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Question
Calculate ΔH° for the reaction between ethene and water to form ethyl alcohol from the
following data:
ΔcH° C2H5OH(l) = -1368 kJ
ΔcH° C2H4(g) = -1410 kJ
Does the calculated ΔH° represent the enthalpy of formation of liquid ethanol?
Solution
Given:
The standard enthalpy of combustion of C2H5OH(l) i.e. Δc H° C2H5OH(l) = –1368 kJ
The standard enthalpy of combustion of ethene i.e. Δc H° C2H4(g) = –1410 kJ
To find: ΔH° for the enthalpy of formation of liquid C2H5OH
Calculation: Given equations are,
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ; ΔcH° = –1368 kJ ...(1)
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) ; ΔcH° = –1410 kJ ...(2)
The required equation is,
C2H4(g) + H2O(l) → C2H5OH(l),
To get required equation, reverse equation (1) and add to equation (2).
The calculated ΔH° = – 42 kJ is not the enthalpy of formation of liquid ethanol because the reaction does not involve the formation of liquid ethanol from its constituent elements.
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