800. K N2(g) + 3F2(g) → 2NF3(g) 0.021 0.063 0.48 atm ΔG°

800. K N2(g) + 3F2(g) → 2NF3(g) 0.021 0.063 0.48 atm ΔG°

Consider the following reaction at 800. K:

N2(g) + 3F2(g) → 2NF3(g)

An equilibrium mixture contains the following partial pressures:

P_N2 = 0.021 atm, P_F2 = 0.063 atm, and P_NF3 = 0.48 atm.

Calculate ΔG° for the reaction at 800. K.

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N2(g) + 3F2(g) ⇌ 2NF3(g)

Kp = (P_NF3)^2 / ((P_N2) (P_F2)^3)

= (0.48)^2 / ((0.021) (0.063)^3)

= 43877.5

ΔG° = –RTlnK

( 참고 https://ywpop.tistory.com/6519 )

= –(8.314 J/mol•K) (800. K) ln(43877.5)

= –71095.7 J/mol

= –71.1 kJ/mol

답: –71.1 kJ/mol

[키워드] 자유에너지와 평형상수 기준, 평형상수와 자유에너지 기준, G와 K 기준, K와 G 기준