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Topic: ICe table calculation  (Read 14950 times)

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princessyana

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ICe table calculation
« on: June 07, 2005, 07:15:17 PM »
 I've been working on this problem for a long time and i just don't get how to solve it...
...@ 25 degrees C, Keq=0.145 for the following rxn:
2Br...Br2 +Cl2   Initial Concentrations of Br2 and Cl2 are both 0.025, what will their equilibrium concentrations be?
this is what I've got

       2BrCl  <-->  Br2     +     Cl2
I        0            0.025       0.025
C    +2x             -x             -x
E       2x           0.025-x     0.025-x                     0.145=???(i only know it's  
                                                                                products over reactants)
« Last Edit: June 07, 2005, 08:28:23 PM by hmx9123 »

Offline hmx9123

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Re:ICe table calculation
« Reply #1 on: June 07, 2005, 08:25:29 PM »
The first thing to do is to set up your Keq expression correctly.  You say that it is products over reactants... what are your products in this case?  Your reactants?  You seem to know this already, but what would you put for such a case?  If having the chemical formulas there confuses you, what about a simpler case, like this:

A + B <--> C

Your reactants are A and B, and your product is C.  Set up the equilibrium expression as products over reactants, and see what you get.  Come back here, and tell us what you got.  If you're having trouble with this, let us know what you're thinking and we'll go from there.

Once you have the Keq set up, you only need to substitute in your mathematical expressions you derived in your ICE table (which is correct, good job).  BTW, I edited your post to make it a little more clear by adding subscripts and an arrow.
« Last Edit: June 07, 2005, 08:31:27 PM by hmx9123 »

princessyana

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Re:ICe table calculation
« Reply #2 on: June 07, 2005, 09:12:32 PM »
Well, the Keq= [c]/[a]
so it would be 0.145= [Cl2][Br2]/[2BrCl]^2???? so 0.145=[0.025-x]^2/[2x]^2
is the 2BrCl squared???  

Offline hmx9123

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Re:ICe table calculation
« Reply #3 on: June 07, 2005, 11:45:26 PM »
The expresssion that you put for the Keq of the A + B <--> C reaction is correct.  You're almost right with the Keq for the chlorine/bromine reaction as well, and you've got the right idea with the exponents.  However, take a good look at what your products are.  The equation you've written is reverse of the way you'd normall see it, i.e., the normal way to write this is:

Br2 + Cl2 <--> 2 BrCl

Why?  Well, the initial concentration of BrCl is zero, so it is being formed, thus we consider it to be the 'products' even though the whole process is in equilibrium.  If the initial concentration of the bromine and chlorine gasses were zero, then they would be considered the products.

You also ask about the exponent.  You did correctly place the coefficient as an exponent on the BrCl term.  You can think of this as a simplified reaction; all the products are multiplied together, so you can take this equation:

A + B <--> 2 C

and re-write it as:

A + B <--> C + C

Now you can see where the exponents come from.  The Keq would look like this:

        [ C ][ C ]
Keq = [ A ][ B ]

Of course, you can simplify that into:

            [ C ] 2
Keq = [ A ][ B ]

So, your equation is going to look like this:

              [ BrCl ] 2
Keq = [ Br2 ][ Cl2 ]

Remember that you're looking for the total concentration of BrCl, not double, so be careful not to double it, as I see you had 2BrCl written in there.  It is true that there are 2 mols of BrCl to one of either Br2 or Cl2, but don't figure out what the concentration of BrCl is and then double it, or you'll wind up with a 4:1 molar ratio instead of the 2:1 that you're looking for.  I hope that wasn't too confusing.  Now that you've got the equation, you can just substitute in the appropriate algebraic expressions (x, 2x, whatever) and you'll have a nice quadratic equation to solve.  If you have any more questions, please drop us another post.
« Last Edit: June 07, 2005, 11:46:41 PM by hmx9123 »

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