# as chem combustion question help??

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#1
Given the following data:
CH4(g) +2O2(G) --> CO2(g) +2H20 (L) ENTHALPY CHANGE= -890kjmol-1

CO(g) + 1/2 O2 (g) ---> CO2(g) ENTHALPY CHANGE= -284kjmol-1

C(s)+O2(g)---> CO2(g) ENTHALPY CHANGE= -393kjmol-1

H2 (g) + 1/2 O2 (g) ----> H20 (l) ENTHALPY CHANGE= -286kjmol-1

Calculate:
1) The enthalpy of formation of methane
2)The enthalpy of formation of carbon monoxide
3) the enthalpy change when methane is brunt in limited oxygen to form carbon monoxide and water.

I need help with Q3. I wrote the formula so it will be:
CH4(g) +1.5O2(G)----> co2(g) + 2H20 (L)

Bc this formula is the combustion of methane does that mean we set up a cycle (with formation) so the formula underneath when drawing the cycle would be C(s) +1.5O2 +2H2

Since we set up a formation cycle its P-R so i get 606 kj/mol but the answers say -606. Pls can someone help with this??
0
2 months ago
#2
your formula for Q3 produces CO2 not CO like the question says
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2 months ago
#3
1. -393-286x2-(-890)=-75
2. -393-(-284)=-109
3. -109-286x2-(-75)=-606
Last edited by Äries; 2 months ago
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#4
(Original post by Äries)
1. -393-286x2-(-890)=-75
2. -393-(-284)=-109
3. -109-286x2-(-75)=-606
where did you get the values of -75 and -109 from?
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2 months ago
#5
(Original post by idk__21)
where did you get the values of -75 and -109 from?
These are the answers of parts (1) and (2).
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#6
(Original post by Äries)
These are the answers of parts (1) and (2).
oh rightt thanks
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