The Kc for the following reaction at 100 °C is 2.0. If the equilibrium mixture contains 2.0 M NO and 1.0 M Br2, what is the molar concentration of NOBr?

2NOBr (g) ⇌ 2NO(g) + Br2 (g)

The answer to your question is [NObr] = 1.41

Explanation:

Data

Kc = 2

[NO] = 2 M

Br₂ = 1.0 M

NOBr = ?

Process

1.- Write the Equation of the Kc

                    Kc =

2.- Solve for [NOBr]

                   [NOBr]² =

3.- Substitution

                   [NOBr]² =

4.- Simplification

                    [¨NOBr]² =

                    [NOBr]² = 2

5.- Result

                    [NOBr] =

                    [NOBr] = 1.41

The molar concentration of NOBr is 1.414 M

Explanation:

Step 1: Data given

Kc = 2.0

The equilibrium mixture contains:

2.0 M NO

1.0 M Br2

Step 2: The balanced equation

2NOBr (g) ⇌ 2NO(g) + Br2 (g)    Kc = 2.0

Step 3: Define Kc

Kc = [Br2]*[NO]²/[NOBr]²

⇒ with Kc = 2.0

⇒with [Br2] = 1.0M

⇒with [NO] = 2.0 M

⇒with [NOBr] = TO BE DETERMINED

Step 4: Calculate the molar concentration of NOBr

Kc = [Br2]*[NO]²/[NOBr]²

2.0 = (1.0*2.0²) / NOBr²

NOBr² = 4.0 / 2.0 = 2.0

NOBr = √2.0 = 1.414 M

The molar concentration of NOBr is 1.414 M

it is stable as it is in group 0 which have full outer shells of electrons