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

Consider the reaction at `300 K, C _(6) H _(6) (l) + (15)/(2) O _(2) (g) to 6CO_(2)(g) + 3 H _(2) O (l) , Delta H = -327 1 kJ`
What is `Delta U` for the combustion of 1.5 mole of benzene at `27^(@)C`?

A

`-3667.25 kJ`

B

` - 4900.88 kJ`

C

`- 4806.5 kJ`

D

`-3274.75 kJ`

Text Solution

Verified by Experts

For mole of combustion of benzene : `Delta n _(g) = - 1. 5`
`DeltaH = Delta U + Delta n _(g) RT implies - 3271= Delta U - ( 1. 5 xx 8. 314 xx 300)/(1000) implies Delta U =- 3267 . 25 kJ`
For `1.5 ` mole of combustion of benzene, `Delta U =- 3267. 25 xx 1. 5 =-4900 . 88 kJ`
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