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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: ssssss on July 23, 2004, 12:07:00 PM

Title: Equivalent concept
Post by: ssssss on July 23, 2004, 12:07:00 PM

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I have a problem in understanding the equivalent concept used in stochiometry.This is what i know about it....

Equivalent concept is similar to mole concept that means in any chemical reaction the reactants react in same equivalents to give same equivalents of products.

BaCl2+AL2[SO4]3--->BaSo4 + AlCl3
a b 0 0 [Before reaction]
0 b-a a a [after rection]

Where a and b are the equivalents.Now if we take the balance reaction:

3BaCl2+Al2[So4]3--->3BaSo4 + 2AlCl3
p q 0 0 [Before reaction]
p-3q 0 3q 2q [After reaction]

Where p and q are moles.

Now if we solve the problem with any of the above two methods the RESULT IS THE SAME.CAN YOU EXPLAIN ME WHY?

WHY EQUIVALENT CONCEPT AND MOLE CONCEPT GIVE SAME RESULTS?

Title: Re:Equivalent concept
Post by: movies on July 23, 2004, 01:36:09 PM
I'll try to help, but the way you have posed the question is a little confusing to me, but here goes:

First, I think you've made an error in your "after reaction" part of the second formula.  I think it should read like this:

0 q-3p 3q 2q

Anyway, I'm not sure how you are getting the same result for the two.  It's not obvious to me how that could happen.  Furthermore you can't really solve using the first formula because it isn't balanced.*

I would say that the main point of this problem is that you get 2 equivalents of Al(3+) and 3 equivalents of SO4(2-) for each mole of Al2[SO4]3.

* If you were to balance the equation from the first formula and just plug in the numbers, then you would arrive at the corrected formula I put above.  If that is what you did it shouldn't surprise you that the two give the same answer since all you have really done is change the order of your operations.
Title: Re:Equivalent concept
Post by: Mitch on July 23, 2004, 03:29:20 PM
ssssss: Don't post the same question all over the internet. ::)

The chemical education community is a small community.  8)
Title: Re:Equivalent concept
Post by: ssssss on July 24, 2004, 04:05:27 AM
I'll try to help, but the way you have posed the question is a little confusing to me, but here goes:

First, I think you've made an error in your "after reaction" part of the second formula.  I think it should read like this:

0 q-3p 3q 2q

Anyway, I'm not sure how you are getting the same result for the two.  It's not obvious to me how that could happen.  Furthermore you can't really solve using the first formula because it isn't balanced.*

I would say that the main point of this problem is that you get 2 equivalents of Al(3+) and 3 equivalents of SO4(2-) for each mole of Al2[SO4]3.

* If you were to balance the equation from the first formula and just plug in the numbers, then you would arrive at the corrected formula I put above.  If that is what you did it shouldn't surprise you that the two give the same answer since all you have really done is change the order of your operations.


You should know that when we are dealing with equivalent we do not need to balance an equation.We usually solve equations with equivalents when the equations are too weird to balance.

And i want to put the respective moles,equivalents under the reactants and products respectively but made a little mistake.I apologize if it was confusing.


Secondly,can you tell me how to find Equivalent weights for a big compound.E.g  Zn2Cr3[Fe{CN}4]2.
Title: Re:Equivalent concept
Post by: movies on July 26, 2004, 07:29:03 PM
I guess I am not familiar with the "equivalent" method.  I'm not sure how it could work and still give you the right answers.  Any chemical equation can be balanced, it just might take some time.  That's the way that I have always done it.

It's not obvious to me how you could solve a question like that _without_ balancing the equation.
Title: Re:Equivalent concept
Post by: ssssss on July 27, 2004, 11:21:34 AM
In equivalent concept you use equivalents instead of moles.

Equivalents=mass/Equivalent weight
Title: Re:Equivalent concept
Post by: movies on July 27, 2004, 12:36:59 PM
Wouldn't that calculation just give you the number of moles of whatever equivalent you were calculating for?

It doesn't seem to be so much a different method as a shortcut to do the same thing.
Title: Re:Equivalent concept
Post by: ssssss on July 30, 2004, 01:52:45 AM
Yes its very similar to the mole concept.You only got to stick to one thing that is equal number of equivalents will react to give equal number of equivalents.

Equivalents=weight/Equivalent weight

Where equivalent weight=Molecular weight/n factor

Here n factor depends upon the nature of reagent.
For H2SO4 n=2[number of replacable H+ ions]
For NaOH n=1[number of replacable OH- ions]
For elements n=Valency
For salts n=Charge on cationic or anionic part.

This method is Extremely useful as it can be used without knowing the nature of reaction[no need of balancing].

But it can only be used if you know the n factor of the reagents correctly.
Title: Re:Equivalent concept
Post by: movies on July 30, 2004, 02:04:38 AM
That's interesting.  I never thought of it before.

Doesn't that method also require that you know the limiting reagent ahead of time though?
Title: Re:Equivalent concept
Post by: ssssss on July 30, 2004, 02:17:05 AM
Now see this reaction

A + B---->C + D

Suppose A and B are the reactants with equivalents x and y[x>y] then C and D Formed will be y equivalents.It does have a limiting reagent but the good thing is that there is not much difficulty in determining the limiting reagent as no balancing is required.
Title: Re:Equivalent concept
Post by: movies on July 30, 2004, 12:30:20 PM
Heh, I don't know.  I can balance equations pretty fast and then I am always sure that I'm not missing something.

If you can do it without balancing the equations, that's cool too!
Title: Re:Equivalent concept
Post by: ssssss on August 01, 2004, 03:58:13 AM
Sometimes it is really difficult to balance some  reactions.so it is good in those cases.But it is adviced to use the mole concept when the reaction is already balanced.

Secondly,this method is excellent in solving REDOX REACTIONS.