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A reaction that occurs in the internal combustion engine is n2(g) + o2(g) ⇌ 2 no(g) (a) calculate δh o and δs o for the reaction at 298 k. δh o = kj δs o = j/k (b) assuming that these values are relatively independent of temperature, calculate δg o at 55°c, 2570°c, and 3610°c. 55°c: kj 2570°c: kj 3610°c: kj

Respuesta :

Dejanras
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.
TemperatureBig

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.

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