Answered

Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

A 0.50 mol sample of COBr2 is transferred to a 9.50-L flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species.

Sagot :

ajeigbeibraheem

Answer:

Equlibrium concentration for each species ae as follows:

[CO] = 0.043 mol/L

[Br₂] = 0.043 mol/L

[COBr₂] = 0.01 mol/L

Explanation:

Let take a look at the chemical equation taking place at equilibrium

COBr2(g) ⇄ CO(g) + Br2(g)

The concentration of COBr2 i.e.

[COBr2] = no of moles/volume

= 0.50 mol/9.50 L

[COBr2] = 0.0530 mol/L

At standard conditions

Kc for COBr2 = 0.190

Now, the ICE table for the above reaction can be computed as follows:

                       COBr2(g)              ⇄       CO(g)         +           Br2(g)

Initial               0.053                                   0                           0

Change             -x                                       +x                          +x

Equilibrium     (0.053 - x)                              x                           x

[tex]\mathsf{K_c = \dfrac{[CO][Br_2]}{[COBr_2]}}[/tex]

[tex]K_c = \dfrac{(x) (x)}{(0,053 -x)}[/tex]

[tex]0.190= \dfrac{x^2}{(0.053 -x)}[/tex]

x² = 0.190(0.053 - x)

x² = 0.01007 - 0.190x

x² + 0.190x - 0.01007 = 0

Using quadratic formula:

x ≅ 0.043 mol/L

SInce: x = [CO][Br₂] = 0.043 mol/L

[COBr₂] = 0.053 - x

[COBr₂] = 0.053 - 0.043 mol/L

[COBr₂] = 0.01 mol/L

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.