Equlibrium Constant, Keq

[nothing123]

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Hi,

This might sound a little trivial but this expression: if Keq > 1, then it favours products...is that a definite rule or just a general one? Let's say we had a reaction A2 --> 2A and the concentration of A2 at the start was 1M and at the end was 0.4M. Then wouldn't the Keq = (0.6M)^2/0.4 < 1? Sounds a little dumb but I just wanted to double check.

Thanks.

 

[Quantum Chem]

Keq is a reaction specific constant. It does not change. Yes, if K>1 then the products are always in more concentration than reactants. Makes sense too, if Keq = [Products]/[Reactants] then anything above 1 means that there are more products.

Your example is a little off though. I'm going to rewrite the expression in an easier format: A -> 2B. If you start with 1M of A and end with .4M of A then you would have 1.2M of B. Here's why: when doing this you have to convert things into moles and take into account the coefficients. So lets assume this is in one liter of solution. You have 1 mole of A that turns into 2 moles of B. Since you didn't add any volume you still have one liter of solution, thus 1.2M of B.

Now do the ratio and you get K = [1.6]^2/[.4] = 6.4

 

[nothing123]

Yep you're right. I hate to beat a dead horse here but what if you had AB -> A + B. Now wouldn't K = (0.6)(0.6)/0.4? I just don't seem to be convinced of that hard and fast rule when we're dealing with decimals and coefficients.

 

[Quantum Chem]

Yes, in this case you would have [.6][.6]/[.4] because for each mole of AB that is degraded on mole of A and B are produced. But this isn't to say that all reactions that have one molecule turn into two different molecules that the Keq < 1. It is less than one for this particular reaction because the Keq is reaction specific.

 

[minhaj]

Yep you're right. I hate to beat a dead horse here but what if you had AB -> A + B. Now wouldn't K = (0.6)(0.6)/0.4? I just don't seem to be convinced of that hard and fast rule when we're dealing with decimals and coefficients.

I would think that in this case K<1 and favors the reactant.

 

[nothing123]

Yes, in this case you would have [.6][.6]/[.4] because for each mole of AB that is degraded on mole of A and B are produced. But this isn't to say that all reactions that have one molecule turn into two different molecules that the Keq < 1. It is less than one for this particular reaction because the Keq is reaction specific.

Yes, I completely understand that. I was just being picky and came up with a hypothetical example that countered that rule. In this case, there are still more products right? But the K is less than one. So that goes back to my original question, when K > 1 or vice versa, that is simply a general rule right?

Thanks for your help.

 

[hdu]

General rule:
Starting with all components at 1 M

If Keq>1 the reaction proceeds spontaneously forward
If Keq is 1 the reaction is in equilibrium
If Keq<1 the reaction proceeds in reverse

I hope it helps

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