The following equilibrium constant is Kc=[C][D]/[A][B]=2.6 A)Initially, only and

Question

The following equilibrium constant is Kc=[C][D]/[A][B]=2.6

A)Initially, only and are present, each at 2.00 . What is the final concentration of once equilibrium is reached?

B)What is the final concentration of at equilibrium if the initial concentrations are = 1.00 and = 2.00 ?

Explanation / Answer

The "REVERSIBLE" chemical reaction- A + B C + D Kc = [C][D] / [A][B] = 6.8 Concentration at the start and after equilibrium- [A] = 2.00 M --> (2.00 - X) M [B] = 2.00 M --> (2.00 - X) M [C] = 0.00 M --> X M [D] = 0.00 M --> X M 6.8 = X^2 / (2.00 - X)^2 take the square root of both sides- 2.6077 = X / (2.00 - X) 5.2154 - 2.6077 X = X 5.2154 = 3.6077 X X = 1.4456 at equilibrium- [A] = 0.55 M [B] = 0.55 M [C] = 1.45 M [D] = 1.45 M

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