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06/10/2020 1:01 pm
Topic starter
The enthalpy change (ΔH) for the reaction,
N2(g) + 3H2(g) → 2NH3(g) is -92.38 kJ at 298 K. The internal energy change ΔU at 298 K is
(a) -92.38kJ
(b) -87.42 k.J
(c) -97.34kJ
(d) -89.9kJ
1 Answer
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06/10/2020 1:03 pm
(b) -87.42 k.J
Explanation:
ΔE = ΔH - ΔnRT
= -92.38 - (-2 x 8.31 x 298/1000)
= - 92.38 + 4.95 = - 87.43 kJ