Q.

The enthalpy change H for the reaction. N2(g)+3H2(g)2NH3(g) is-92.38kj at 298K. The internal energy change U at 298K is

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a

-89.9kJ

b

-87.42kJ

c

97.34kJ 
 

d

-92.38kJ

answer is B.

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Detailed Solution

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H=U+nRT-92.38×103=U+(-2)8.314×298

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The enthalpy change ∆H for the reaction. N2(g)+3H2(g)→2NH3(g) is-92.38kj at 298K. The internal energy change ∆U at 298K is