Hess's Law Paper

Question

Calculate the standard enthalpy of formation for cyclohexanol (C₆H₁₁OH) provided the following

C₆H₁₁OH (l) ∆Hc° = -3727 kJ mol^-1

C (s) ∆H_c° = -393.5 kJ mol^-1

H₂ (g) ∆H_c° = -285.5 kJ mol^-1

Answer 1

6C + 6H₂ + 1/2 O₂ ==> C₆H₁₁OH TARGET EQUATION

Given:

(1). C₆H₁₁OH (l) + 17/2 O₂ ==> 6CO₂ + 6H₂O ... ∆Hc° = -3727 kJ mol^-1

(2) C (s) + O₂ ==> CO₂ ... ∆H_c° = -393.5 kJ mol^-1

(3) 2 H₂ (g) + O₂ ==> 2 H₂O ... ∆H_c° = -285.5 kJ mol^-1

Procedure:

eq 2 x 6: 6C + 6 O₂ ==> 6 CO₂ ... ∆H = -22,363 kJ

eq 3 x 3: 6 H₂ + 3 O₂ ==> 6 H₂O ... ∆H = -857 kJ

reverse eq 1: ==> 6 CO₂ + 6H₂O ==> C₆H₁₁OH + 17/2 O₂ ... ∆H = +3727 kJ

----------------------------------------------------------- add them together and combine/cancel like terms to get....

6C + 6 O₂ + 6 H₂ + 3 O₂ + 6CO₂ + 6H₂O ==> 6 CO₂ + 6 H₂O + C₆H₁₁OH + 17/2 O₂

6C + 6O₂ + 6H2 + 1/2 O₂ ==> C₆H₁₁OH TARGET EQUATION

∆H_f = 3727 - 857 - 22,363 = 19,493 kJ