# How do you calculate the enthalpy change for the reaction 2C_8H_18(l) + 17O_2(g) -> 16CO(g) + 18H_2O(l)? DeltaH^o"_(rxn) =?

## Given; $2 {C}_{8} {H}_{18} \left(l\right) + 17 {O}_{2} \left(g\right) \to 16 C O \left(g\right) + 18 {H}_{2} O \left(l\right)$ DeltaH^o"_(rxn) = -11,020 kJ $2 C O \left(g\right) + {O}_{2} \left(g\right) \to 2 C {O}_{2} \left(g\right)$ DeltaH^o"_f = -566.0 kJ

Dec 11, 2016

You want $\Delta H$ for:
$2 {C}_{8} {H}_{18} \left(l\right) + 17 {O}_{2} \left(g\right) \rightarrow 16 C O \left(g\right) + 18 {H}_{2} O \left(l\right)$, and then you give us the thermodynamic reaction for precisely this equation. I suspect you want the parameters for the complete combustion of octane:
${C}_{8} {H}_{18} \left(l\right) + \frac{25}{2} {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 9 {H}_{2} O \left(l\right)$