2NOBr(g) ⇌ 2NO(g) + Br2(g) 25℃ 34% 0.25 atm Kp Kc

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