Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Rutherford on April 11, 2013, 02:15:44 PM
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Determine ΔH for the reaction: 4CO + 8H2 ---> 3CH4 + CO2 + 2H2O given the following data: (attached).
By what logic can I solve this problem when more than one possible ways of solving it exist, but only one is correct?
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By what logic can I solve this problem when more than one possible ways of solving it exist, but only one is correct?
If there's two valid ways they must both give the same answer.
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One rigorous way would be assume coefficients a,b,c,d and e. One for each reaction.
And then balance out each of the 5 species in your target equation.
5 equations in 5 unknowns. Solve.
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There are more than one answer. If I started, e.g. by multiplying (b) by 2 and then deducting 2*(a) from it, I can adjust the other equations to get the target equation. If I started by multiplying (b) by 5 and then adding (a), I can adjust the other equations to get the target equation. There are, I think infinite number of ways to get the target equation, but the values for ΔH would be always different. What logic could help me here, or is this a hole in Hess' law?
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There are more than one answer. If I started, e.g. by multiplying (b) by 2 and then deducting 2*(a) from it, I can adjust the other equations to get the target equation. If I started by multiplying (b) by 5 and then adding (a), I can adjust the other equations to get the target equation. There are, I think infinite number of ways to get the target equation, but the values for ΔH would be always different. What logic could help me here, or is this a hole in Hess' law?
ΔH should come the same each way.
Out of these infinite ways you think there are show me just two.
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1° ΔH=-2(a)+2(b)+4(c)+2(d)-(e)
2° ΔH=-(a)+3(b)+6(c)+(d)-2(e)...
*The letters represent the equations.
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1° ΔH=-2(a)+2(b)+4(c)+2(d)-(e)
2° ΔH=-(a)+3(b)+6(c)+(d)-2(e)...
*The letters represent the equations.
Did you calculate numerical values? Are we sure they don't end up the same?
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Ok, fine I did the heavy lifting for you. I get -747.2 both ways.
So, what's wrong? I don't see any problems.
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Hmm, now I see that these answers are the same, but when I first attempted this, I got on two different ways two different answers, so I concluded what I did. I will check my papers to see whether I mistaken somewhere.
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What's insane about this problem is that it gives you 5 equations when only need 3. That's just wrong.
Rxn(desired)=4*Rxn(2)+8*Rxn(3)-3*Rxn(5)
ΔHr°=-747.1 kJ·mol-1
I'm sure your answer will come out like this however you do it.
One rigorous way would be assume coefficients a,b,c,d and e. One for each reaction.
And then balance out each of the 5 species in your target equation.
5 equations in 5 unknowns. Solve.
I'm not sure what you're doing here ... could you write maybe 2 of those equations so I can understand?
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What's insane about this problem is that it gives you 5 equations when only need 3. That's just wrong.
Not at all. One skill is to know what's useful data. Extraneous data is a good distraction tool.
When I apply Hess's Law in practice the Handbook has maybe 2000 equations of which I need 4.
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What's insane about this problem is that it gives you 5 equations when only need 3. That's just wrong.
Not at all. One skill is to know what's useful data. Extraneous data is a good distraction tool.
When I apply Hess's Law in practice the Handbook has maybe 2000 equations of which I need 4.
Fair enough. What's the method using simultaneous equations? I like to apply algebra to these things.
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I'm sorry, I've actually mistaken when I attempted this for the first time. One change of sign lead me to perplexity :-[.
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Fair enough. What's the method using simultaneous equations? I like to apply algebra to these things.
I haven't tried but my idea was something like:
a *(C + 0.5 O2 -CO ) +b *( CO + 0.5 O2 -CO2) + .= 4 CO + 8H2 -3 CH4 -CO2 -2 H2O
now on left side collect terms containing CO, O2, H2 etc.
Eventually the expression for CO equates to 4, for H2 8 and so on.
For O2 etc. equate to zero since they are absent in the target. You should get 7 equations in 7 unknowns (if my calculations are right!)
That ought to give you all potential solutions.
Again, I haven't tried it but since you seem so interested try it out and let me know. I hope your enthusiasm can motivate you to do the number crunching?
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I'm sorry, I've actually mistaken when I attempted this for the first time. One change of sign lead me to perplexity :-[.
It happens. :)
Next time be more patient before calling a problem insane. You could have hurt Hess's feelings.
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Fair enough. What's the method using simultaneous equations? I like to apply algebra to these things.
I haven't tried but my idea was something like:
a *(C + 0.5 O2 -CO ) +b *( CO + 0.5 O2 -CO2) + .= 4 CO + 8H2 -3 CH4 -CO2 -2 H2O
now on left side collect terms containing CO, O2, H2 etc.
Eventually the expression for CO equates to 4, for H2 8 and so on.
For O2 etc. equate to zero since they are absent in the target. You should get 7 equations in 7 unknowns (if my calculations are right!)
That ought to give you all potential solutions.
Again, I haven't tried it but since you seem so interested try it out and let me know. I hope your enthusiasm can motivate you to do the number crunching?
The presence of multiple solutions tells us there is more than one way of fitting all of the equations we could write. The method will work to generate the equations quite nicely though (e.g. -a*Rxn(1)+b*Rxn(2)=4 according to the method and indeed any solution we come up with has to fit this).
Sorry, it was infinitely easier to do it by guesswork :p, and I can't see how a single solution will be produced anyway so I won't try and solve it myself.
When you use your 2000-equation handbook you need not actually work out the reaction manipulation surely - just use enthalpies of formation (products minus reactants) or combustion (reactants minus products) for each reactant/product?
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The presence of multiple solutions tells us there is more than one way of fitting all of the equations we could write.
In this question yes. Because Raderford showed us two solutions. In the general case how do you know?
There could be a unique solution. There could be multiple solutions. There could even be no solutions.
Sorry, I don't see how you can be so sure about the "presence of multiple solutions".
Sorry, it was infinitely easier to do it by guesswork :p, and I can't see how a single solution will be produced anyway so I won't try and solve it myself.
You asked Mister I-like-to-apply-algebra-to-these-things.. :) I will concede though that Chemistry is much more comfortable as a spectator sport; sadly I'm guilty of that often myself.
I've been burnt often enough though to realize that unless I work through it or see it worked in detail even the most reasonable and obvious sounding analysis and strategies are often wrong in hindsight. Devil's in the details. Ergo, I'm still not sure if or not my suggested strategy will work here. Perhaps. Perhaps not.
This is a common pattern though: The easiest strategy to solve small cases by hand is never the same as the best way to solve a large system and using a computer. Often reducing your problem to linear algebra helps becauase the toolkit there is so rich and efficient.
When you use your 2000-equation handbook you need not actually work out the reaction manipulation surely - just use enthalpies of formation (products minus reactants) or combustion (reactants minus products) for each reactant/product?
Nope. Real life isn't that clean unfortunately.
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In this question yes. Because Raderford showed us two solutions. In the general case how do you know?
There could be a unique solution. There could be multiple solutions. There could even be no solutions.
Sorry, I don't see how you can be so sure about the "presence of multiple solutions".
I was talking about this particular case - and explaining why I won't solve it by hand! Raderford came up with 2 solutions and I came up with a different one, and I'm sure there are more than 3 ... so either your simultaneous equations must somehow boil out to a cubic or higher degree polynomial (they won't as they're all linear) or they will only form a list of conditions that must be fulfilled, not guarantee a solution.
You asked Mister I-like-to-apply-algebra-to-these-things.. :) I will concede though that Chemistry is much more comfortable as a spectator sport; sadly I'm guilty of that often myself.
I've been burnt often enough though to realize that unless I work through it or see it worked in detail even the most reasonable and obvious sounding analysis and strategies are often wrong in hindsight. Devil's in the details. Ergo, I'm still not sure if or not my suggested strategy will work here. Perhaps. Perhaps not.
This is a common pattern though: The easiest strategy to solve small cases by hand is never the same as the best way to solve a large system and using a computer. Often reducing your problem to linear algebra helps becauase the toolkit there is so rich and efficient.
No I love solving the problems. But I have yet to come across one (except one particular example which had 10 equations ...) which I can't do by guessing. I do like algebra, but this method takes too long for most examples. I will keep it in mind in case I come across any "insane" questions with reaction manipulation I can't solve by guesswork. A bit like simultaneous equations - if it looks easy, just guess!
Nope. Real life isn't that clean unfortunately.
Huh ... so you must be working with chemicals that don't have well-known enthalpies of formation/combustion?
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Huh ... so you must be working with chemicals that don't have well-known enthalpies of formation/combustion?
Sure. Fairly often.
Other times the known values are for the wrong allotrope. Or known at a temperature different from that of interest.
You'd say, fine, use specific heats. Those may not be known. Lot's of problems. Industrial problems are messy and forget the technical complications. Most often lack of data is the biggest hurdle.