The equilibrium constant K_c for the reaction: is 7.90 times 10^-5. The initial

Question

The equilibrium constant K_c for the reaction: is 7.90 times 10^-5. The initial composition of the reaction mixture is [C] = [D] = [E] = 1.50 times 10^-3 M. What is the equilibrium concentration of C, D, and E? We need to create an ICE chart to monitor the concentrations of both the products and reactants. By substituting the equilibrium concentrations into the equilibrium constant expression, we can solve for the change in concentration and determine the equilibrium concentration of everything.

Explanation / Answer

For the reaction,

Kc = [D][E]/[C]

let x be the change in concentration at equilibrium

7.90 x 10^-5 = (1.50 x 10^-3 - x)(1.50 x 10^-3 - x)/(1.50 x 10^-3 + x)

1.185 x 10^-7 + 7.90 x 10^-5x = 2.25 x 10^-6 - 3 x 10^-3x + x^2

x^2 - 3.08 x 10^-3x + 2.013 x 10^-6 = 0

x = 9.41 x 10^-4 M

So the equilibrium concentrations would be,

[C] = 2.44 x 10^-3 M

[D] = 5.59 x 10^-4 M

[E] = 5.59 x 10^-4 M

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