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The enthalpy change (DeltaH) for the rea...

The enthalpy change `(DeltaH)` for the reaction
`N_2(g) + 3H_2 (g) to 2NH_3(g)`
is -92.38 kJ at 298 K. What is `DeltaU` at 298 K ? `(R = 8.314 j K^(-1) mol^(-1))`

A

`-92.38 kJ`

B

`-87.42kJ`

C

`-97.34kJ`

D

`-89.9kJ`

Text Solution

Verified by Experts

The correct Answer is:
b
`Delta=DeltaU+Delta nRT`
[where `Deltan`= no. of molecules of products - no. of molecules of reactants]
`or . -92.38xx1000= DeltaU-2xx8.314xx298`
or , `Delta U= -87424 J= -87.421 KJ`
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