Δng=Gaseous moles products−Gaseous moles reactants
=2−3=-1 mole
ΔH=ΔE+ΔngRT
=−1364.47−1×8.314×10−3×298 [The value of R must be converted into kJ, so it is multiplied by 10−3]
=−1364.47−2.49
=−1366.96 kJ m−1
JEE Main 2014 — Chemistry Physical Chemistry
For complete combustion of ethanol,
C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l),
the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol−1 at 25℃. Assuming ideality the Enthalpy of combustion, ΔcH, for the reaction will be: (R=8.314 kJ mol−1)
Held on 6 Apr 2014 · Verified 6 Jul 2026.
−1366.95 kJ mol−1
−1361.95 kJ mol−1
−1460.50 kJ mol−1
−1350.50 kJ mol−1
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