Q.

Consider the reaction at 300K 
C6H6(l)+152O2(g)6CO2(g)+3H2O(l):ΔH=3271KJ what is U for the combustion of 1.5 mole of benzene at 27°C:

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a

-3274.75 KJ

b

-4900.88 KJ

c

-3267.25 KJ

d

-4906.5 KJ

answer is B.

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Detailed Solution

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ΔH=ΔU+ΔnRT Δn=ngPngR =6152=32 3271=ΔU+32×8.314×300×1033271+3.741=ΔUΔU=3267.259KJ/ mole  For 1.5 mole =1.5×(3267.259)=4900.88

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