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`?

Options

A. `– 3667.25 kJ`
B. `-4900.88 kJ`
C. `- 4906.5 kJ`
D. `-3274.75 kJ`

Answer

B

Explanation

For 1 mole of combustion of benzene, `Deltan _(g) =- 1. 5`
`Delta H = DeltaU + Delta n _(g) RT implies - 32 71 = Delta U - (1. 5 xx 8. 314 xx 300)/(1000) implies Delta U =- 32 67. 25 kJ `
For 1. 5 mole of combustion of benzene, `Delta U = - 3267. 23 xx 1. 5 =- 4900. 88 kJ`