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The enthalpy change for the reaction, ...

The enthalpy change for the reaction,
`N_2(g)+3H_2(g)rightarrow2NH_3(g)` Is -92.38 kJ at 298 K.
What is `DeltaU` at 298 K?

A

`-92.38 kJ`

B

`-87.42 kJ`

C

`-97.34 kJ`

D

`-89 . 9 kJ`

Text Solution

Verified by Experts

`Delta n _(g) = 2 - 4 =- 2, Delta H = Delta U + Delta n _(g) RT`
`Delta U . = Delta H - Delta n _(g) RT = - 92. 38 xx 1000 _ (-2) xx 8.314 xx 298 =- 87424 J =- 87 . 424 kJ`
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