Question

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??

Answer 1

`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"`