Question

The enthalpy change (◇H) for the reaction N2 (g)+3H2 (g)+ 2NH3 (g) in -92.38 kJ at 298K.

what is ◇U at 298K?

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Answer 1

◇H =◇U + ◇nRT

-92.38=◇U-2×298×0.0083

=◇U-4.95

◇U=-87.43 kJ mol^-1

Answer 2

Explanation:

plzz refer to the attached file..

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