136k views
3 votes
Hydrogen bromide decomposes when heated to 437C according to the equation: 2HBr(g) H2(g) Br2(g). If the reaction starts with 2.15 mol of hydrogen bromide in 1.0 liter, and decomposes to 36.7%, what is the equilibrium constant of the decomposition of hydrogen bromide

User Chromatix
by
8.4k points

1 Answer

4 votes

Answer:

the equilibrium constant of the decomposition of hydrogen bromide is 0.084

Step-by-step explanation:

Amount of HBr dissociated


2.15 \ mole * (36.7)/(100) \\\\=0.789 \ mole

2HBr(g) ⇆ H2(g) + Br2(g)

Initial Changes 2.15 0 0 (mol)

- 0.789 + 0.395 + 0.395 (mol)

At equilibrium 1.361 0.395 0.395 (mole)

Concentration 1.361 / 1 0.395 / 1 0.395 / 1

at equilibrium (mole/L)


K_c=([H_2][Br_2])/([HBr]^2) \\\\=((0.395)(0.395))/((1.361)^2) \\\\=(0.156025)/(1.852321) \\\\=0.084

Therefore, the equilibrium constant of the decomposition of hydrogen bromide is 0.084

User Tsatiz
by
8.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.