Question

Calculate the standard enthalpy change for the reaction:

2C8H18(l) + 17O2(g) --> 16CO(g) + 18H2O(l)

Given:
2C8H18(l) + 25O2(g) --> 16CO2(g) + 18H2O(l): ∆Hº = -11,020 kJ/mol

2CO(g) + O2(g) --> 2CO2(g): ∆Hº = -566.0 kJ/mol

Answer 1

Add equation 1 to the reverse of equation 2. I did this in my head so write it down and confirm it before proceeding.