Question

The equilibrium constant Kc for the reaction below is is 7.355×10-5 .

C->D+E
The initial composition of the reaction mixture is [C] = [D] = [E] = 1.5110×10-3 M.
What is the equilibrium concentration of C,D, and E?

Answer 1

To find the equilibrium concentrations of C, D, and E in a reaction with Kc = 7.355x10^-5 and initial concentrations of [C] = [D] = [E] = 1.5110x10^-3 M, we can use the equilibrium constant expression and substitute the given values to solve for the equilibrium concentrations. The equilibrium concentration for D and E can be found by calculating the square root of the product of their concentrations.

Step-by-step explanation:

The equilibrium constant, Kc, for the reaction C -> D + E is 7.355x10^-5. The initial composition of the reaction mixture is [C] = [D] = [E] = 1.5110x10^-3 M. To find the equilibrium concentrations of C, D, and E, we can use the equilibrium constant expression: Kc = [D][E] / [C]. We can substitute the given values into this expression to solve for the equilibrium concentrations of C, D, and E.

If we substitute the given initial concentrations into the expression, we get: Kc = [D][E] / [C] = (x)(x) / (1.5110x10^-3) = x^2 / (1.5110x10^-3) = 7.355x10^-5.

Rearranging the equation, we find x^2 = (7.355x10^-5)(1.5110x10^-3). Taking the square root of both sides gives us x = √[(7.355x10^-5)(1.5110x10^-3)]. We can then calculate the numerical value of x, which represents the equilibrium concentration of both D and E since their concentrations are equal.