Calculating heat of formations

[Sconesnatcher]

Heres the question

 

Calculate the heat of formation of methane from the following data:

C + O2 CO2 DH = -393kJ/mol

H2 + 1/2O2 H2O DH = -280kJ/mol

CH4 + 2O2 CO2 + 2H2O DH = -895kJ/mol

 

This is not homework its the kind of question that will be on the test so I need to know how to do it. I know what Hess' law is but I just have no idea how to go about solving this. I'm assuming the final reaction can be reversed and it will be endothermic with a DH of +895kJ/mol but apart from that I'm lost.

[CaptainPanic]

Let's keep it all in one thread, shall we?

Q2) Calculate the heat of formation of methane from the following data:

C + O2 -> CO2 DH = -393kJ/mol

H2 + 1/2O2 -> H2O DH = -280kJ/mol

CH4 + 2O2 -> CO2 + 2H2O DH = -895kJ/mol

Not sure what this one means. WHat have those reactions got to do with methane?

(I added the arrows (->) in the quote above)

CH4 = methane.

You have to combine the reactions so that you make the formation of methane form its elements. Then also recombine the heats of reactions to form the heat of formation.

 

Because you already got an answer in the previous thread, I really would expect a new question... be more specific. Repeating the same question doesn't help you.

Right now, it seems to me that you haven't really tried, and you're just hoping that somebody will provide you with the answer. On this forum, you don't simply get the answers, instead you will learn.

[Sconesnatcher]

Im not looking for answers Im trying to figure out how its done. How do I combine the reactions? I tried adding the enthalpy changes but what I got wasn't the heat of formation of methane.

[hermanntrude]

OK the first step you need to take is to find an equation which actually has methane in it. The only candidate is the third one:

 

[ce]CH4 + 2O2 -> CO2 + 2H2O[/ce] [math]\Delta H = -895 kJmol^{-1}[/math]

 

The only problem with this equation is it has methane as a reactant rather than a product. You need to turn it around. What happens to the enthalpy of a reaction when you run it backwards?

 

[ce]CO2 + 2H2O -> CH4 + 2O2[/ce] [math]\Delta H[/math] = ???

 

I'll leave it for you to find out what to do to get the enthalpy for this reaction.

 

Then you need to find equations which have carbon and hydrogen as reactants (remember that a heat of formation is the enthalpy associated with the formation of a substance from its elements, in this case, carbon and hydrogen).

 

The first two equations look good to me:

 

[ce] H2 + 1/2O2 -> H2O[/ce] [math]\Delta H = -280 kJmol^{-1}[/math]

 

and

 

[ce] C + O2 -> CO2[/ce] [math]\Delta H = -393 kJmol^{-1}[/math]

 

The only problem now is that we want 4 moles of hydrogen and we've only got two. So we need to double the first equation:

 

[ce] 2H2 + O2 -> 2H2O[/ce] [math]\Delta H = ???[/math]

 

Again, i'll leave it to you to figure out what needs to be done to the enthalpy when you double the equation.

 

Finally, you can add them all together to get the equation we've been looking for:

 

[ce] 2H2 + O2 -> 2H2O[/ce] [math]\Delta H = ???[/math]

[ce] C + O2 -> CO2[/ce] [math]\Delta H = -393 kJmol^{-1}[/math]

[ce]CO2 + 2H2O -> CH4 + 2O2[/ce] [math]\Delta H[/math] = ???

 

total:

 

[ce]2H2 + 2O2 + C + CO2 + 2H2O -> 2H2O + CO2 + 2O2 + CH4[/ce]

 

[ce]2O2[/ce], [ce]CO2[/ce], and [ce]H2O[/ce] all appear on both sides, so they can be cancelled:

 

[ce]2H2 + C -> CH4[/ce]

 

You can get the enthalpy for this by simply adding the enthalpies of the equations you added together to make it. This is an example of Hess's law, which I recommend you read up on.

 

I hope this helps. If not, please ask me a specificly worded question IN THIS THREAD (not a new one). Thanks.

[Sconesnatcher]

Thanks alot. That explained the whole thing. Didn't know you were meant to literally add up every reactant and product in the reactions.