# Question #fdae0

##### 1 Answer

#### Answer:

#### Explanation:

For starters, notice that **equilibrium concentrations** of the reactants, **higher** than the equilibrium concentrations of the products,

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

#"A " + " B " rightleftharpoons " C " + " D"#

Notice that all the chemical species are present in **mole ratios**. If you take **consumed** when the two substances react, you can say that, at **equilibrium**, you have

#["A"] = "2.00 M" - x -># the concentration of#"A"# decreasedby#x#

#["B"] = "2.00 M" - x -># the concentration of#"B"# decreasedby#x#

#["C"] = 0 + x -># the concentration of#"C"# increasedby#x#

#["D"] = 0 + x -># the concentration of#"D"# increasedby#x#

Now all you have to do is plug these values into the expression of the *equilibrium constant*,

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

You will end up with

#K_c = (x * x)/((2.00 - x)(2.00 - x))#

#3.5 = x^2/(2.00 - x)^2#

Rearrange this to quadratic equation form

#3.5 * (2.00 - x)^2 = x^2#

#2.5x^2 - 14x + 14 = 0#

This quadratic equation will produce two *positive values*

#{(x_1 = 4.30), (x_2 = 1.30) :}#

Since the equilibrium concentrations of **all species** involved in the reaction **must be**

#x = 1.30#

This means that the equilibrium concentration of

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