For N₂(g) + 3H₂(g) → 2NH₃(g) at 298 K, ΔU = −87.4 kJ. What is ΔH (R = 8.314 J K⁻¹ mol⁻¹)?
Correct answer D. −92.4 kJ
Explanation
ΔH = ΔU + ΔnRT with Δn = 2 − 4 = −2. So ΔH = −87.4 + (−2 × 8.314 × 298 ÷ 1000) = −87.4 − 4.96 = −92.4 kJ.
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