Respuesta :
Answer: The enthalpy change of the combustion of ethane is -3122 kJ.
Explanation:
Combustion reaction is defined as the reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water molecules.
Enthalpy change 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]
For the given chemical reaction:
[tex]2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(4\times \Delta H^o_f_{(CO_2)})+(6\times \Delta H^o_f_{(H_2O)})]-[(2\times \Delta H^o_f_{(C_2H_6)})+(7\times \Delta H^o_f_{(O_2)})][/tex]
We are given:
[tex]\Delta H^o_f_{(H_2O)}=-286kJ/mol\\\Delta H^o_f_{(O_2)}=0kJ/mol\\\Delta H^o_f_{(C_2H_6)}=-85kJ/mol\\\Delta H^o_f_{(CO_2)}=-394kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(4\times (-394))+(6\times (-286))]-[(2\times (-85))+(7\times 0)]\\\\\Delta H^o_{rxn}=-3122kJ[/tex]
Hence, the enthalpy change of the combustion of ethane is -3122 kJ.
