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Benzene burns according to the following...

Benzene burns according to the following equation at `300K (R=8.314J mol e^(-1)K^(-1))`
`2C_(6)H_(6)(l)+15O_(2)(g)rarr12CO_(2)(g)+6H_(2)O(l) " "DeltaH^(@)=-6542kJ //mol`
What is the `DeltaE^(@)` for the combustion of `1.5 mol` of benzene

A

`-3271 KJ`

B

`-9812 KJ`

C

`-4906.5 `KJ

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
D
From given reaction `Deltan_(g) =12 -15=-3`
so `DeltaE^(@) =DeltaH^(@)-Deltan_(g)RT =-6542 + 3RT`
for 1.5 mole , `Delta E^(@) =(1.5)/(2) {-6542 + 3RT} = 4900 KJ`
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