Question

In the chemical reaction:Br2(g) + Cl2(g) ⇌ 2 BrCl(g) KP = 0.150If there is initially 0.500 atm of BrCl and nothing else in a container, what would be the equilibrium concentration of BrCl?

Answer 1

Answer:

The partial pressure of BrCl at equilibrium is 0.08 atm.

Explanation:

The equilibrium constant of the reaction is given by =$K_p=0.150$

$Br_2(g) + Cl_2(g)\rightleftharpoons 2 BrCl(g)$

initial

0       0         0.500 atm

At equilbrium

p      p       (0.500-2p)

The equilibrium constant's expression of the reaction is given by ;

 $K_p=\frac{[BrCl]^2}{[Br_2][Cl_2]}$

$0.150=\frac{(0.500-2p)^2}{p\times p}$

Solving for p:

p = 0.21 atm

The partial pressure of BrCl at equilibrium is:

(0.500-2p) = (0.500 - 2 × 0.21 )atm = 0.08 atm[/tex]

Answer 2

The equilibrium concentration of BrCl at equilibrium is 0.08 atm.

Chemical reaction:

Br₂(g) + Cl₂(g) ⇌ 2 BrCl(g)

Equilibrium constant:

The equilibrium constant of the reaction is given by:

Initial                   0       0         0.500 atm

At equilibrium      p      p       (0.500-2p)

The equilibrium constant's expression of the reaction is given by ;

$K_p=\frac{[BrCl]^2}{[Br_2][Cl]_2}\\\\0.150=\frac{(0.500-2p)^2}{p*p}$

(0.500-2p) = (0.500 - 2 * 0.21 ) atm

(0.500-2p) = 0.08 atm

p = 0.21 atm

The partial pressure or Equilibrium concentration of BrCl at equilibrium is: 0.21 atm

Find more information about Equilibrium concentration here:

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