What will be the standard heat of reaction c2h6 + 7/2o2?
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Although the reactions are not written out the ΔHc° is given and this is the standard enthalpy of combustion or the change in enthalpy when the compound undergoes combustion. Since combustion is a reaction with oxygen you can write out each of the reaction equations with their corresponding ΔHc° values.
C2H2(g) + 5/2O2(g) → 2CO2(g) + H20(l) ΔH° =-1300kJ
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H20(l) ΔH°=-1560kJ
H2(g) + 1/2O2(g) →H2O(l) ΔH°= -286kJ
The equations are balanced so that only 1 mole of each of the compounds is combusted because the standard enthalpy of combustion is per mole of the given compound (this is why there are fractional stoichiometric coefficients).
Now that you have the three equations you can use Hess's Law to determine the enthalpy change for the hydrogenation of ethyne to ethane. Because ethane has to be on the products side you flip the second equation meaning that the ΔH°will now be +1560kJ. In order to have two moles of hydrogen gas on the reactants side, the third equation must be multiplied by 2 meaning that the ΔH° will now be -572kJ. Now if you were to add up the three equations you would get the equation for the hydrogenation of ethyne to ethane so the ΔH° values can be added together ΔH°=-312kJ/mol
C2H2(g) + 5/2O2(g) → 2CO2(g) + H20(l) ΔH° =-1300kJ
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H20(l) ΔH°=-1560kJ
H2(g) + 1/2O2(g) →H2O(l) ΔH°= -286kJ
The equations are balanced so that only 1 mole of each of the compounds is combusted because the standard enthalpy of combustion is per mole of the given compound (this is why there are fractional stoichiometric coefficients).
Now that you have the three equations you can use Hess's Law to determine the enthalpy change for the hydrogenation of ethyne to ethane. Because ethane has to be on the products side you flip the second equation meaning that the ΔH°will now be +1560kJ. In order to have two moles of hydrogen gas on the reactants side, the third equation must be multiplied by 2 meaning that the ΔH° will now be -572kJ. Now if you were to add up the three equations you would get the equation for the hydrogenation of ethyne to ethane so the ΔH° values can be added together ΔH°=-312kJ/mol
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