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Enthalpy Of Combustion ( Calculation ) Guide Needed

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iAdoreNaz Male
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  #1 Old 01-10-2011 Default Enthalpy Of Combustion ( Calculation ) Guide Needed

I'm not an STPM student, but I have an STPM revision books and syllabus that I'm currently learning is pretty much similar to STPM syllabus. I need your help on this question though. It's about Enthalpy Of Combustion . Thanks a bunch! .. Need the answers ASAP!

1. Methanol can be made in a two-step process.
Step 1 : CH4(g) + H2O(g) <--> (reversible arrow) CO(g) + 3H2(g)
Step 2 : CO2(g) + 3H2(g) <--> (reversible arrow) CH3OH (g)

Use the following enthalpies of combustion to calculate the enthalpies of reaction for each of the steps :

Enthalpy of combustion of CH4 ( g ) = - 890kJmol-1
Enthalpy of combustion of H2 = - 245 kJmol-1
Enthalpy of combustion of CO = - 283kJmol-1
Enthalpy of combustion of CH3OH = - 671kJmol-1

Thanks a bunch!
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  #2 Old 01-10-2011 Default Re: Enthalpy Of Combustion ( Calculation ) Guide Needed

Quote:
Originally Posted by iAdoreNaz View Post
I'm not an STPM student, but I have an STPM revision books and syllabus that I'm currently learning is pretty much similar to STPM syllabus. I need your help on this question though. It's about Enthalpy Of Combustion . Thanks a bunch! .. Need the answers ASAP!

1. Methanol can be made in a two-step process.
Step 1 : CH4(g) + H2O(g) <--> (reversible arrow) CO(g) + 3H2(g)
Step 2 : CO2(g) + 3H2(g) <--> (reversible arrow) CH3OH (g)

Use the following enthalpies of combustion to calculate the enthalpies of reaction for each of the steps :

Enthalpy of combustion of CH4 ( g ) = - 890kJmol-1
Enthalpy of combustion of H2 = - 245 kJmol-1
Enthalpy of combustion of CO = - 283kJmol-1
Enthalpy of combustion of CH3OH = - 671kJmol-1

Thanks a bunch!
Did you mistype your second equation? It doesn't balance up I took the liberty of using Wikipedia and the page on production of methanol shows you're lacking the product water. I'll base my answer on that equation.

http://en.wikipedia.org/wiki/Methanol#Production

I think I'll leave my answer first before I explain my steps and for you to clarify if the answers are correct because I'm not too sure either. I am using Hess' Law for this. By the way, I don't think it's necessary and confusing to put the reversible arrows. Because, I am going to assume the question wants the enthalpy change of the forward reaction. Anyways, if it's not, it'll just be a matter of flipping the signs in front of the enthalpy values.

1. +373kJmol-1

2. -26kJmol-1
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  #3 Old 01-10-2011 Default Re: Enthalpy Of Combustion ( Calculation ) Guide Needed

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Originally Posted by frostbyte13 View Post
Did you mistype your second equation? It doesn't balance up I took the liberty of using Wikipedia and the page on production of methanol shows you're lacking the product water. I'll base my answer on that equation.

http://en.wikipedia.org/wiki/Methanol#Production

I think I'll leave my answer first before I explain my steps and for you to clarify if the answers are correct because I'm not too sure either. I am using Hess' Law for this. By the way, I don't think it's necessary and confusing to put the reversible arrows. Because, I am going to assume the question wants the enthalpy change of the forward reaction. Anyways, if it's not, it'll just be a matter of flipping the signs in front of the enthalpy values.

1. +373kJmol-1

2. -26kJmol-1
Oh yah, thanks for noticing ..
It's suppose to be
CO(g) + 2H2(g) <--->(reversible arrow) CH3OH(g)

Surprisingly , there is no answer given for this question. My lecturer for chemistry doesn't provided us an exact answer. He said " you have to be confident with your answer" . HAHA . It's kinda a good and a bad thing.
I don't blame him.

By the way, I've just entered this chapter, and I really don't know how to solve this. I know how to do basic Enthalpy of Combustion and Formation by using Hess Law, but with the question that I've posted , I really don't know where to start.

Please help me =) ~ Thanks!

+
I've tried to do the question again and again. And I couldn't get the same answer as yours. And one more thing, it asked about the reaction for each steps and not CO2 + 3 H2 → CH3OH + H2O . Right? Hmm~.. I'm confuse now. xD..

Last edited by iAdoreNaz; 01-10-2011 at 10:06 PM. Reason: Automerged Doublepost
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  #4 Old 02-10-2011 Default Re: Enthalpy Of Combustion ( Calculation ) Guide Needed

I think I know what frostebyte13 did, but i don't entirely agree. There shouldn't be a need to reverse enthalpies to give values to the "combustion" of H2O, since H2O is present in both the top eqn and bottom eqn.

Anyway, I'm not entirely sure either, but this is my working and the subsequent answers.

CH4 + H2O -> CO + 3H2

CO2 + H2O

(Arrow from CH4 to bottom eqn, arrows from bottom eqn to CO and 3H2)

Enthalpy of rxn: -890 + 283 + 3(245) = +128

-------------------------------------------------------------
CO + 2H2 -> CH3OH

CO2 + H2O

(Arrows from CO and 2H2 to bottom eqn, arrow from bottom eqn to CH3OH)

Enthalpy of rxn: -283 + 2(-245) + 671 = -102

-----------------------------------------------------------

Based on the equations you gave, you can actually get an overall equation and you should be able to check the values you calculated (unless I'm mistaken).

CH4 + H2O -> CO + 3H2
CO + 2H2 -> CH3OH
By adding the equations together (the same way you would add redox eqns), you get
CH4 + H2O -> H2 + CH3OH

Calculate the enthalpy of this rxn (using Hess' law, the same way it was done above):
-890 + 245 + 671 = +26

If you add the first two enthalpies calculated earlier, you also get +26.

Based on this, I assume this answer to be right, IF I did not make any recurring errors in calculation or method.
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Last edited by lxy; 02-10-2011 at 02:09 AM.
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  #5 Old 02-10-2011 Default Re: Enthalpy Of Combustion ( Calculation ) Guide Needed

Thanks Ixy. I got the same calculation as yours. So, I bet it's correct?

By the I have another question, I just got this from Yahoo! Answers .

"When Benzoic Acid is burned, the enthalpy of combustion is -3223.6 kJ/mol. Use this information and tabulated values of the standard enthalpies of formation of liquid water and carbon dioxide to calculate the standard enthalpy of formation of benzoic acid"

I started with:
C7H6O2 + 15/2O2 --> 7CO2 + 3H2O deltaH= -3223.6 kJ
2H2 + O2 --> 2H2O deltaH= -571.6 kJ
C + 1/2 O2 --> CO2 deltaH= -393.5 kJ

p.s - My answer for this question is -388.3kJmol-1 . I'm not sure though. Please help ^^.. ~

Last edited by iAdoreNaz; 02-10-2011 at 03:01 PM.
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  #6 Old 02-10-2011 Default Re: Enthalpy Of Combustion ( Calculation ) Guide Needed

OK, I think you'd better take lxy's working more seriously than mine since she is a senior compared to me haha. I might have made a mistake somewhere is the use of the enthalpies provided. Please ask your chemistry lecturer for an answer. That "be confident in your answer" concept doesn't buy me one single bit. If you are confident in your answer yet it's wrong and you don't know it, what's the point of learning. Learning requires the mindset of one humble enough to admit mistakes. Being confident in answers kinda thrashes that sometimes. Or it could be he is just plain lazy
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  #7 Old 02-10-2011 Default Re: Enthalpy Of Combustion ( Calculation ) Guide Needed

Quote:
Originally Posted by frostbyte13 View Post
OK, I think you'd better take lxy's working more seriously than mine since she is a senior compared to me haha. I might have made a mistake somewhere is the use of the enthalpies provided. Please ask your chemistry lecturer for an answer. That "be confident in your answer" concept doesn't buy me one single bit. If you are confident in your answer yet it's wrong and you don't know it, what's the point of learning. Learning requires the mindset of one humble enough to admit mistakes. Being confident in answers kinda thrashes that sometimes. Or it could be he is just plain lazy
That could be the case, but all this while, he's been correct.. ~
So, yeah =)..
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  #8 Old 02-10-2011 Default Re: Enthalpy Of Combustion ( Calculation ) Guide Needed

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Originally Posted by iAdoreNaz View Post
Thanks Ixy. I got the same calculation as yours. So, I bet it's correct?

By the I have another question, I just got this from Yahoo! Answers .

"When Benzoic Acid is burned, the enthalpy of combustion is -3223.6 kJ/mol. Use this information and tabulated values of the standard enthalpies of formation of liquid water and carbon dioxide to calculate the standard enthalpy of formation of benzoic acid"

I started with:
C7H6O2 + 15/2O2 --> 7CO2 + 3H2O deltaH= -3223.6 kJ
2H2 + O2 --> 2H2O deltaH= -571.6 kJ
C + 1/2 O2 --> CO2 deltaH= -393.5 kJ

p.s - My answer for this question is -388.3kJmol-1 . I'm not sure though. Please help ^^.. ~

Er. I hope so. hahaha.

For your 2nd question:
I would use the enthalpy of formation of one mole H2O, not two. So that would be -285.8
And I think the eqn for formation of CO2 should be C + O2 -> CO2

C6H5COOH + 15/2 O2 -> 7CO2 +3H2O

O2 + H2 + C

(Arrows from bottom eqn to benzoic acid, CO2 and H2O)

Enthalpy of formation of benzoic acid is the arrow from the bottom eqn to the benzoic acid, so reverse the values for the other two arrows.

Enthalpy of formation of benzoic acid = -3223.6 + 7(393.5) + 3(285. = +388.3.

Can't quite figure out why my answer is positive while yours is negative. =/
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  #9 Old 02-10-2011 Default Re: Enthalpy Of Combustion ( Calculation ) Guide Needed

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Er. I hope so. hahaha.

For your 2nd question:
I would use the enthalpy of formation of one mole H2O, not two. So that would be -285.8
And I think the eqn for formation of CO2 should be C + O2 -> CO2

C6H5COOH + 15/2 O2 -> 7CO2 +3H2O

O2 + H2 + C

(Arrows from bottom eqn to benzoic acid, CO2 and H2O)

Enthalpy of formation of benzoic acid is the arrow from the bottom eqn to the benzoic acid, so reverse the values for the other two arrows.

Enthalpy of formation of benzoic acid = -3223.6 + 7(393.5) + 3(285. = +388.3.

Can't quite figure out why my answer is positive while yours is negative. =/
This is how i got it
* When Benzoic Acid is burned, the enthalpy of combustion is -3223.6 kJ/mol
==> C6H5COOH + 15/2 O2 -> 7CO2 +3H2O deltaH -3223.6kjMol-1

the equation for formation of benzoic acid should be
==> 7CO2 + 3H20 -> 15/2 O2 + C6H5COOH deltaH +3223.6kJMol-1

thus the standard enthalpy of formation of benzoic acid would be
+3223.6 = -7(-393.5) - 3(-285. + C6H5COOH
C6H5COOh = -388.3 .

~ what do you think? im not too sure either..
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  #10 Old 02-10-2011 Default Re: Enthalpy Of Combustion ( Calculation ) Guide Needed

Quote:
Originally Posted by iAdoreNaz View Post
This is how i got it
* When Benzoic Acid is burned, the enthalpy of combustion is -3223.6 kJ/mol
==> C6H5COOH + 15/2 O2 -> 7CO2 +3H2O deltaH -3223.6kjMol-1

the equation for formation of benzoic acid should be
==> 7CO2 + 3H20 -> 15/2 O2 + C6H5COOH deltaH +3223.6kJMol-1

thus the standard enthalpy of formation of benzoic acid would be
+3223.6 = -7(-393.5) - 3(-285. + C6H5COOH
C6H5COOh = -388.3 .

~ what do you think? im not too sure either..

Oh yes, you're right. Haha. Sorry, I got the arrows mixed up.
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