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Question 84 of 115

Q.Calculate ΔH∘\Delta H^\circ for the reaction between ethene and water to form ethyl alcohol from the following data: ΔcH∘ C2H5OH(l)=−1368 kJ\Delta_cH^\circ\ C_2H_5OH_{(l)} = -1368\ kJ; ΔcH∘ C2H4(g)=−1410 kJ\Delta_cH^\circ\ C_2H_{4(g)} = -1410\ kJ. Does the calculated ΔH∘\Delta H^\circ represent the enthalpy of formation of liquid ethanol?

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2017Subjective· 3mImportance★★★★★
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Apply Hess's law: since both C2H4C_2H_4 and C2H5OHC_2H_5OH burn to the same products (CO2+H2OCO_2+H_2O), their combustion enthalpies can be combined directly to give the hydration reaction's ΔH\Delta H.

The target reaction is the hydration of ethene: C2H4(g)+H2O(l)→C2H5OH(l)C_2H_4(g) + H_2O(l) \rightarrow C_2H_5OH(l).

By Hess's law, since both C2H4C_2H_4 and C2H5OHC_2H_5OH combust ultimately to CO2CO_2 and H2OH_2O (with water itself common to both sides), the enthalpy of this reaction equals the combustion enthalpy of the reactant minus the combustion enthalpy of the product:

ΔHrxn∘=ΔcH∘(C2H4)−ΔcH∘(C2H5OH)\Delta H^\circ_{rxn} = \Delta_cH^\circ(C_2H_4) - \Delta_cH^\circ(C_2H_5OH)

ΔHrxn∘=(−1410)−(−1368)=−1410+1368=−42 kJ\Delta H^\circ_{rxn} = (-1410) - (-1368) = -1410 + 1368 = -42\ kJ

So the reaction of ethene with water to give ethanol is exothermic, releasing 42 kJ per mole.

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