Amixture initially contains a, b, and c in the following concentrations: [a] = 0.700 m , [b] = 0.850 m , and [c] = 0.300 m . the following reaction occurs and equilibrium is established: a+2b⇌c at equilibrium, [a] = 0.570 m and [c] = 0.430 m . calculate the value of the equilibrium constant, kc.

The value of Equilibrium constant is  2.16

Explanation:

Given-

From the chemical reaction ,

ICE table can be written as -

                              A         +        2B       ⇄        C

initial moles       0.700           0.850              0.300

At equilibrium    0.700 - x     0.850 - 2x      0.300 + x

From question , at equilibrium the concentration of A = 0.570 M

The concentration of A ( ICE table ) = concentration (given in question )

0.700 - x  = 0.570

x = 0.700 - 0.570

x = 0.13

Putting the value of x in ICE table , to obtain the concentration of B

[B] = 0.850 - 2x

[B] = 0.850 - 2*0.13

[B] = 0.850 - 0.26

[B] = 0.59 M

[C] = 0.430 M       ( Given )

Equilibrium Constant ( Kc )

The value of equilibrium constant is given as the concentration of the  products each raised to the power of their respective stoichiometry by the concentration of reactants each raised to the power of their respective stoichiometry.

Kc = [C] / [A] [B]²

putting the value of concentration terms at equilibrium ,

Kc = [0.430] / [0.570][0.59]²

Kc = 2.16

answer

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