determine the enthalpy for the reaction

Question

Using the equations 2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.6 kJ/mol C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol Determine the enthalpy (in kJ/mol) for the reaction 6 C (s) + 3 H₂ (g) → C₆H₆ (l).

Answer 1

Hess' Law

Targe equation: 6C(s) + 3H₂(g) ==> C₆H₆(l)

Given:

eq. 1: 2C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g) ... ∆H° = -6271 kJ/mol

eq. 2: 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ... ∆H° = -483.6 kJ/mol

eq. 3: C (s) + O₂ (g) → CO₂ (g) ... ∆H° = -393.5 kJ/mol

Procedure:

multiply eq. 3 by 6: 6C(s) + 6 O2(g) ==> 6 CO2(g) ... ∆H = 6x-393.5 = -2361 kJ

reverse eq. 1 & divide by 2: 6 CO2(g) + 3 H2O(g) ==> C6H6(l) + 7.5 O2(g) ... ∆H = +6271/2 = 3136 kJ

multiply eq. 2 by 1.5: 3H2(g) + 1.5 O2(g) ==> 3H2O(g) ... ∆H = -483.6 / 1.5 = -322 kJ

Add all 3 equations to get....

6C(s) + 6 O2(g) + 6 CO2(g) + 3 H2O(g) + 3H2(g) + 1.5 O2(g) ==> 6 CO2(g) + C6H6(l) + 7.5 O2(g) + 3H2O(g)

Cancel like items on each side of equation to get ...

6C(s) + 3H2(g) ==> C6H6(l) ... TARGET EQUATION

Now, add up the ∆H values to obtain...

-2361 + 3136 + (-322)

∆Hrxn = 453 kJ