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At 27oC for the combustion of two mole of benzene, H-E is equal to

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By Expert Faculty of Sri Chaitanya
a
-3.74 kJ
b
-7.48 kJ
c
+1.25 kJ
d
-1.25 kJ
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detailed solution

Correct option is B

2C6H6(l)+15O2(g)→ 12CO2(g)+6H2O(l); ∆ng=12-15=-3; ∆H-∆E=∆nRT;                  =-3(8.314×10-3KJmol-1K-1)(300 K) ;                  =-7.48kJ ;


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