更新1:
N2(g) + 2O2(g) → 2NO2(g) ΔH⁰298 = 66.4 kJ 2NO(g) + O2(g) → 2NO2(g) ΔH⁰298 = −116.2 kJ
Calculate the standard enthalpy of formation of NO(g) (in kJ/mol) from the following data.?
回答 (1)
2017-11-07 10:04 am
Rewrite the given thermochemical equations as :
(1/2)N₂(g) + O₂(g) → NO₂(g) …... ΔH°₂₉₈ = (1/2)(+66.4 kJ) = +33.2 kJ
NO₂(g) → NO + (1/2)O₂(g) …… ΔH°₂₉₈ = -(1/2)(-116.2 kJ) = +58.1 kJ
Add the above two thermochemical equations, and cancel NO₂(g) and (1/2)O₂ on the both sides.
(1/2)N₂(g) + (1/2)O₂(g) → NO(g) …… ΔH°₂₉₈ = (+33.2) + (58.1) kJ = +91.3 kJ
Hence, standard enthalpy of formation of NO(g) = +91.3 kJ
(1/2)N₂(g) + O₂(g) → NO₂(g) …... ΔH°₂₉₈ = (1/2)(+66.4 kJ) = +33.2 kJ
NO₂(g) → NO + (1/2)O₂(g) …… ΔH°₂₉₈ = -(1/2)(-116.2 kJ) = +58.1 kJ
Add the above two thermochemical equations, and cancel NO₂(g) and (1/2)O₂ on the both sides.
(1/2)N₂(g) + (1/2)O₂(g) → NO(g) …… ΔH°₂₉₈ = (+33.2) + (58.1) kJ = +91.3 kJ
Hence, standard enthalpy of formation of NO(g) = +91.3 kJ
收錄日期: 2021-04-18 17:59:53
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