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# C2H5OH (l) + 3O2(g)->2CO2 (g) + 3H2O (l) Delta H -1367kJ H2 (g) + 1/2 O2-> H2O (l) Delta H -286kJ C(graphite) + O2 (g) -> CO2 (g) Delta H -394 Calculate the change in enthalpy for this equation 2C (g) + 3H2 (g) + 1/2 O2-> C2H5OH (l) Delta H??

##ΔH = "-279 kJ"##

This is a Hess's Law problem.

Our target equation is

2C(graphite) + 2H₂(g) + ½O₂(g) → C₂H₅OH(l); ##ΔH = "?"##

We have the following information:

**1.** C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O (l); ##ΔH= "-1367 kJ"##**2.** H₂(g) + ½O₂(g) → H₂O(l); ##ΔH = "-286 kJ"##**3.** C(graphite) + O₂(g) → CO₂(g); ##ΔH = "-394 kJ"##

Our target equation has 2C(graphite), so we multiply Equation **3** by 2:

**4.** 2C(graphite) + 2O₂(g) → 2CO₂(g); ##ΔH = "-798 kJ"##

That means that we also multiply ##ΔH## by 2.

Our target equation has no CO₂, so we reverse equation **1** to cancel the CO₂.

**5.** 2CO₂(g) +3H₂O(l) → C₂H₅OH(l) + 3O₂(g) ; ##ΔH = "+1367 kJ"##

That means that we also change the sign of ##ΔH##.

Our target equation has no H₂O, so we multiply Equation 2 and its ##ΔH## by 3 to cancel the H₂O:

**6.** 3H₂(g) + ³/₂O₂(g) → 3H₂O(l); ##ΔH = "-858 kJ"##

Now we add equations **4**, **5**, and **6** and their ΔH values.

This gives

2C(graphite) + 2H₂(g) + ½O₂(g) → C₂H₅OH(l); ##ΔH = "-279 kJ"##