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A reaction that occurs in the internal combustion engine is
N2(g) + O2(g) ⇌ 2 NO(g)
(a) Calculate ΔHo and ΔSo for the reaction at 298 K.
(b) Assuming that these values are relatively independent of temperature, calculate ΔGo at 93°C, 2590°C, and 3450°C.


Sagot :

Answer:

Explanation:1) ΔrH = 2mol·ΔfH(NO) - (ΔfH(O₂) + ΔfH(N₂)).

ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).

ΔrH = 180.6 kJ.

2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).

ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).

ΔS = 24.8 J/K.

3) ΔG = ΔH - TΔS.

55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.

2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.

3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.