# Question #09aa6

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

Your equilibrium reaction looks like this

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

You know that the equilibrium constant for this reaction is

#K_c = 1.7 * 10^(-3)#

Right from the start, the fact that *significantly* higher concentrations of nitrogen gas and oxygen gas.

In other words, at this particular temperature, the equilibrium lies **to the left**.

Set up your **ICE table** and plug in your values to find the equilibrium concentrations of the three species

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

By definition, the equilibrium constant will be equal to

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

In this case, you will have

#K_c = (color(red)(2)x)^color(red)(2)/((2.24 - x) * (0.55 - x)) = (4x^2)/((2.24 - x)(0.55 - x))#

Now, because the value of

#2.24 - x ~~ 2.24" "# and#" "0.55 - x ~~ 0.55#

This will get you

#1.7 * 10^(-3) = (4x^2)/(2.24 * 0.55)#

#x = sqrt( (2.24 * 0.55 * 1.7 * 10^(-3))/4) = 0.0223#

The equilibrium concentrations of the three chemical species will thus be

#["N"_2] = 2.24 - 0.0223 = "2.2 M"#

#["O"_2] = 0.55- 0.0223 = "0.53 M"#

#["NO"] = color(red)(2) xx 0.0223 = "0.045 M"#

The answers are rounded to two **sig figs**.

As predicted, the equilibrium concentrations of the reactants are significantly higher than the equilibrium concentration of the product.

**SIDE NOTE** *You can test the validity of the approximations by solving the quadratic*

#4x^2 - 1.7 * 10^(-3) * (1.232 - 2.79x + x^2) = 0#

#4x^2 + 0.004743 - 0.0020944 = 0#

*This quadratic equation will produce two values, a positive one and a negative one. The positive one is*

#x ~~ 0.0022297#

*This shows that the approximation holds because*

#|(0.022297 - 0.00223)/0.0022297| xx 100% " << " 5%#