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posted by  ss83 on 7/10/2008 8:11:30 PM  |  status: Live  

Equilibrium and the Equilibrium Constant Expression

Course Textbook Chapter Problem
General Chemistry N/A 14 N/A
Question Details:
Use the reactions below and their equilibrium constants to predict the equilibrium constant for the reaction, 2{\rm{A}}( s ) \rightleftharpoons 3{\rm{D}}( g ).
{\rm{A}}( s ) \rightleftharpoons {\textstyle{1 \over 2}}{\rm{B}}( g ) + {\rm{C}}( g )\; K_1 = 0.0334
3{\rm{D}}( g ) \rightleftharpoons {\rm{B}}( g ) + 2{\rm{C}}( g )\; K_2 = 2.35
  K  =?
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posted by Ainala8055 on 7/10/2008 8:24:06 PM  |  status: Live
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ss83's comment:
"Im so confused on this one, this answer was not correct"
Response Details:
2{\rm{A}}( s ) \rightleftharpoons 3{\rm{D}}( g )
K value is not for solids,but6 only for gases in case of Kp and for liquids and gases for Kc.
K= K2
   =2.35
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posted by Red Raiders on 7/10/2008 8:34:19 PM  |  status: Live
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ss83's comment:
"please stop submitting answers. You have not helped me once."
Response Details:
I calculated and got K=2.35
 
rate me livesaver!!!
 
lemme me know anything
s
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posted by AJ <3 on 7/10/2008 10:43:41 PM  |  status: Live
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ss83's comment:
"wow AJ, thank you for taking the time to show me how its done!!"
Response Details:
This might be a little complicated to understand, so I'll try my best to explain.
We are given:
{\rm{A}}( s ) \rightleftharpoons {\textstyle{1 \over 2}}{\rm{B}}( g ) + {\rm{C}}( g )\; K_1 = 0.0334
3{\rm{D}}( g ) \rightleftharpoons {\rm{B}}( g ) + 2{\rm{C}}( g )\; K_2 = 2.35 
And we need to combine these two reaction too like this:
2{\rm{A}}( s ) \rightleftharpoons 3{\rm{D}}( g )
Carefully examine this reaction's formula. It needs to have 2A on the Left hand side. From the following two equations, how can we get 2A on the Left hand side?
{\rm{A}}( s ) \rightleftharpoons {\textstyle{1 \over 2}}{\rm{B}}( g ) + {\rm{C}}( g )\; K_1 = 0.0334
3{\rm{D}}( g ) \rightleftharpoons {\rm{B}}( g ) + 2{\rm{C}}( g )\; K_2 = 2.35 
Do you see the first equation? It has 1A on the Left hand side. If we multiply the equation by 2, then we get a new formula of:
 
But we don't stop there. We also have to multiply K1 by 2.

Ok, so now we have 2A on the Left hand side. No we need to get a 3D on the Right hand side. So ask yourself, how can we get 3D on the Right hand side. Let's look at what we have to work with now:
  
3{\rm{D}}( g ) \rightleftharpoons {\rm{B}}( g ) + 2{\rm{C}}( g )\; K_2 = 2.35 
Look at equation 2 carefully. We have 3D, but it's on the Left hand side, so we need to get it on the Right hand side. Well, we can get 3D on the Right hand side by reversing the equation to get:
But we can't stop there yet. Now we have to change the value of K2. Whenever you reverse the reaction, the K value must be divided by 1 to get the new K value. This is the work for it:

Ok so now we pretty much have done everything we need to get the form 2{\rm{A}}( s ) \rightleftharpoons 3{\rm{D}}( g ).
Lets combine our two new equations:
 
To get:
 +  +
Cancel out the B(g)'s and the 2C(g)'s:
And we have our equation, which means we have the right K1 and K2 values.
To solve for K you have to multiply the two K1 and K2 values together:
 
 =

And that's all. This was a basic problem of this type and just be prepared for more complicated problems. Remember, what you do to the whole equation, you must apply to the K value.

Added correction by electropositive:
which is multiplied by 2 in this problem, the K value is raised to the power of that number (not just multiplied).

so in this case, I calculated the change of K1 as
(0.0443)2=0.00111556

Therefore, when you take

K=K1K2
K=(0.00111556)(0.425531915)
k=4.75 x 10-4   answer
 
Sorry If I made this more confusing that it was suppose to be.
 -AJ

Please ask for further assistance.
Please ask for further assistance. AJ
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posted by electropositive on 7/10/2008 11:01:36 PM  |  status: Live
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ss83's comment:
"Thank you very much for showing me the "short cut" version :) This will help during a timed exam :)"
Response Details:
according to my book, when you multiply a reaction by a number, in this case:

{\rm{A}}( s ) \rightleftharpoons {\textstyle{1 \over 2}}{\rm{B}}( g ) + {\rm{C}}( g )\; K_1 = 0.0334 

which is multiplied by 2 in this problem, the K value is raised to the power of that number (not just multiplied).

so in this case, I calculated the change of K1 as
(0.0443)2=0.00111556

Therefore, when you take

K=K1K2
K=(0.00111556)(0.425531915)
k=4.75 x 10-4   answer


Am I correct about this?

Otherwise, AJ <3, your step by step was more clearly explained than my textbook.
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posted by AJ <3 on 7/10/2008 11:04:34 PM  |  status: Live
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Response Details:
Haha, yea you're most likely right, I was basing all my solutions on memory. The book knows best =D
Please ask for further assistance. AJ
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