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Question

Benzene bums in O2 according to the equation:
C6H6(l)+(15/2)O2(g)3H2O(g)+6CO2(g)
If standard heat enthalpies of C6H6(l),H2O(g)andCO2(g) are 11.7, -68.3 and -94.0 kcal respectively. The amount of heat liberated by burning 0.5 g benzene in kcal is :

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Solution

Standard heat enthalpy of a compound (H)= Standard heat of formation of that compound (ΔfH)

6C+3H2C6H6(l);ΔH=11.7kcal ...( i)

H2+12O2H2O(g);ΔH=68.3kcal ...(ii)

C+O2CO2(g);ΔH=94.0kcal ...(iii)

By inspection method 3×(ii)+6×(iii)(i) gives:

C6H6=(15/2)O26CO2+3H2O;

ΔH=780.6kcal

78gC6H6producesheat=780.6kcal

0.5gC6H6producesheat=(780.6×0.5)/78
=5kcal

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