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Question
The enthalpy changes (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
The enthalpy changes (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
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Solution
The correct option is B −87.42 kJ
We k now that,
ΔE=ΔH−ΔnRT
=−92.38−(−2×8.31×2981000)
=−92.38+4.95=−87.42 kJ
Option (B) is correct.
We k now that,
ΔE=ΔH−ΔnRT
=−92.38+4.95=−87.42 kJ
Option (B) is correct.
=−92.38−(−2×8.31×2981000)
=−92.38+4.95=−87.42 kJ
Option (B) is correct.
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