2NOBr(g) ⇌ 2NO(g) + Br2(g) 25℃ 34% 0.25 atm Kp Kc
Consider the following equilibrium.
2NOBr(g) ⇌ 2NO(g) + Br2(g)
If nitrosyl bromide (NOBr) is 34% dissociated at 25°C
and the total pressure is 0.25 atm,
calculate Kp and Kc for the dissociation at this temperature.
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2NOBr(g) ⇌ 2NO(g) + Br2(g) 또는
NOBr(g) ⇌ NO(g) + 1/2 Br2(g)
NOBr의 초기 압력 = p 라 가정하면,
평형에서,
> P_NOBr = 0.66p
> P_NO = 0.34p
> P_Br2 = (0.34/2)p
전체 압력 = 0.25 atm 이므로,
0.66p + 0.34p + (0.34/2)p = 0.25
1.17p = 0.25
p = 0.25 / 1.17 = 0.214 atm
평형에서,
각 기체의 분압은
> P_NOBr = 0.66 × 0.214 = 0.141 atm
> P_NO = 0.34 × 0.214 = 0.073 atm
> P_Br2 = (0.34/2) × 0.214 = 0.036 atm
2NOBr(g) ⇌ 2NO(g) + Br2(g)
Kp = (P_NO)^2 (P_Br2) / (P_NOBr)^2
= (0.073)^2 (0.036) / (0.141)^2
= 0.00965
Kp = Kc (RT)^Δn
( 참고 https://ywpop.tistory.com/6523 )
Kc = Kp / (RT)^Δn
= 0.00965 / [0.08206 × (273.15+25)]^(3–2)
= 0.000394
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