For the complete combustion of ethanol, C2HsOH(l)+3O2(g)→+2CO2+2H2O(l) the amount of heat produced as measured inthe bomb calorimeter is 1364.47 kJ mol−1 at 25∘C. Assuming ideality, the enthalpy of combustion ΔcH for the reaction will be [ R = 8.314 JK−1mol−1 ]
–1366.95kJmol−1
C2H5OH(l)+3O2(g)→+2CO2+3H2O(l)
Amount of heat produced in the bomb calorimeter = ΔU =- 1364.47 kJ mol−1
Enthalpy of a combustion reaction is: ΔH=ΔU+ΔngRT
Where , Δ U = internal energy
Δng = moles of gas (products -reacants), R = Gas constant=, T = Temperature in K
As per the equation,
Δng=2−3=−1
T=25∘C=25+273K
⇒T=298K
Thus, ΔH=−1364.47+[(−1)×8.314×2981000]
ΔH=−1364.47−2.477=−1366.947 kJ mol−1
Hence, they enthalpy of combustion , ΔcH for the given reaction will be - 1366.95 kJ mol−1