The reversible chemical reaction A+B⇌C+D has the following equilibrium constant: Kc=[C][D][A][B]=1.7 What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M

The final concentration of D at equilibrium is 0.7422 M.

Explanation:

You know:

A + B ⇔ C + D

You can see that according to the stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) the relationship between A, B, C and D is 1.

On the other hand, the chemical equilibrium is established when there are two opposite reactions which take place simultaneously at the same speed. It is represented by the equilibrium constant, which is defined as:

where a, b, c and d are the stoichiometric coefficients, which in this case have a value of 1. Then the equilibrium constant is:

"x" is considered as the concentration of A that is consumed by the reaction. Then, following the stoichiometry set out above, "x" concentration of B will also be consumed, and "x" concentration of C and B will occur.

Then,  if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M,  you can raise:

[A]= (1.00 - x) M

[B]= (2.00 - x) M

[C]= x M

[D]= x M

Replacing in the exprision of the equilibrium constant stated previously:

Resolving:

The solutions are x=6.5434 and x=0.7422

If you observe the expression of concentration of A, which is the reagent that has the lowest initial concentration, the solution of x = 6.5434 makes no sense because the value of x is greater than 1. This would indicate that the concentration after the reaction would be negative and consume more than available. And this is not possible. So the solution of x is 0.7422.

If [D]=x, then the final concentration of D at equilibrium is 0.7422 M.

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the answer to this question is 3.