0.140 mol NO, 0.060 mol H2, 0.260 mol H2O 2.0 L 330 K

0.140 mol NO, 0.060 mol H2, 0.260 mol H2O 2.0 L 330 K

 

 

NO 0.140 mol, H2 0.060 mol, H2O 0.260 mol이 330 K의 2.0 L 용기에 담겨 있다. 다음 평형이 이루어졌다고 한다.

2NO(g) + 2H2(g) ⇌ N2(g) + 2H2O(g)

평형에서 [H2] = 0.010 M이다.

(a) NO, N2, H2O의 평형 농도를 계산하시오.

(b) Kc를 계산하라.

 

 

A mixture of 0.140 mol of NO, 0.060 mol of H2, and 0.260 mol of H2O is placed in a 2.0 L vessel at 330 K. Assume that the following equilibrium is established:

2NO(g) + 2H2(g) ⇌ N2(g) + 2H2O(g)

At equilibrium [H2] = 0.010 M.

(a) Calculate the equilibrium concentrations of NO, N2, and H2O.

(b) Calculate Kc.

 

 

 

각 물질의 초기 농도를 계산하면,

[NO] = 0.140 mol / 2.0 L = 0.070 M

 

[H2] = 0.060 mol / 2.0 L = 0.030 M

 

[H2O] = 0.260 mol / 2.0 L = 0.130 M

 

 

 

ICE 도표를 작성하면,

......... 2NO(g) + 2H2(g) ⇌ N2(g) + 2H2O(g)

초기(M): 0.070 ... 0.030 ... 0 ... 0.130

변화(M): –2x ... –2x ... +x ... +2x

평형(M): (0.070–2x) ... (0.030–2x) ... x ... (0.130+2x)

 

 

 

평형에서 [H2] = 0.010 M 이므로,

[H2] = (0.030–2x) = 0.010

 

2x = 0.030 – 0.010 = 0.020

 

x = 0.020 / 2 = 0.010 M

 

 

 

평형에서 각 물질의 농도를 계산하면,

[NO] = 0.070–2x

= 0.070 – 2(0.010) = 0.050 M

 

[H2] = 0.010 M

 

[N2] = 0.010 M

 

[H2O] = 0.130+2x

= 0.130 + 2(0.010) = 0.150 M

 

 

 

2NO(g) + 2H2(g) ⇌ N2(g) + 2H2O(g)

 

Kc = [N2] [H2O]^2 / [NO]^2 [H2]^2

( 참고 https://ywpop.tistory.com/7136 )

 

= (0.010 × 0.150^2) / (0.050^2 × 0.010^2)

= 900

 

 

 

 

[키워드] 2NO(g) + 2H2(g) ⇌ N2(g) + 2H2O(g) 평형 기준