The correct option is B 3.74 kJ
Combustion of benzene is represented by following reaction;
C6H6(l)+152O2(g)→6CO2(g)+3H2O(l)
The difference in gaseous moles of product and reactant is :
△ng=6−152=−3/2
And we have relation,
△H=△U+△ngRT
so,
△H−△U=△ngRT
△H−△U=−32×8.314×300
=3741.3 J=−3.741 kJ
Hence the difference between △H and △U for the combustion of liquid benzene is 3.74 kJ.