# 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

Mar 4, 2018

$\text{0.651 M}$

#### Explanation:

You know that this reaction

$\text{A " + " B " rightleftharpoons " C " + " D}$

consumes $\text{A}$ and $\text{B}$ in a $1 : 1$ mole ratio and produces $\text{C}$ and $\text{D}$ in a $1 : 1$ mole ratio.

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

${K}_{c} = \left(\left[\text{C"] * ["D"])/(["A"] * ["B}\right]\right) = 1.8$

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

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

${K}_{c} = \frac{x \cdot x}{\left(1.00 - x\right) \left(2.00 - x\right)}$

$1.8 = {x}^{2} / \left(\left(1.00 - x\right) \left(2.00 - x\right)\right)$

Rearrange to quadratic equation form to get

$0.2 {x}^{2} + 5.40 \cdot x - 3.60 = 0$

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 $\text{D}$ will be equal to

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

The answer is rounded to three sig figs.