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Question

The Heat Of Combustion Of Benzene At 270°C Found By Bomb Calorimeter I.E. For The Reaction C6H6(l)+712O2(g)6CO2(g)+3H2O(l) Is 780K.Calmol-1. The Heat Evolved On Burning 39g Of Benzene In An Open Vessel Will Be


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Solution

Heat of combustion:

  • The term "heat of combustion" refers to the amount of heat needed to totally burn down a chemical species when oxygen is present.

Step 1: Finding change in internal energy

  • C6H6 has a molar mass of = 6×12+6×1=78grams.
  • The energy content of 78gramsof benzene is 780K.Calmol-1
  • Energy for 39gof benzene is equal to 780×3978=390.
  • Therefore, U=390K.Calmol-1

Step 2: Finding change in number of moles of gaseous species

  • T=27+273KT=300K
  • The indicated response is
  • C6H6(l)+712O2(g)6CO2(g)+3H2O(l)
  • Change in mole number (n)=6-7.5=-1.5

Step 3: Finding change in heat

  • R=2×10-3Kcal
  • Having said that,
  • ΔH=ΔU+ΔnRTΔH=390+(1.5×2×10-3×300)ΔH=3900.9ΔH=389.1Kcal

39g of benzene burned in an open vessel will produce 389.1Kcal of heat.


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