Is this right? chem heat of combustion

[joejo]

Hi guys can you please check if this is right...thanks in advance

The heat released by the reaction 2C + 2H2 = CH4 is 74.8kJ/mol. Using this information and a table of standard enthalpy of formation values, determine the heat of combustion for the burning of methane to form CO2 and H2O(g).
C02 = -393.3 kJ/mol
H20 = -286.2 kJ/mol

2C + 2H2 = CH4 H1 = 74.8
C + O2 = CO2 H 2= -393.3

H2 +1/2 O2 = H2O H3 = -286..2


CH4 +3O2 = CO2 + 2H2O -H1 + H2 +2H3 = -1040.5 Kj/mol

 

[joejo]

can anyone help me out?

 

[The Bob]

I may be incorrect but I got -754.3 kJ mol^-1.

The formation of CH₄ is 74.8 kJ mol^-1.
The formation of CO₂ and H₂O is -679.5 kJ mol^-1.
The formation of O is not counted due to it not changing.

This means that -74.8 kJ mol^-1 + (-679.5 kJ mol^-1) = -754.3 kJ mol^-1.

The Bob (2004 ©)

 

[VietDao29]

Bob, you are wrong.
joejo, the H1 is negative -74.8 kJ/mol.
It should be -890.9 kJ.
Viet Dao,

 

[The Bob]

Bob, you are wrong.
joejo, the H1 is negative -74.8 kJ/mol.
It should be -890.9 kJ.
Viet Dao,

-74.8 kJ mol^-1 + (-679.5 kJ mol^-1) = -754.3 kJ mol^-1.

It was. See?

The Bob (2004 ©)

 

[The Bob]

Now I have had another look, I am going to agree with joejo. Why? because he is right. It is -1040.5 kJ mol^-1.

Sorry

The Bob (2004 ©)

P.S.

-1040.5 Kj/mol

Just as a small hate of mine, can you please use the correct capitals?? Cheers.

 

[VietDao29]

The heat released by the reaction 2C + 2H2 = CH4 is 74.8kJ/mol.

. That means $\Delta H_1 = -74.8 \mbox{ kJ/mol}$.
So $-\Delta H_1 +\Delta H_2 + 2 \Delta H_3 = -890.9 \mbox{ kJ}$.
Am I missing something?
Viet Dao,

 

[The Bob]

Bob, you are wrong.

Although this did not affect what I thought or wrote, it normally gets people's defenses up and does not allow them to see errors. May I ask that you address the issue in a lighter-hearted way?

Am I missing something?

Nothing is missed. You are correct. I did the missing. However, both of us have done something wrong. joejo has (perhaps) lost some experience here because we both told him what had to be done. Although I was wrong, as was joejo, our methods were correct. What joejo really needed was some aimed help, not the answer. Of this we are both guilty and here is a point where we must learn to help (not tell the answer and method), not to argue and to have a bit of fun along the way.

I will take responsibilty if joejo is confused (not insult intended) due to my incorrect posts.

The Bob (2004 ©)

 

[VietDao29]

May I ask that you address the issue in a lighter-hearted way?

I just want my posts to be neat, short, but also contain enough information. So others can finish it in a glance. . Anyway, I'll take your advice. Thanks
Viet Dao,

 

[Oerg]

Exothermic reactions should always be negative enthalpy.

 

[The Bob]

Exothermic reactions should always be negative enthalpy.

Absolutely correct. So what is a reaction with a positive enthaply?

The Bob (2004 ©)

 

[NewScientist]

Endothermic!

 

[The Bob]

Endothermic!

Another correct answer.

The Bob (2004 ©)

 

[joejo]

i think its still wrong...


shouldn't it be... -802.3kJ

 

[The Bob]

shouldn't it be... -802.3kJ

How?

The Bob (2004 ©)

 

[joejo]

By definition the burning of methane to make CO2 and H2O is...

CH4 + 2 O2 --> CO2 + 2 H2O

The heat for this reaction is 2(-286.3) + (-393.3) -(-74.8) - 0 = -891.1 kJ.


right??

 

[joejo]

can anyone help me out?

 

[The Bob]

Your answer keeps on changing.

Which is it?

The Bob (2004 ©)

 

[joejo]

the bob...read above...i'll post it below just in case u never saw it.

By definition the burning of methane to make CO2 and H2O is...

CH4 + 2 O2 --> CO2 + 2 H2O

The heat for this reaction is 2(-286.3) + (-393.3) -(-74.8) - 0 = -891.1 kJ.


right??

 

[The Bob]

the bob...read above...i'll post it below just in case u never saw it.

I did not miss it. I saw it. I was commenting on this:

By definition the burning of methane to make CO2 and H2O is...

CH4 + 2 O2 --> CO2 + 2 H2O

The heat for this reaction is 2(-286.3) + (-393.3) -(-74.8) - 0 = -891.1 kJ.

and

i think its still wrong...

shouldn't it be... -802.3kJ

These ARE two different answers, both posted by you.

The next problem is also by you. Your original post states:

H2 +1/2 O2 -> H2O is -286.2

yet you said in your last post:

the bob...read above...i'll post it below just in case u never saw it.

By definition the burning of methane to make CO2 and H2O is...

CH4 + 2 O2 --> CO2 + 2 H2O

The heat for this reaction is 2(-286.3) + (-393.3) -(-74.8) - 0 = -891.1 kJ.

You have got a difference for the formation of water. You said it was 286.2 and now you say it is 286.3. Please make up your mind.

If the value is 286.2 then me and VietDao29 are right. If it is 286.3 then you are right.

This is why I asked. Ok?

The Bob (2004 ©)

P.S. May I ask you not to be sarcastic in future. It makes me, if no one else, very on edge and will explain all of the boldness in this post. I was annoyed due to your sarcasm.

 

[joejo]

i wasn't being sarcastic...sorry for the misunderstanding...so was my second answer right...

 

[The Bob]

i wasn't being sarcastic...sorry for the misunderstanding...so was my second answer right...

In which case I apologise. I did misunderstand.

Your answer is correct if the Enthalpy change of Formation for water is -286.3 kJ mol^-1. If it is -286.2 kJ mol^-1 then VietDao29 and me are correct.

The Bob (2004 ©)

P.S. Apologise again.

 

[joejo]

thank you...just to verify...my second answer (-891.1 kJ.) right?? sorry I am getting confused and I can imagine you are too! thanks again

 

[The Bob]

thank you...just to verify...my second answer (-891.1 kJ.) right?? sorry I am getting confused and I can imagine you are too! thanks again

Yes, if the enthaply change of formation of water is -286.3 kJ mol^-1.

The Bob (2004 ©)

 

[joejo]

thanks again! :)