Question

for the reaction below, kc = 7.2 × 10⁻⁴. note kc is sometimes called k. what is the equilibrium concentration of nof if the reaction begins with 0.92 m no and 0.92 m f₂? 2 no (g) f₂ (g) ⇌ 2 nof (g)

Answer 1

To find the equilibrium concentration of NOF, use the equilibrium constant expression and an ICE table. The equilibrium concentration of NOF is approximately 0.0064 M.

Step-by-step explanation:

To find the equilibrium concentration of NOF, we can use the equilibrium constant expression. For the reaction 2 NO (g) + F₂ (g) ⇌ 2 NOF (g), the equilibrium constant (Kc) is 7.2 × 10⁻⁴. We can set up an ICE table to calculate the equilibrium concentration of NOF:

1. Initial concentration of NOF = 0 M
2. Change in concentration of NOF = 2x
3. Equilibrium concentration of NOF = 2x

Since the initial concentrations of NO and F₂ are both 0.92 M, the initial concentration of NOF is 0 M. By substituting the equilibrium concentrations into the equilibrium constant expression, we get:

Kc = [NOF]² / ([NO]² [F₂])

7.2 × 10⁻⁴ = (2x)² / ((0.92)² (0.92))

Solving for x, we find that the equilibrium concentration of NOF is approximately 0.0064 M.

Answer 2

The equilibrium concentration of the NOF is 0.024 M

An equilibrium's position can be determined using the equilibrium constant. When K is large, the equilibrium is in favor of the formation of products; when K is small, the reactants are in favor of the equilibrium. At equilibrium, the concentrations of reactants and products are similar if K is near 1.

We know that;

K =
`[NOF]^2`/
`[NO]^2 [F_2]`

7.2 × 10⁻⁴
`[NO]^2 [F_2]` =
`[NOF]^2`

7.2 × 10⁻⁴
`(0.92)^2`(0.92)=
`[NOF]^2`

[NOF] = √7.2 × 10⁻⁴
`(0.92)^2`(0.92)

= 0.024 M

Therefore the concentration is 0.024 M.