i) C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l),∆H = -330 kCal
ii) C(s) + O2(g) → CO2 (g) , ∆H = -94.3 kCal
iii) H2(g) + ½O2(g) → H2O (l) , ∆H = -68.5 kCal
Calculate enthalpy change for the reaction , 2C(s) + 3H2(g) +½O2(g) → C2H5OH (l)
Answers
Answer:
Δ
H
=
-279 kJ
This is a Hess's Law problem.
Our target equation is
2C(graphite) + 2H₂(g) + ½O₂(g) → C₂H₅OH(l);
Δ
H
=
?
We have the following information:
1. C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O (l);
Δ
H
=
-1367 kJ
2. H₂(g) + ½O₂(g) → H₂O(l);
Δ
H
=
-286 kJ
3. C(graphite) + O₂(g) → CO₂(g);
Δ
H
=
-394 kJ
Our target equation has 2C(graphite), so we multiply Equation 3 by 2:
4. 2C(graphite) + 2O₂(g) → 2CO₂(g);
Δ
H
=
-798 kJ
That means that we also multiply
Δ
H
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) ;
Δ
H
=
+1367 kJ
That means that we also change the sign of
Δ
H
.
Our target equation has no H₂O, so we multiply Equation 2 and its
Δ
H
by 3 to cancel the H₂O:
6. 3H₂(g) + ³/₂O₂(g) → 3H₂O(l);
Δ
H
=
-858 kJ
Now we add equations 4, 5, and 6 and their ΔH values.
This gives
2C(graphite) + 2H₂(g) + ½O₂(g) → C₂H₅OH(l);
Δ
H
=
-279 kJ