Question

The combustion of pentane, C5H12, occurs via the reaction
C5H12(g)+8O2(g)→5CO2(g)+6H2O(g)
with heat of formation values given by the following table:
Substance ΔH∘f
(kJ/mol)
C5H12 (g) -119.9
CO2(g) −393.5
H2O(g) −241.8
Calculate the enthalpy for the combustion of 1 mole of pentane.

Answer 1

The enthalpy of the reaction in an aqueous solution can be determined by taking the difference between the summation of enthalpies of the products multiplied to their respective stoichiometric coefficient and the summation of enthalpies of the reactants multiplied to their respective  stoichiometric coefficient. In this case, the equation is -241(6) + -393.5 (5) -[-119.9] equal to 641.4 kJ 

Answer 2

Answer : The enthalpy for the combustion of 1 mole of pentane is -3024.8 kJ

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$C_5H_{12}(g)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_5H_{12})}\times \Delta H^o_f_{(C_5H_{12})})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(C_5H_{12}(g))}=-119.9kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(5\times -393.5)+(6\times -241.8)]-[(1\times -393.5)+(8\times 0)=-3024.8kJ$

Therefore, the enthalpy for the combustion of 1 mole of pentane is -3024.8 kJ