Question

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

The reversible chemical reaction
A+B⇌C+D

has the following equilibrium constant:
Kc=[C][D][A][B]=1.8

Answer 1

`"0.651 M"`

Explanation:

You know that this reaction

`"A " + " B " rightleftharpoons " C " + " D"`

consumes `"A"` and `"B"` in a `1:1` mole ratio and produces `"C"` and `"D"` in a `1:1` mole ratio.

Moreover, you know that the equilibrium constant that describes this equilibrium is equal to

`K_c = (["C"] * ["D"])/(["A"] * ["B"]) = 1.8`

Now, if you take `x` `"M"` to be the concentration of `"A"` that is consumed by the reaction, you can say that the reaction will also consume `x` `"M"` of `"B"` and produce `x` `"M"` of `"C"` and `x` `"M"` of `"D"`.

So you can say that the equilibrium concentrations of the four chemical species will be

`["A"] = (1.00 - x) quad "M"`

`["B"] = (2.00 - x) quad "M"`

`["C"] = x quad "M"`

`["D"] = x quad "M"`

This means that the expression of the equilibrium constant will take the form

`K_c = (x * x)/((1.00-x)(2.00-x))`

`1.8 = x^2/((1.00-x)(2.00-x))`

Rearrange to quadratic equation form to get

`0.2x^2 + 5.40 * x - 3.60 = 0`

This quadratic equation will produce two solutions, one positive and one negative. Since we've taken `x` to represent concentration, you can discard the negative solution and say that

`x = 0.651`

This means that the equilibrium concentration of `"D"` will be equal to

`["D"] = color(darkgreen)(ul(color(black)("0.651 M")))`

The answer is rounded to three sig figs.