Question

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.300 M , [B] = 1.40 M , and [C] = 0.700 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.140 M and [C] = 0.860 M . Calculate the value of the equilibrium constant, Kc.

Answer 1

Following is the - complete Answer -&- Explanation, for the given Question, in......typed format....

Answer:

Value of the Equilibrium Constant:   K_c = 5.2665 ( approx. )  ,i.e. of the given reaction.

Explanation:

Following is the complete Explanation, for the above Answer...

• Given:

1. Initial molar concentration of compound A = [A]_o = 0.300 M (mol/L )
2. Initial molar concentration of compound B = [B]_o = 1.40 M (mol/L )
3. Initial molar concentration of compound C = [C]_o = 0.700 M  (mol/L )
4. balanced chemical equation:   A + 2B C ----------------------------- Equation - 1
5. At Equilibrium molar concentration of compound A = [A]_eq = 0.140 M  (mol/L )
6. At Equilibrium molar concentration of compound C = [C]_eq = 0.860 M (mol/L )

• ​​​​​​​Step - 1:

​​​​​​​According to Equation - 1, we will find the following expression/ formula, for determining the value of the Equilibrium Constant, K_c  , of the given reaction :

Equilibrium constant: K_c = [C]_eq / ( [A]_eq x [B]_eq2 ) ------------------------------ Equation - 2

According to Equation - 1, and using the given molar concentrations, let's find the value of the Equilibrium concentrations, of the reactants, and the products, of the given reaction, using the following ICE Table, and then let's use them to calculate the value of the Equilibrium constant ( i.e. K_c ) :   as the following ----

• Step - 2:

​​​​​​​Following is the ICE Table, we have prepared, using the given values ( i.e. according to Equation - 1 ) :

	[A], M ( mol/L )	[B], M  ( mol/L )	[C], M ( mol/L )
Initial conc. ( mol/L)	0.300	1.40	0.700
Change in conc. ( mol/L )	- X	- 2X	+ X
Equilibrium conc. ( mol/L )	( 0.300 - X )	( 1.40 - 2X )	( 0.700 + X )

• Step - 3:

​​​​​​​Since, we are given with the following values, i.e. :

1. [A]_eq = 0.140 M  (mol/L ) =  0.300 - X
2. [C]_eq = 0.860 M (mol/L ) = 0.700 + X

​​​​​​​Using the above values, we will get the following value of ' X ' , as it is shown below:

0.300 - X =  0.140

X = ( 0.300 - 0.140 ) = 0.16 M   ( mol/L )

• Step - 4:

​​​​​​​Using the above value of 'X'  , we will find the following value of the Equilibrium conc. of 'B' -----

[B]_eq =   ( 1.40 - 2 x ( 0.16 ) ) =   1.08 M ( mol/L )

• Step - 5:

​​​​​​​Now, plugging in values in Equation - 2, we will get the following :

  K_c = [C]_eq / ( [A]_eq x [B]_eq2 ) ------------------------------ Equation - 2

  K_c = ( 0.860 M )   / ( ( 0.140 M ) x ( 1.08 M )² )   

  K_c = 5.2665 ( approx. ) = Equilibrium Constant of the reaction .

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