375℃ N2 1 mol H2 3 mol NH3 0.21 Kp 4.31×10^(-4)

375℃ N2 1 mol H2 3 mol NH3 0.21 Kp 4.31×10^(-4)

One mole of N2 and three moles of H2

are placed in a flask at 375℃.

Calculate the total pressure of the system at equilibrium

if the mole fraction of NH3 is 0.21.

The Kp for the reaction is 4.31×10^(-4).

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ICE 도표를 작성하면,

............................. N2 . + . 3H2 .. ⇌ .. 2NH3

초기(mol): 1 ................. 3 .................... 0

변화(mol): –x ......... –3x ............ +2x

평형(mol): 1–x ......... 3–3x ........ 2x

전체 몰수 = (1–x) + (3–3x) + 2x = 4 – 2x

NH3의 몰분율 = 2x / (4 – 2x) = 0.21

2x = 4(0.21) – 2x(0.21)

2x + 2x(0.21) = 4(0.21)

(2 + 0.42)x = 0.84

x = 0.84 / (2 + 0.42)

= 0.347

전체 몰수 = 4 – 2(0.347) = 3.306 mol

> N2의 몰수 = 1 – 0.347 = 0.653 mol

> H2의 몰수 = 3 – 3(0.347) = 1.959 mol

> NH3의 몰수 = 2(0.347) = 0.694 mol

> N2의 몰분율 = 0.653 / 3.306 = 0.1975

> H2의 몰분율 = 1.959 / 3.306 = 0.5926

> NH3의 몰분율 = 0.21

from 돌턴의 분압법칙,

( 참고 https://ywpop.tistory.com/48 )

> P_N2 = 0.1975×P_tot

> P_H2 = 0.5926×P_tot

> P_NH3 = 0.21×P_tot

N2 + 3H2 ⇌ 2NH3

Kp = (P_NH3)^2 / [(P_N2) (P_H2)^3]

Kp = (0.21×P_tot)^2 / [(0.1975×P_tot) (0.5926×P_tot)^3]

Kp = (0.21)^2 (P_tot)^2 / [(0.1975) (P_tot) (0.5926)^3 (P_tot)^3]

Kp = (0.21)^2 (P_tot)^2 / [(0.1975) (0.5926)^3 (P_tot)^4]

4.31×10^(-4) = (0.21)^2 / [(0.1975) (0.5926)^3 (P_tot)^2]

(P_tot)^2 = (0.21)^2 / [(0.1975) (0.5926)^3 (4.31×10^(-4))]

P_tot = (2489.48)^(1/2)

= 49.8947 atm

답: 50 atm

[키워드] N2 + 3H2 ⇌ 2NH3 Kp 기준