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.

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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 

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