Calculate Δ H rxn for the reaction: 5 C ( s ) + 6 H 2 ( g ) → C 5 H 12 ( l ) Use the following reactions and given ΔH’s: C 5 H 12 ( l ) + 8 O 2 ( g ) → 5 CO 2 ( g ) + 6 H 2 O ( g ) Δ H = − 3244.8 kJ C ( s ) + O 2 ( g ) → CO 2 ( g ) Δ H = − 393.5 kJ 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g ) Δ H = − 483.5 kJ

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

[tex]5C(s)+6H_2(g)\rightarrow C_5H_{12}(l)[/tex] [tex]\Delta H=?[/tex]

The intermediate balanced chemical reaction will be,

(1) [tex]C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(g)[/tex] [tex]\Delta H_1=-3244.8kJ[/tex]

(2) [tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex] [tex]\Delta H_2=-393.5[/tex]

(3) [tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex] [tex]\Delta H_3=-483.5kJ[/tex]

Reversing (1) , Multiply (2) by 5 , (3) by 3 and add

[tex]\Delta H=\Delta H_1+5\times \Delta H_2+3\times \Delta H_3[/tex]

[tex]\Delta H=(+3244.8)+(5\times -393.5)+(3\times -483.5)[/tex]

[tex]\Delta H=-173.2kJ[/tex]

Therefore, the enthalpy of reaction is, -173.2 kJ