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Consider the reacting at 300K C(6)H(6)...

Consider the reacting at `300K`
`C_(6)H_(6)(l)+(15)/(2)O_(2)(l)rarr 6CO_(2)(g)+3H_(2)O(l),DeltaH=-3271 kJ`
What is `DeltaU` for the combustion of `1.5` mole of benzene at `27^(@)C` ?

A

`-3267.25 kJ`

B

`-4900.88 kJ`

C

`-4906.5 kJ`

D

`-3274.75 kJ`

Text Solution

Verified by Experts

The correct Answer is:
B
For combustion of 1 mole of benzene
`" "Deltan_(g)=-1.5`
`" "DeltaH=DeltaU+Deltan_(g)RT`
`implies -3271 = DeltaU-(1.5 xx 8.314 xx 300)/(1000)`
`implies" "DeltaU=-3267.25 " kJ"`
For 1.5 mole of combustion of benzene
`DeltaU=-3267.25 xx1.5`
`=-4900.88 " kJ"`
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