Well, let's clown around with Hess's law!
First, notice that the first equation is doubled, so let's double that enthalpy change:
2 C2H4O(g) + 10 O2(g) → 8 CO2(g) + 8 H2O(l) ΔH = −5224.4 kJ
Next, we need to reverse the second equation, so the sign of the enthalpy change will change:
−(HOCH2CH2OH(l) + 5/2 O2(g) → 2 CO2(g) + 3 H2O(l)) ΔH = 1189.8 kJ
Now we add these two equations together:
2 C2H4O(g) + 10 O2(g) → 8 CO2(g) + 8 H2O(l) ΔH = −5224.4 kJ
−(HOCH2CH2OH(l) + 5/2 O2(g) → 2 CO2(g) + 3 H2O(l)) ΔH = 1189.8 kJ
2 C2H4O(g) + 10 O2(g) − HOCH2CH2OH(l) − 5/2 O2(g) → 8 CO2(g) + 8 H2O(l) − 2 CO2(g) − 3 H2O(l)
ΔHtotal = −5224.4 kJ + 1189.8 kJ
Simplifying the equation:
2 C2H4O(g) + 5 O2(g) + HOCH2CH2OH(l) → 6 CO2(g) + 5 H2O(l)
ΔHtotal = -4034.6 kJ
So, the enthalpy change for the reaction is approximately -4034.6 kJ. Now that's one hot and cold reaction!