For the reaction \(\mathrm{N_2(g)+3H_2(g)\rightarrow2NH_3(g)}\) at 298 K, \(\Delta H-\Delta U\) is approximately
A. \(-4.96\) kJ
B. \(-2.48\) kJ
C. 0
D. \(+4.96\) kJ
1 answer
Best answer
Answer: A. \(-4.96\) kJ
\(\Delta n_g=2-4=-2\); \(\Delta H-\Delta U=\Delta n_gRT=-2\times8.314\times298\approx-4955\) J.
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· 23 Oct 2026
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