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

Answer:

the equilibrium constant of the decomposition of hydrogen bromide is 0.084

Explanation:

Amount of HBr dissociated

$2.15 \ mole \times \frac{36.7}{100} \\\\=0.789 \ mole$

                                  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)

$K_c=\frac{[H_2][Br_2]}{[HBr]^2} \\\\=\frac{(0.395)(0.395)}{(1.361)^2} \\\\=\frac{0.156025}{1.852321} \\\\=0.084$

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