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Topic: Enthalpy Assistance Please  (Read 19287 times)

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lisabella3686

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Enthalpy Assistance Please
« 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.
« Last Edit: July 04, 2005, 06:12:19 PM by lisabella3686 »

Offline lemonoman

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Re:Enthalpy Assistance Please
« Reply #1 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  :'(

lisabella3686

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Re:Enthalpy Assistance Please
« Reply #2 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.


lisabella3686

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Re:Enthalpy Assistance Please
« Reply #3 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?

lisabella3686

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Re:Enthalpy Assistance Please
« Reply #4 on: July 04, 2005, 06:32:39 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?

arnyk

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Re:Enthalpy Assistance Please
« Reply #5 on: July 04, 2005, 06:34:10 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.  

lisabella3686

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Re:Enthalpy Assistance Please
« Reply #6 on: July 04, 2005, 06:45:16 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?

arnyk

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Re:Enthalpy Assistance Please
« Reply #7 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

lisabella3686

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Re:Enthalpy Assistance Please
« Reply #8 on: July 04, 2005, 06:58:10 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.
« Last Edit: July 04, 2005, 07:02:39 PM by lisabella3686 »

arnyk

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Re:Enthalpy Assistance Please
« Reply #9 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 = ?
« Last Edit: July 04, 2005, 07:15:36 PM by arnyk »

lisabella3686

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Re:Enthalpy Assistance Please
« Reply #10 on: July 04, 2005, 07:14:28 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

Offline lemonoman

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Re:Enthalpy Assistance Please
« Reply #11 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

lisabella3686

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Re:Enthalpy Assistance Please
« Reply #12 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?

lisabella3686

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Re:Enthalpy Assistance Please
« Reply #13 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.

arnyk

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Re:Enthalpy Assistance Please
« Reply #14 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" ;))
« Last Edit: July 04, 2005, 08:14:56 PM by arnyk »

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