# 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??

##### 1 Answer

This is a Hess's Law problem.

Our target equation is

2C(graphite) + 2H₂(g) + ½O₂(g) → C₂H₅OH(l);

We have the following information:

**1.** C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O (l);

**2.** H₂(g) + ½O₂(g) → H₂O(l);

**3.** C(graphite) + O₂(g) → CO₂(g);

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

**4.** 2C(graphite) + 2O₂(g) → 2CO₂(g);

That means that we also multiply

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

**5.** 2CO₂(g) +3H₂O(l) → C₂H₅OH(l) + 3O₂(g) ;

That means that we also change the sign of

Our target equation has no H₂O, so we multiply Equation 2 and its

**6.** 3H₂(g) + ³/₂O₂(g) → 3H₂O(l);

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

This gives

2C(graphite) + 2H₂(g) + ½O₂(g) → C₂H₅OH(l);