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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: lisabella3686 on July 04, 2005, 05:56:30 PM

Title: Enthalpy Assistance Please
Post by: lisabella3686 on July 04, 2005, 05:56:30 PM: Hello Everyone!

I am so extactic to have found this forum!!!

I am taking a 1st Year University Chemistry course over the summer and I have an exam tomorrow.

I am having trouble understanding the concepts involved with enthalpy calculations.

I do not know even where to begin with a question such as this:

The standard enthalpies of combustion of CH3OH(l), C(s) and H2(g) are -726, -394 and -286 kJ/mol. What is the standard enthalpy of formation of CH3OH(l)?

I am aware that you would like me to have attempted the question, but I simply am lost.
Please *delete me*!

I also do not understand how to use the "Bohr-Haber Process" aka. Lattice Enthalpy.
If anyone can explain this, it would be wonderful!

*Modification* I understand the basics of enthalpy and Hess' Law, but the connection between formation and combustion is confusing me. I also know that in some instances I can subtract enthalpies of the reactants from the products.
Title: Re:Enthalpy Assistance Please
Post by: lemonoman on July 04, 2005, 06:14:25 PM: Well they've given you the enthalpies of combustion for each of "CH3OH(l), C(s) and H2(g)"

So you have the following 3 reactions:

? CH₃OH + ? O₂ --> ? CO₂ + ? H₂O
? C + ? O₂ --> ? CO₂ + ? H₂O
? H₂ + ? O₂ --> ? CO₂ + ? H₂O

Which you'll have to find out on your own.

Now, you WANT the standard enthalpy of formation of CH3OH...ie. for the reaction

? C + ? H₂ + ? O₂ --> CH₃OH

Now, you can rearrange the first three, and add them together in some multiples, etc...and use enthalpy rules and stuff (like if you reverse a reaction, what happens to the enthalpy?)..

Hopefully this helps. This should definitely get you started - and if you need anything else, let us know.

P.S. If, in the future, you can answer others people's questions and give back to the forum we will love you more ;D than if you just pass by and use us :'(
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 04, 2005, 06:23:58 PM: haha, I didn't plan on any using. I only hope that I would understand things well enough to explain them to someone else.

I'm still not clear on the connection between formation and combustion. I was under the impression that there was a relationship between these types of reactions, in which, the forward reaction enthalpy was the negative enthalpy of the forward reaction for the reverse. Am I way in left field?

thanx for your help.
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 04, 2005, 06:28:09 PM: I am also unsure where to begin with this question. I have tried to use some sort of q=mcdt and pv=nrt, but I am just not sure.

If the standard enthalpy of combustion of C2H2(g), is -1300kJ/mol, how many litres of C2H2 measured at 25°C and 2.62 atm must be burned to liberate 10.0x10^6 kJ of heat?
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 04, 2005, 06:32:39 PM: Quote from: lemonoman on July 04, 2005, 06:14:25 PM

So you have the following 3 reactions:

? CH₃OH + ? O₂ --> ? CO₂ + ? H₂O
? C + ? O₂ --> ? CO₂ + ? H₂O
? H₂ + ? O₂ --> ? CO₂ + ? H₂O

How can I balance the 2nd and 3rd equations without the presence of H and C in the reactants?
Title: Re:Enthalpy Assistance Please
Post by: arnyk on July 04, 2005, 06:34:10 PM: Quote from: lisabella3686 on July 04, 2005, 06:28:09 PM
I am also unsure where to begin with this question. I have tried to use some sort of q=mcdt and pv=nrt, but I am just not sure.

If the standard enthalpy of combustion of C2H2(g), is -1300kJ/mol, how many litres of C2H2 measured at 25°C and 2.62 atm must be burned to liberate 10.0x10^6 kJ of heat?

Ok so if you want 10 x10^6 kJ of heat, and the heat released is 1300 kJ/mol, how many moles of C2H2 would you need?

Knowing the moles you need, determine the volume.
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 04, 2005, 06:45:16 PM: Quote from: arnyk on July 04, 2005, 06:34:10 PM
Ok so if you want 10 x10^6 kJ of heat, and the heat released is 1300 kJ/mol, how many moles of C2H2 would you need?

Knowing the moles you need, determine the volume.
So,

Q=mcdt

10x10^6 - 1300 = 9.99 x 10^9 J

9.99 x 10^9 J = m * C (C2H2?) * 25°C

m/MM = n

ratio = 2:4

2 C2H2 + 602 -> 4C02 + 4H2O

V= (n/2) (0.0820578) (298K)/ 2.62atm

is this right?
Title: Re:Enthalpy Assistance Please
Post by: arnyk on July 04, 2005, 06:52:41 PM: I'm not sure why you subtracted?

You want to find how many moles you need for 10 x10^6 kJ of energy.

1 mol = 1300 kJ

? mol = 10 x10^6 kJ
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 04, 2005, 06:58:10 PM: Quote from: arnyk on July 04, 2005, 06:52:41 PM
I'm not sure why you subtracted?

You want to find how many moles you need for 10 x10^6 kJ of energy.

1 mol = 1300 kJ

? mol = 10 x10^6 kJ

I subtracted because I was just fishing for some sort of understanding. I always seem to take the hard way of doing things. I am terrible at math and I can't seem to make these kinda connections. Do you have a simple, step by step, method for doing these kind of questions?
I deeply appologize for being a pain. haha, but I do understand the way you did it, once you showed me the basics.

*Modification* 10* 10^6/ 1300 = 7692.30mol = 72 x 10^4 L

thank you. I'm just a wee bit stressed out because I couldn't find any help until today, and my exam is pending.
Title: Re:Enthalpy Assistance Please
Post by: arnyk on July 04, 2005, 07:08:56 PM: *Lil' Moddy of my Own* Great Job! ;)

Alrighty then, we'll work through it until someone who actually knows what they're doing comes along ;).

1 mol = 1300 kJ

? mol = 10 x10^6 kJ

Just a little ratio here.

(10 x10^6 kJ) / (1300 kJ.mol) = 7692.3 mol

So you need 7692.3 mol of C2H2 to release 10 x10^6 kJ of energy.

Now you know that:

n = 7692.3 mol
P = 2.62 atm or 265.4 kPa
R = 8.31 kPaL/molK <-- (just remembered the units ;))
T = 25°C or 298K
V = ?
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 04, 2005, 07:14:28 PM: Quote from: arnyk on July 04, 2005, 07:08:56 PM
*Lil' Moddy of my Own* Great Job! ;)

Alrighty then, we'll work through it until someone who actually knows what they're doing comes along ;).

1 mol = 1300 kJ

? mol = 10 x10^6 kJ

Just a little ratio here.

(10 x10^6 kJ) / (1300 kJ.mol) = 7692.3 mol

So you need 7692.3 mol of C2H2 to release 10 x10^6 kJ of energy.

Now you know that:

n = 7692.3 mol
P = 2.62 atm or 265.4 kPa
R = 8.31 X 10^-3 (units...bleh...)
T = 25°C or 298K
V = ?

thank you for the tedious response. I appologise, but I meant the general enthalpy questions; like the one i asked earlier. I should have indicated that. I'm sorry.
Anything I can help with to make it better? haha I am majoring in Biology and German
Title: Re:Enthalpy Assistance Please
Post by: lemonoman on July 04, 2005, 07:40:29 PM: back to the FIRST question, is where I'm going to go. I'm going to use {H} for delta H

Reaction 1: 2 CH₃OH + 3 O₂ --> 2 CO₂ + 4 H₂O -- {H} = -726 kJ/mol
Reaction 2: C + O₂ --> CO₂ + 0 H₂O -- {H} = -394 kJ/mol
Reaction 3: 2 H₂ + O₂ --> 0 CO₂ + 2 H₂O -- {H} = -286 kJ/mol

Throwing the H2O in as a product was a red herring. Sorry bout that, I just saw it myself.

Anyways, check this out:

We want the enthalpy for C + 2 H₂ + 1/2 O₂ --> CH₃OH -- THIS is what makes it enthalpy of formation - it is the the reaction where all the reactants are the elements in their 'natural' form (as they are found in nature) and the only product is the compound you want.

Anyways, if you take Reaction 1, and reverse in, AND halve it...you get

Reaction 4: 1 CO₂ + 2 H₂O --> CH₃OH + 3/2 O₂ -- {H} = -(-726) kJ/mol = 726 kJ/mol

Now we took the negative of the {H} because we reversed the reaction. We DIDN'T divide by two...because the units on {H} are already PER MOLE...

Now, having done this a couple times, I realize that the water is going to cancel out in the final equation (if you don't get this the first time, no worries. Do a couple more, see if it comes to you, and if it doesn't, let us know, we'll explain)...Anyways it's like an 'instinct' thing.

So yeah. Let's take Reaction 3:

Reaction 5: 2 H₂ + O₂ --> 0 CO₂ + 2 H₂O -- {H} = -286 kJ/mol

And in the same way we won't do anything to reaction 2:

Reaction 2: C + O₂ --> CO₂ + 0 H₂O -- {H} = -394 kJ/mol

Now, let's recap

Reaction 2: C + O₂ --> CO₂ + 0 H₂O -- {H} = -394 kJ/mol
Reaction 5: 2 H₂ + O₂ --> 0 CO₂ + 2 H₂O -- {H} = -286 kJ/mol
Reaction 4: 1 CO₂ + 2 H₂O --> CH₃OH + 3/2 O₂ -- {H} = -(-726) kJ/mol = 726 kJ/mol

We can add all these together and get:

1 CO₂ + 2 H₂O + 2 H₂ + O₂ + C + O₂ --> CO₂ + 0 H₂O + 0 CO₂ + 2 H₂O + CH₃OH + 3/2 O₂

Which cancels out to:

C + 2 H₂ + 1/2 O₂ --> CH₃OH

Which is the formation reaction for CH₃OH !

So we added reactions 2, 5, and 4 together to get the formation reaction and we'll use Hess' Law to get the Enthalpy of Formation (sum the reactions --> sum the enthalpies)

So {H}_f = (726 kJ/mol) + (-394 kJ/mol) + (-286 kJ/mol) = 46 kJ/mol

And there you have it.

P.S. Normally, we don't go around solving entire questions for people like this. We like to guide people along. But you seemed to understand the other guy's thing when he explained it all out, hopefully this does the same.

P.P.S. It was murder to type that out. Now I know why very few of the 'experienced' people don't use subscripts. It almost made me cry :'( ... LOL
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 04, 2005, 07:46:39 PM: Thank you soooo much! You are fantastic and I greatly appreciate all of your hard work.

I think I was just thrown off by the extra products and I was a little discouraged. I understand that you just like to give guidance, and I respect that.

So if I follow these steps for any question with similar wording, I shouldn't have a problem?

1: take the combustion values and create combustion equations with each reactant.
2. Using Hess' law, use each step to create one lone equation for the desired value
3. Calculate
4. Presto, Answer, hopefully correct

riight?
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 04, 2005, 07:51:45 PM: wait a tick, I got 46kJ/mol when I first tried it, but aparently that's not the answer.
My sample midterm exam gives -240 as the answer.
Title: Re:Enthalpy Assistance Please
Post by: arnyk on July 04, 2005, 08:13:45 PM: Well there's a simple way to tell which is right. Is the reaction exothermic or endothermic?

PS: Finally found another Ontarian! (although he referred his fellow Ontarian as "the other guy" ;))
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 04, 2005, 08:17:06 PM: Quote from: arnyk on July 04, 2005, 08:13:45 PM
Well there's a simple way to tell which is right. Is the reaction exothermic or endothermic?

PS: Finally found another Ontarian! (although he referred his fellow Ontarian as "the other guy" ;))

YAY! Ontario! Is this site mostly american or canadian?
Title: Re:Enthalpy Assistance Please
Post by: Mitch on July 04, 2005, 08:22:51 PM: Its an international forum, although most of the admins live in California.
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 04, 2005, 08:28:38 PM: Nice! international is good.

I am going to go study for my exam. Thank you again for your help. I shall return later.

Good Evening
Title: Re:Enthalpy Assistance Please
Post by: lemonoman on July 04, 2005, 08:31:32 PM: Yay Ontario. lol...

I'm going to Tim Hortons to help some girls with Analytical Chemistry.

I live to give. ;D

I'll be back to answer any questions on this thread. It's my favorite. :)
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 04, 2005, 08:37:45 PM: Yay Waterloo!
I go to Western, but I am taking this course through Athabasca U. through independant study.

If you guys can band together and try to find out how the answer is -240, I would greatly appreciate it.

thanx mucho.
I'll check back in a while
Title: Re:Enthalpy Assistance Please
Post by: arnyk on July 04, 2005, 08:58:24 PM: Sweet I work at Timmie's! Although I might be selling out if this Staples thing goes through. ;)
Title: Re:Enthalpy Assistance Please
Post by: lemonoman on July 05, 2005, 03:01:28 AM: Okay Lisa, you owe me BIG on this :D

But here we go. I made a mistake ;D

Go to my huge-*ss post above, with all the reactions, and you'll notice,

Reaction 5: 2 H2 + O2 --> 0 CO2 + 2 H2O -- {H} = -286 kJ/mol

This isn't really true. It should be -286 kJ/mol of H₂...so...

Reaction 5b: H2 + 1/2 O2 --> 0 CO2 + 1 H2O -- {H} = -286 kJ/mol H₂
But since we need 2 mols of H2, then
Reaction 5c: 2 H2 + O2 --> 0 CO2 + 2 H2O -- {H} = -572 kJ/2 mol H₂ = -572 kJ/mol O₂

Note how we just slid in a single O₂ for 2 H₂. We can do that, because it's in the equation (like a titration)

Reaction 2: C + O2 --> CO2 + 0 H2O -- {H} = -394 kJ/mol
is fine. One mole of O2. We're cool.

Now comes a tricky part again :P
Reaction 4: 1 CO2 + 2 H2O --> CH3OH + 3/2 O2 -- {H} = -(-726) kJ/mol = 726 kJ/mol CH3OH

So {H} = 726 kJ/(1 mol CH3OH) = 726 kJ/(3/2 mol O2) = 484 kJ/mol O2

That each of the equations has O2 in it. So we'll use O2 as the main compound thing. Units are important, keep that in mind.

Anyways, we now have
Reaction 2: C + O2 --> CO2 + 0 H2O -- {H} = -394 kJ/mol O₂
Reaction 5: 2 H2 + O2 --> 0 CO2 + 2 H2O -- {H} = -572 kJ/mol O₂
Reaction 4: 1 CO2 + 2 H2O --> CH3OH + 3/2 O2 -- {H} = 484 kJ/mol O₂

TOTALLY notice that each is in terms of kJ/mol O₂ - before we had the one reaction in terms of kJ/mol H₂ and the other one...and that's not the same unit. So you can't add them together.

NOW when you add up the numbers you get

(484 kJ/mol O₂) + (-572 kJ/mol O₂) + (-394 kJ/mol O₂) = -482 kJ/mol O₂

Which is for C + 2 H2 + 1/2 O2 --> CH3OH

And so, again with the conversion:

-482 kJ/(mol O₂) = -482 kJ/(2 mol CH3OH) = -241 kJ/mol CH3OH

Which is your answer.

P.S. It is less important to note that: (726) + (-394) + (-286) + (-286) = -240
But this is NOT the correct way to do it.

P.P.S. Did you know that here at Waterloo, we have T-Shirts that say, "Friends don't let friends go to Western"? ;D - They might be all over the place. Maybe even at Western :P
Title: Re:Enthalpy Assistance Please
Post by: Mitch on July 05, 2005, 01:44:57 PM: Quote from: arnyk on July 04, 2005, 08:58:24 PM
Sweet I work at Timmie's! Although I might be selling out if this Staples thing goes through. ;)

Did you know that those of us who are south of the border don't know what Tim Bits are, but I do. ;)
Title: Re:Enthalpy Assistance Please
Post by: arnyk on July 05, 2005, 03:00:19 PM: Well then you must be very cultured indeed eh? I think Wendy's bought Tim Hortons a couple years back though so once again an American company takes over a successful Canadian enterprise. ;)
Title: Re:Enthalpy Assistance Please
Post by: lemonoman on July 05, 2005, 03:18:48 PM: From the Wendy's Internation Website:

"There are more than 2,400 Tim Hortons restaurants in Canada and more than 250 in the U.S. Founded in 1964, Tim Hortons merged with Wendy's International, Inc. in 1995."

Let's see...32 million people in Canada...2400 Tim Hortons...that's one for every 13000 people. Seems like a lot more though :P

AND there's 250 Tim Hortons in the US? Since when?

It won't be long before Walmart buys them all out anyways.
Title: Re:Enthalpy Assistance Please
Post by: arnyk on July 05, 2005, 03:25:10 PM: Funny how they say "merged" when Ron Joyce clearly completely sold out for $450 million (CDN ;)).
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 06, 2005, 09:58:18 PM: Thank you so much!
I should have read it over more carefully.
No biggie. I think my exam went ok...but I'm not sure either way. A lot of it felt pretty simple, but that usually means that I missed a co-efficient multiplier or something.

haha..Tim Hortons is a big conglomeration, as is Walmart. We have 16 or 17 Tim Hortons on campus. It's crazy

Friends may not let friends go to Western, but atleast our library isn't sinking because we didn't calculate the weight including the books.

Q: How do you get a Waterloo grad off your front porch?

A: Pay him for the pizza.
Title: Re:Enthalpy Assistance Please
Post by: lemonoman on July 07, 2005, 02:11:23 AM: Snap.
Title: Re:Enthalpy Assistance Please
Post by: lemonoman on July 07, 2005, 02:24:27 AM: All of the UWO jokes revolve around how loose women are.

Definition (Slang)
============
Western -- v; to obtain commitment-free sex.
Usage : We're going out to the bar to see if we can Western tonight

Or,

Q: How many Western students does it to change a lightbulb?
A: One to change the lightbulb and four to find the perfect JCREW outfit to wear for the occasion.

Q. Why don't they have Christmas at Western?
A. They can't find a virgin and three wise men.

Good times. ;D
Title: Re:Enthalpy Assistance Please
Post by: arnyk on July 07, 2005, 09:18:24 AM: Here's a zing on both of ya:

University of Waterloo:
Waterlosers. Being the top school in Canada, Waterloo students are smug and
arrogant believing that they are the best around! Damn you all for being right. Waterloosers have absolutely nothing to do all day but study. Well that and booze it up. The transit system sucks as does the night life. These guys come back to T.O. as often as possible to party it up. Oh yeh and they are even snobbier than Queens and St. George! This campus is ruled by Chinese peeps(in all things business) and by all those kids who you never really thought were THAT smart in school b/c they were so damn quiet. But if you're here for anything other than biomed, engineering and accounting you need to get over yourself! It's not THAT good a school! Trivia: the library was designed by Waterloo students and is currently sinking into the ground b/c they forgot to account for the weight of the books in the library. Way to go Waterlosers way to go!

University of Western Ontario:
It should just be called Western. It's not a university. It's Western. An anomaly. A night club/ resort that somehow got mislabelled as a university. Western doesn't count. Period. More so than the rest. At the other sucky universities you're really going to college. Here you're not even going to college, you're going to a non stop party. Your average intelligence is actually decreasing. Students here carry backpacks full of beer and those booze caps with the long straws attached. But hey there is justice in the world. A degree from Western is worth less than that 2 4 ! Side Note: If you're in business (post undergrad) the above doesn't apply to you at all.

Oh and the sinking library gag: That's a really common urban legend at many universities. Just google it. ;)
Title: Re:Enthalpy Assistance Please
Post by: arnyk on July 07, 2005, 09:24:01 AM: Oh and here's a couple nice one-liners for you since Lisa ain't here to fire back. ;)

Waterloo -- v; to fart in a tub/pool.
Usage : Ahh gross! Did you just Waterloo!?!

>>Who does the Waterloo Engineering Society fear the most?
>>
>>- Immigration.

::)
Title: Re:Enthalpy Assistance Please
Post by: lemonoman on July 07, 2005, 10:29:30 AM: You stole those from the same site I took my Western Jokes from. Like the paz-him-for-the-pizza joke can be used with ANY university..and IS :P

To round things off, let's have a new twist on an old favorite:

Q: What does a Western student call a Waterloo student after graduation?
A: Boss

I like that one best. Classy. :bubbles: :nep: :Flush:

Those were the classiest smilies I could find....

and here's one that's particularly appropriate...

":offtopic2:"
Title: Re:Enthalpy Assistance Please
Post by: lisabella3686 on July 07, 2005, 05:07:50 PM: off topic- schmoff topic

I really don't think that Western is anymore of a party school than many other Universities. Unless you are looking for a party, it's not like you just walk around campus and then BAM, party breaks out.

Spend some time at Western and you'll see. I think the only difference between Western and Waterloo party scene is that London's transit is better and allows more access to bars. Richmond st. has many restaurants and such, all in one location.