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2.50 mol NOCl was placed in a 2.50 L reaction vessel at 400ºC. After equilibrium was established, it was found that 28% of the NOCl had dissociated according to the equation 2NOCl(g) 2NO(g) + Cl 2(g). Calculate the equilibrium constant, K c, for the reaction.

Respuesta :

Answer:

Explanation:

2NOCl(g)    ⇄      2NO(g) + Cl 2(g)

C ( 1 - .28 )            .28 C         .14 C

Kc = [ NO ]²  x [ Cl₂ ] / [ NOCl ]²

= (.28 C )² x .14 C / C² ( 1 - .28 )²

= .021173 x C

C = concentration of reactant

= 2.5 / 2/5 = 1 M

Kc = .021173 x 1

= 211.73 x 10⁻⁴ M .

Cricetus

The equilibrium constant will be "2.117×10⁻²".

Given:

  • Number of moles = 2.50 mol
  • Volume of solution = 2.5 L

At equilibrium,

  • Concentration of NO = 0.28 M
  • Concentration of Cl₂ = 0.14 M

Now,

The concentration of NOCl will be:

= [tex]\frac{Number \ of \ moles}{Volume \ of \ solution}[/tex]

= [tex]\frac{2.5}{2.5}[/tex]

= [tex]1 \ M[/tex]

At equilibrium,

The concentration of NOCl will be:

= [tex]1-0.28[/tex]

= [tex]0.72 \ M[/tex]

hence,

The equilibrium constant,

→ [tex]K_c =\frac{ [NO]^2 [Cl_2]}{[NOCl]^2}[/tex]

By substituting the values, we get

        [tex]= \frac{(0.28)^2\times (0.14)}{(0.72)^2}[/tex]

        [tex]= 2.117\times 10^{-2}[/tex]

Thus the above answer is right.

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