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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

0.5

C

1.5

D

-1.5

Text Solution

Verified by Experts

The correct Answer is:
D
Combustion of benzene `C_(6)H_(6(l))` is
`C_(6)H_(6(l))+(15)/(2)O_(2(g))rarr 6CO_(2(g))+3H_(2)O_((l))`
`Delta n=(6+0)-(0+(15)/(2))=-(3)/(2)`
`Delta n =-1.5`
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