contestada

Carbonyl fluoride, cof2, is an important intermediate used in the production of fluorine-containing compounds. for instance, it is used to make the refrigerant carbon tetrafluoride, cf4 via the reaction 2cof2(g)⇌co2(g)+cf4(g), kc=6.40 if only cof2 is present initially at a concentration of 2.00 m, what concentration of cof2 remains at equilibrium?

Respuesta :

kayceeports
Let's apply the ICE or the Initial-Change-Equilibrium approach. The solution is as follows:

               2 cof2(g) ⇌ co2(g) + cf4(g)
I                    2              0           0
C                 -2x            +x         +x
-----------------------------------------------
E              2-2x              x            x

According to the reaction, the expression for Kc is

Kc = [CF₄][CO₂]/[COF₂]²
Substituting the E values and finding x,
6.40 = [x][x]/[2-2x]²
x = 0.835 m

Since E for COF₂ is 2-2x, the amount at equilibrium is equal to:
E = 2 - 2(0.835) = 0.33 m

The concentration of [tex]\rm \bold{COF_2}[/tex] at equilibrium has been 0.33 M.

The balanced chemical equation for the given reaction has been:

[tex]\rm 2\;COF_2\;\rightarrow\;CO_2\;+\;CF_4[/tex]

Since the initial concentration of [tex]\rm COF_2[/tex] has been 2 M, by applying ICE, the concentration can be:

         [tex]\rm COF_2[/tex]           [tex]\rm CO_2[/tex]          [tex]\rm CF_4[/tex]        

I          2                  0               0

C       -2x               +x              +x

E      2 - 2x             x                 x

The value of [tex]\rm k_c[/tex] has been the ratio of the product concentration to reactant concentration.

[tex]\rm k_c[/tex] for the given reaction will be:

[tex]\rm k_c[/tex] = [tex]\rm \dfrac{[CO_2]\;[CF_4]}{[COF_2]^2}[/tex]

6.40 = [tex]\rm \dfrac{(x)\;(x)}{(2 - 2x)^2}[/tex]

By simplifying the above equation,

x = 0.835 M.

The concentration of [tex]\rm COF_2[/tex] at equilibrium has been 2 - 2x

2 - 2x = 2 - 2 (0.835)

The concentration of [tex]\rm COF_2[/tex] = 0.33 M.

The concentration of [tex]\rm \bold{COF_2}[/tex] at equilibrium has been 0.33 M.

For more information about equilibrium concentration, refer to the link:

https://brainly.com/question/16645766