correcting people from nearly seven years ago about their misunderstandings of LCP.

I bet they will have finished their degrees, got jobs and possibly had children by now.

I don't think your lecturing will bother them too much.
1
4 years ago
#20
(Original post by Plato's Trousers)
When you say concentration, are you now referring to gas phase reactions? If so, you need to use Kp for that (and use partial pressures in your equilibrium equation). In that case, the effect of pressure will depend on the number of molecules of reactants compared to products.

If you have the same number of reactant molecules as product molecules, then pressure will have no effect (as pressure increases/decreases the same in the numerator and denominator).

If you have more molecules of product than reactants, then increasing pressure will reduce Kp (and vice versa)

If you have fewer molecules of product than reactants, then increasing pressure will increase Kp (and vice versa)
(Original post by Zishi)
Wow, now what's that? All the chem books of my A level course tell that Kp/Kc are only effected by temperature and nothing else...Is that only for my level? =(
Kp not changing with pressure is explained very well here https://www.chemguide.co.uk/physical...ia/change.html
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