## Respuesta :

**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 [tex]\Delta H^o[/tex]

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

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

**The equilibrium reaction follows:**

[tex]C_5H_{12}(g)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(g)[/tex]

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

[tex]\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)})][/tex]

**We are given:**

[tex]\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[/tex]

Putting values in above equation, we get:

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

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