Quantitative relationships involving Hess law and ΔH

Question

Calculate ΔH _rxn for the reaction:

5 C(s) +6H₂(g) = C₅H₁₂(l)

given these reaction and their ΔH

C₅H₁₂(l)+ 8O₂(g) = 5CO₂(g) + 6H₂O(g)

ΔH = - 327.5 kJ

C(s) + O₂(g) = CO₂(g). ΔH= -393.5kj

2H₂(g) + O2 (g) = 2H₂O(g). ΔH = -483.5 kJ

Answer 1

Hess' Law allows us to use known values for equations to calculate certain unknown values for another equation.

5 C(s) + 6H₂(g) ==> C₅H₁₂(l) ... TARGET EQUATION

Given:

(1) C₅H₁₂(l) + 8O₂(g) ==> 5CO₂(g) + 6H₂O(g) ... ∆H = -327.5 kJ

(2) C(s) + O₂(g) ==> CO₂(g) ... ΔH= -393.5kJ

(3) 2H₂(g) + O2 (g) ==> 2H₂O(g) ... ΔH = -483.5 kJ

Procedure:

reverse (1): 5CO2(g) + 6H2O(g) ==> C5H12(l) + 8O2(g) ... ∆H = +327.5 kJ

copy (2)x5: 5C(s) + 5O2(g) ==> 5CO2(g) ... ∆H = -1967.5 kJ

copy(3)x3: 6H₂(g) + 3O2 (g) ==> 6H₂O(g) ... ∆H = -1450.5 kJ

Add up the 3 equations and combine and cancel like items to obtain...

5CO2(g) + 6H2O(g) + 5C(s) + 5O2(g)+ 6H₂(g) + 3O2 (g) => C5H12(l) + 8O2(g) + 5CO2(g) + 6H₂O(g)

5C(s) + 6H2(g) ==> C5H12(l) ... TARGET EQUATION

∆H = 327.5 kJ + -1967.5 kJ + -1450.5 kJ = -3090.5 kJ = -3091 kJ