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What is Delta n for combustion of 1 mole...

What is `Delta n` for combustion of 1 mole of benzene, when both the reactants and the products are gas at 298 K?

A

0

B

`3//2`

C

`-3//2`

D

`1//2`

Text Solution

Verified by Experts

The correct Answer is:
D
`C_(6)H_(6(g))+(15)/(2)O_(2(g))rarr6CO_(2(g))+3H_(2)O_((g))`
`Deltan=6+3-1-(15)/(2)=+(1)/(2)`.
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