The reaction between ethene and water to form ethyl alcohol is given below.
CH2=CH2(g)+H2O(l)→C2H5OH(l)
The above equation can be obtained from the following two equations.
C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)ΔcHoC2H5OH(l)=−1368kJ......(1)
CH2=CH2(g)+3O2(g)→2CO2(g)+2H2O(l)ΔcHoC2H4(g)=−1410kJ......(2)
(2) -(1) gives
CH2=CH2(g)+H2O(l)→C2H5OH(l)ΔcHoC2H4(g)=−1410kJ−(−1368kJ)=−42kJ
→ΔHo=−1368kJ
The calculated ΔHo does not represent the enthalpy of formation of liquid ethanol.
The
enthalpy of formation of liquid ethanol will involve a equation showing
formation of liquid ethanol from elements in standard state. The equation formation of liquid ethanol is given below.
2C(graphite)+3H2(g)+12O2(g)→C2H5OH(l)