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);
