Question

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)

Answer 1

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