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Benzene burns according to the following...

Benzene burns according to the following equations at 300K (R=8.314 J `"mole"^(-1)K^(-1))`
`2C_(6)H_(6)(l) + 15O_(2)(g) rarr 12 CO_(2)(g) + 6H_(2)O(l) " " DeltaH^(@) =- 6542 KJ`
What is the `DeltaE^(@)` for the combustion of 1.5 mol of benzene

A

`-3271` kJ

B

`-9813` 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, `DeltaE^(@) = 1.5/2 (-6542 + 3RT) = -4900.9` kJ
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