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Chem 1a03 Advice ?!

 
Old 12-12-2010 at 09:12 PM   #31
bcars
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Quote:
Originally Posted by ~*Sara*~ View Post
HAHA. That's horrible! What about all the other instructors !! Waiiit, you took chem, in commerce.. am I missing something? Please don't tell me that was your elective :/!
Yep, that was my minor actually, I wanted to get into the Pharmaceutical industry. I took it until second year when I just couldn't stand Jeff Landry any longer and switched my minor to Econ.

~*Sara*~ likes this.
Old 12-12-2010 at 09:40 PM   #32
SilentWalker
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If anyone's doing/done the 2009 exam, can you please post your answers so others can compare? I'm working on it right now. Thanks.

EDIT: Can someone explain how to do this problem please?



EDIT 2: I've also noticed that almost none of the required formulas were provided.

Last edited by SilentWalker : 12-12-2010 at 09:55 PM.
Old 12-12-2010 at 09:56 PM   #33
bcars
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Quote:
Originally Posted by SilentWalker View Post
If anyone's doing/done the 2009 exam, can you please post your answers so others can compare? I'm working on it right now. Thanks.

EDIT: Can someone explain how to do this problem please?



EDIT 2: I've also noticed that almost none of the required formulas were provided.
Answering. I will edit it into this post.
Old 12-12-2010 at 10:03 PM   #34
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Quote:
Originally Posted by SilentWalker View Post
If anyone's doing/done the 2009 exam, can you please post your answers so others can compare? I'm working on it right now. Thanks.

EDIT: Can someone explain how to do this problem please?



EDIT 2: I've also noticed that almost none of the required formulas were provided.
Use the equation for the formation of HCl, that's the enthalpy you want, your target equation. You have a number of equations given, with their enthalpy values, you can use Hess' law to manipulate the equations/enthalpy values to reach the target.
Old 12-12-2010 at 10:06 PM   #35
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Btw, I have a question, say you have the enthalpy of formation for H2O, (i.e. delta Hf H2O), that only applies to H2 + 1/2 O2 --> H2O, right?

how is that value related to the enthalpy of the reaction 2H2 + O2 --> H2O ?

edit:
rxn 1) H2 + 1/2 O2 --> H2O deltaH1
rxn 2) 2H2 + O2 --> H2O delta H2

is deltaH2 just twice deltaH1? if that's the case, then i feel stupid posting this question
Old 12-12-2010 at 10:08 PM   #36
SilentWalker
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Quote:
Originally Posted by waldo92 View Post
Use the equation for the formation of HCl, that's the enthalpy you want, your target equation. You have a number of equations given, with their enthalpy values, you can use Hess' law to manipulate the equations/enthalpy values to reach the target.
Thanks, but I know that much. I just don't know the combination. What's the concept behind it? To end up ONLY with HCl(g)? Or to start ONLY from H2(g) and Cl2(g)?

I think it's the latter, because it's an enthalpy of formation (formation of the substance from it's elements at standard conditions right?). If so, then how do I get rid of the N2(g)?

I really don't understand the basic concepts, which is what the profs didn't teach!! They spent so much time on application and more complex examples without teaching the fundamentals behind it.
Old 12-12-2010 at 10:10 PM   #37
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Quote:
Originally Posted by SilentWalker View Post
Thanks, but I know that much. I just don't know the combination. What's the concept behind it? To end up ONLY with HCl(g)? Or to start ONLY from H2(g) and Cl2(g)?

I think it's the latter, because it's an enthalpy of formation (formation of the substance from it's elements at standard conditions right?). If so, then how do I get rid of the N2(g)?
i think the reaction would be 1/2 H2 + 1/2 Cl2 --> HCl
the reaction that defines the deltaH formation makes only one mole of whatever is being formed, you can get rid of N2 from the other given equations if i remember this question right

edit: i used 1/2 H2 + 1/2 Cl2 --> HCl as the target equation, and got C, don't know if it's the right answer or not, fairly confident though, unless i'm missing something

Last edited by waldo92 : 12-12-2010 at 10:17 PM.
Old 12-12-2010 at 10:22 PM   #38
REPLEKIA/.
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Quote:
Originally Posted by SilentWalker View Post
If anyone's doing/done the 2009 exam, can you please post your answers so others can compare? I'm working on it right now. Thanks.

EDIT: Can someone explain how to do this problem please?



EDIT 2: I've also noticed that almost none of the required formulas were provided.
Alrighty then. lets name the reactions 1,2, and 3.

We want to end up with:
(1/2)H2 + (1/2)Cl2 -> HCl

So we know eq 1 needs to be reversed since it is the only equation with HCl. We'll then need to reverse equation 2 and multiply by 1/2, to cancel out NH3. Equation 3 must be multiplied by 1/2 and we are left with the desired equation by cancelling products/reactants in the resultant reaction.

(delta)Hrxn=(-1)(delta)H1+(-1/2)(delta)H2+(1/2)(delta)H3
(delta)Hrxn=(-1)(-175.00)+(-1/2)(-99.22)+(1/2)(-628.86)]
(delta)Hrxn=-89.82 KJ

Last edited by REPLEKIA/. : 12-12-2010 at 10:26 PM.

SilentWalker says thanks to REPLEKIA/. for this post.
Old 12-12-2010 at 10:36 PM   #39
bcars
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I wasn't expecting people to post after me, so here's how I got the solution.



How I did this:

1. First, you've got to identify the unique compounds that don't have a partner. That is HCl, and Cl2. They are outlined with a rectangle.

2. Next, make everything even in number of mols. I did this by simply multiplying the top equation by 2.

3. You have got to have things you wish to eliminate by having them on opposite sides of the equation. Instead of switching the 1st and 2nd formula (which you should do), I made it easier to understand my solution by switching the 3rd one.

4. Eliminate common variables. 2NH3, 2NH4Cl, and N2 are all outright eliminated. There are 4 mols of H2 against 3 mols of H2, which leaves you with 1 mol of H2 on the right side of the equation.

5. Resulting equation is H2 + Cl2 --> 2HCl, but if you look at my equations above you realize I did it backwards. I'll address that in a second.

6. Resulting H is (-350) + (-99.22) + 628.86 = 179.64.

7. IMPORTANT, WHAT MANY PEOPLE DON'T EVEN THINK OF. Make sure you divide this by two, because you are only looking for ONE mol of HCl. This gives you 89.82.

8. I simply made it negative (or multiply by -1) because I did the equations earlier backwards. Result is -89.82

I hope that helps SilentWalker!

SilentWalker says thanks to bcars for this post.
Old 12-13-2010 at 03:06 PM   #40
REPLEKIA/.
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Well the exam is over. How did everyone make out? I personally am feeling pretty good about the test. Don't post any specifics though, there are still people who get to write late and we can get in trouble for discussing specific problems.
Old 12-13-2010 at 03:18 PM   #41
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Half my friends say it went good, while the other half (including me ) found it quite difficult...
Old 12-13-2010 at 03:20 PM   #42
jamiebenyovi
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i found it incredibly difficult
Old 12-13-2010 at 03:27 PM   #43
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It was fine.
Old 12-13-2010 at 04:01 PM   #44
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The two mark questions went pretty well. Nailed most of them..however the three mark ones went terribly bad. I guessed on like 4 or 5 of them.
Old 12-13-2010 at 04:02 PM   #45
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Yeah if you think 1A03 is hard... better step it up for 1AA3!
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