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So this is my last question on my hw and I spent like 2 hours on this one problem and I cannot figure it out. I tried google and found these two links:
http://answers.yahoo.com/question/index?qid=20080501200654AAgYX37
http://answers.yahoo.com/question/index?qid=20080502202336AAJsb3h
I followed both websites step by step and got .490909 but I forgot that mine is 10 liters and theirs is 1 liter. So I changed it to .0490909 but it's still wrong. I don't understand what I'm doing wrong and if anyone out there is good at chemistry can help me out would be greatly appreciated.
Here is the Question:
An error in the code for this problem was fixed at 1 PM on F, 12/4/2009. Sorry about that.
11. [2pt]
At a particular temperature assume that the equilibrium constant for the reaction:
H2(g) + F2(g) <==> 2HF(g)
is K = 100.0. A reaction mixture in a 10.00 liter flask contains 3.9 moles each of hydrogen and fluorine gases plus 1.5 moles of HF. What will be the concentration of hydrogen when this mixture reaches equilibrium?
Answer: Last Answer: .07 M
Incorrect, tries 30/99.
Hint: First you will need to determine which direction the reaction must go. Then solve as usual. The `x' you solve for will probably be the CHANGE in concentration, not the final concentration.
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[H2]=0.0775 Mol/L
Heh, chemistry major, gotta' go with the sweet name.
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^he'll finish the problem before me, but lol @ name.
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On May 06 2010 13:01 Molybdenum wrote:
[H2]=0.0775 Mol/L
Heh, chemistry major, gotta' go with the sweet name.
OMGGGGGGGGGGGGGGGGGGGG THANK YOU SOOOO MUCH. WHAT DO YOU WANT A CAR?? SPENT LIKE 2 HOURS ON THIS AND U JUST DO IT IN 5 MINUTES..guess whatever they did in the two links is wrong? How did you get the answer? And thanks again wow.
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On May 06 2010 13:01 Molybdenum wrote:
[H2]=0.0775 Mol/L
Heh, chemistry major, gotta' go with the sweet name.
Do you know the I.C.E. Method?
(Initial, Change, Equilibrium..)
Thats how my teacher taught us, and it works pretty easily
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Sorry, messed up my edit and double posted. Real post below.
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On May 06 2010 13:08 MaRiNe23 wrote:Show nested quote +On May 06 2010 13:01 Molybdenum wrote:
[H2]=0.0775 Mol/L
Heh, chemistry major, gotta' go with the sweet name. OMGGGGGGGGGGGGGGGGGGGG THANK YOU SOOOO MUCH. WHAT DO YOU WANT A CAR?? SPENT LIKE 2 HOURS ON THIS AND U JUST DO IT IN 5 MINUTES..guess whatever they did in the two links is wrong? How did you get the answer? And thanks again wow.
I used an 'ICE chart', I'll post it.
H2 + F2 <==> 2HF K=100.0=([HF]^2)/([H2][F2])
3.9mol 3.9mol 1.5mol (Now divide by 10.00L to get molar)
Initial 0.39M 0.39M .15M
Change -x -x +2x (One mole of H2 and F2 produces 2 moles of HF, K=100, so the reaction will probably go to the right, you could find Q, but this is less work)
Equil. 0.39-x 0.39-x 0.15+2x
Set up K=100.0=([0.15+2x]^2)/([0.39-x][0.39-x])
I solved on my graphic calculator (no approximations), x =0.3125, so [H2] = 0.39-x=0.0775
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wow you guys are very intelligent XD
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Kentor
United States5784 Posts
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This is like grade 11 chemistry - hopefully not for a university level course?
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CA10828 Posts
more like 10th grade but then again i sort of forgot how to do this stuff too lol
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Katowice25012 Posts
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He goes to Texas A&M and is a softmore if I remember correctly.
Or whichever Texas A&M's rival school is.
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EPT
