Question

C8H16(l) + 12O2 ====8CO2 + 8H2O(l)
C8H16(l) + 8O2 ==== 8CO(g) + 8H2O
C8H18(l) + 25/2O2 ==== 8CO2 + 9H2O
C8H18(l) + 17/2O2 ==== 8CO(g) + 9H20

Standard enthalpies of formation of C8H16(l) , C8H18(l) , CO2(g) and H2O are -2840, -2926 , -394 and -286 respectively.Find the standard enthalpy of formation of CO. Using that or any other method find the enthalpy changes of the 4 reactions.
Please Help!

Answer 1

1) C8H16(l) + 12O2 ====8CO2 + 8H2O(l) ΔH1
2) C8H16(l) + 8O2 ==== 8CO(g) + 8H2O ΔH2
3) C8H18(l) + 25/2O2 ==== 8CO2 + 9H2O ΔH3
4) C8H18(l) + 17/2O2 ==== 8CO(g) + 9H20 ΔH4

First you have to calculate the enthalpies of formation for the first (ΔH1) and third reaction (ΔH3), you have the enthalpies of formation of the products and the reactants, remember that ΔHf O2 =0. Then you have to subtract the reverse of the second reaction from the first reaction and the reverse of the forth reaction from the third reaction. These two differences will have as result the same new reaction 8CO + 4 O2 -> 8 CO2 with a ΔH5
So then you will be able to said
5) ΔH1-ΔH2=ΔH3-ΔH4 = ΔH5 = 8ΔHco2-8 ΔHco = -3152 -8 ΔHco
from the second reaction
ΔH2 = 8 ΔH(H2O) + 8 ΔH(CO) - ΔH(C8H16)
ΔH2 = 8 (-286) + 8 ΔHco ) - (-2840)
ΔH2 = 8ΔHco - 2288+2840
ΔH2 = 8ΔHco + 552

replace ΔH2 = 8ΔHco + 552 in the equation 5

ΔH1-ΔH2= -3152-8 ΔHco
ΔH1- 8ΔHco + 552 = -3152-8 ΔHco

and calculate ΔHco

then calculate the ΔH2 and the ΔH4

Answer 2

let me know if you need more help

Answer 3

I think there's a little problem in

ΔH1- [8ΔHco + 552] = -3152-8 ΔHco


8 ΔHco cancels out, right? So cannot find ΔHco

By the way thank you for the support. Please reply.

Answer 4

Yes you are wright the equations are not independent. Let me think another solution. Sorry I will be back to you soon.

Answer 5

You can couple reactions here to find the \(\Delta\)H you need.