# Question #5fb4f

##### 2 Answers

#### Explanation:

So, you know that the *equilibrium constant*,

Right from the start, this tells you that the equilibrium will favor the **reactants**, since **larger** than the equilibrium concentration of the product.

Since the reaction starts with only the product, you can of course expect that the equilibrium concentration of *smaller* than what you started with, i.e. smaller than

You can find the equilibrium concentrations of all the species involved in the reaction by using an **ICE table**

#"N"_text(2(g]) " "+" " "O"_text(2(g]) " "rightleftharpoons" " color(red)(2)"NO"_text((g])#

By definition, the equilibrium constant for this reaction will be

#K_c = (["NO"]^color(red)(2))/(["N"_2] * ["O"_2]) = ((0.0400-color(red)(2)x)""^color(red)(2))/(x * x)#

#K_c = ((0.0400 - 2x)""^2)/x^2 = 0.050#

Rearrange to get

#3.95x^2 - 0.16x + 0.0016 = 0#

This quadratic equation will produce two positive solutions for **positive** as well.

#x_1 = 0.01799" "# and#" "color(red)(cancel(color(black)(x_2 = 0.02252)))#

Therefore, the equilibrium concentrations of the species that took part in the reaction are

#["N"_2] = x = "0.01799 M" = color(green)("0.0180 M")#

#["O"_2] = x = color(green)("0.0180 M")#

#["NO"] = 0.0400 - 2x = 0.0400 - 2 * 0.0180 = color(green)("0.00400 M")#

#### Explanation:

The forward reaction is:

You should know that the

Therefore, the equilibrium concentrations are:

Replacing the concentrations in the expression of

Solve for

Therefore, the equilibrium concentrations are: