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Consider the reaction at 300 K C(6) ...

Consider the reaction at 300 K
`C_(6) H_(6) (l)+(15)/(2)O_(2)(g)rarr6CO_(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

`-3267.25kJ`

B

`-4900.88kJ`

C

`-4906.5kJ`

D

`-3274.75kJ`

Text Solution

Verified by Experts

The correct Answer is:
2
For 1 mole of combustion of benzene
`Deltan=-1.5`
`DeltaH^(@)=DeltaU+Deltan_(9)RT`
`rArr-3271=DeltaU-(1.5xx8.314xx300)/(1000)`
`rArr" "DeltaU=-3267.25xx1.5=-4900.88kJ`
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