# Question fdae0

Oct 24, 2016

["A"] = "0.70 M"

#### Explanation:

For starters, notice that ${K}_{c} > 1$, which should tell you that the equilibrium concentrations of the reactants, $\text{C}$ and $\text{D}$, are higher than the equilibrium concentrations of the products, $\text{A}$ and $\text{B}$.

Now, since you know how to set up the ICE table, I'll just walk you through what's happening without making an ICE table of my own.

The equilibrium reaction given to you is

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

Notice that all the chemical species are present in $1 : 1$ mole ratios. If you take $x$ to be the concentration of $\text{A}$ and $\text{B}$ consumed when the two substances react, you can say that, at equilibrium, you have

• ["A"] = "2.00 M" - x -> the concentration of $\text{A}$ decreased by $x$

• ["B"] = "2.00 M" - x -> the concentration of $\text{B}$ decreased by $x$

• $\left[\text{C}\right] = 0 + x \to$ the concentration of $\text{C}$ increased by $x$

• $\left[\text{D}\right] = 0 + x \to$ the concentration of $\text{D}$ increased by $x$

Now all you have to do is plug these values into the expression of the equilibrium constant, ${K}_{c}$

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

You will end up with

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

$3.5 = {x}^{2} / {\left(2.00 - x\right)}^{2}$

Rearrange this to quadratic equation form

$3.5 \cdot {\left(2.00 - x\right)}^{2} = {x}^{2}$

$2.5 {x}^{2} - 14 x + 14 = 0$

This quadratic equation will produce two positive values

$\left\{\begin{matrix}{x}_{1} = 4.30 \\ {x}_{2} = 1.30\end{matrix}\right.$

Since the equilibrium concentrations of all species involved in the reaction must be $> 0$, you must pick

$x = 1.30$

This means that the equilibrium concentration of $\text{A}$ will be

color(green)(bar(ul(|color(white)(a/a)color(black)(["A"] = "2.00 M" - "1.3 M" = "0.70 M")color(white)(a/a)|)))#