Question

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

Answer 1

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