1.374 g H2 70.31 g Br2 2.00 L 700 K 0.566 g H2

1.374 g H2 70.31 g Br2 2.00 L 700 K 0.566 g H2

A mixture of 1.374 g of H2 and 70.31 g of Br2

is heated in a 2.00 L vessel at 700 K.

These substances react as follows:

H2(g) + Br2(g) ⇌ 2HBr(g)

At equilibrium the vessel is found to contain 0.566 g of H2.

a. Calculate the equilibrium concentrations of H2, Br2 and HBr.

b. Calculate Kc.

H2 1.374 g과 Br2 70.31 g의 혼합물이

2.00 L의 용기 속에서 700 K로 가열되어

다음과 같이 반응한다.

H2(g) + Br2(g) ⇌ 2HBr(g)

평형 상태의 용기 속에는 H2 0.566 g이 있다.

a) H2, Br2, HBr의 평형 농도를 계산하시오.

b) Kc를 계산하시오.

각 반응물의 몰수를 계산하면,

1.374 g / (2.02 g/mol) = 0.6802 mol H2

( 참고 https://ywpop.tistory.com/7738 )

70.31 / 159.81 = 0.43996 ≒ 0.4400 mol Br2

각 반응물의 몰농도를 계산하면,

> [H2] = 0.6802 mol / 2.00 L = 0.3401 M

> [Br2] = 0.4400 mol / 2.00 L = 0.2200 M

ICE 도표를 작성하면,

....................... H2(g) ........ + ........ Br2(g) ........ ⇌ ........ 2HBr(g)

초기(M): 0.3401 ..................... 0.2200 ........................ 0

변화(M): –x ................................ –x ................................... +2x

평형(M): 0.3401–x ............... 0.2200–x ................. 2x

[그림] ICE 도표 및 평형 상수 식.

( 참고 https://ywpop.tistory.com/7136 )

평형에서 H2의 몰농도를 계산하면,

0.566 g / (2.02 g/mol) / 2.00 L

= 0.566 / (2.02) / 2.00

= 0.140 M

= [H2]

0.3401–x = 0.1401 이므로,

x = 0.3401 – 0.1401

= 0.200

[Br2] = 0.2200–x

= 0.2200 – 0.200

= 0.020 M

[HBr] = 2x

= 2(0.200)

= 0.400 M

Kc = [HBr]^2 / [H2] [Br2]

= (0.400)^2 / (0.140 × 0.020)

= 57.14

[ 관련 글 https://ywpop.tistory.com/22578 ]

The equilibrium constant Kc for the reaction

H2(g) + Br2(g) ⇌ 2HBr(g)

is 2.18×10^6 at 730°C.

Starting with 3.20 moles of HBr in a 12.0 L reaction vessel,

calculate the concentrations of H2, Br2, and HBr at equilibrium.

[키워드] H2(g) + Br2(g) ⇌ 2HBr(g) 평형 기준, H2 + Br2 ⇌ 2HBr 평형 기준

A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated to 700 K in a 2.00 L vessel and reacts as follows. There is 0.566 g of H2 in the vessel at equilibrium.