A+B -> C+D has the following equilibrium constant: Kc= [C][D]/[A][B] =2.6 Initia

Question

A+B -> C+D
has the following equilibrium constant:

Kc= [C][D]/[A][B] =2.6

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

What is the final concentration of [D] at equilibrium if the initial concentrations are [A] = 1.00 M and [B]= 2.00 M?

Please show steps and solution! Thank you.

Explanation / Answer

The "REVERSIBLE" chemical reaction- A + B C + D Kc = [C][D] / [A][B] = 2.6 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 2.6 = X^2 / (2.00 - X)^2 take the square root of both sides- 1.61 = X / (2.00 - X) 3.22x - 1.61X = X 3.22 = 2.61 X X = 1.23 at equilibrium- [A] = 0.77M [B] = 0.77 M [C] = 1.23 M [D] = 1.23 M

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