Question

The equilibrium constant for the following reaction: H2(g) + Br2(g) ↔ 2HBr (g) is 1.65 x 103 at a certain temperature. Find the equilibrium pressure of HBr if 7.10 atm of HBr is introduced into a sealed container at this temperature.

Answer 1

Answer:

6.97 atm was the equilibrium pressure of HBr .

Explanation:

The value of the equilibrium constant =$K_p=1.65\times 10^3$

$H_2+Br_2\rightleftharpoons 2HBr$

Initially:

0             0                 7.10 atm

At equilibrium

x              x                 (7.10-2x)

The expression of equilibrium constant can be written as:

$K_p=\frac{p_{HBr}}{p_{H_2}\times p_{Br_2}}$

$1.65\times10^3=\frac{(7.10-2x)}{x^2}$

Solving for x:

x = 0.065

Partial pressure of HBr at equilibrium :(7.10 - 2 × 0.065) atm = 6.97 atm

6.97 atm was the equilibrium pressure of HBr .