Answer to Question #204361 in Biology for Ankush

Question

Use Hess’s law to calculate the heat of reaction for the following:    

2 C6H6(g) + 15 O2(g)   🡪 12 CO2(g)    + 6 H2O(g)

Answer 1

We know the following heat of reactions:

(i) C(s) + 3 H₂(g) → C₆H₆(l) ; ΔH = +45.9kJ 

(ii) H₂​(g) + 1/2 O₂​(g) → H₂O(l) ; ΔH = −285.9kJ
(iii) C(s) + O₂​(g) → CO₂(g) ; ΔH = −393.5kJ

And we have to find heat of reaction of this one:

(iv) C₆H₆(l) + 15/2 O₂(g) → 6 CO₂​(g) + 3 H₂O(l)

Reverse equation (i)

(v) C₆​H₆(l) → 6 C(s) + 3 H₂(g) ; ΔH = −45.9kJ

Multiply equation (ii) with 3.

(vi) 3 H₂(g) + 3/2 O₂(g) → 3 H₂O(l) ; ΔH = −285.9 × 3 = −857.7kJ

Multiply equation (iii) with 6.

(vii) 6 C(s) + 6 O₂(g) → 6 CO₂(g) ; ΔH = −393.5 × 6 = −2361kJ

Add equations (v), (vi) and (vii)

(iv) C₆H₆(l) + 15/2 O₂(g) → CO₂(g) + 3 H₂O(l) ; ΔH = −45.9 − 857.7 − 2361 = −3264.6kJ

The enthalpy of combustion of benzene is −3264.6kJ.