Question 09aa6

Oct 5, 2016

Here's what I got.

Explanation:

Your equilibrium reaction looks like this

${\text{N"_ (2(g)) + "O"_ (2(g)) rightleftharpoons color(red)(2)"NO}}_{\left(g\right)}$

You know that the equilibrium constant for this reaction is

${K}_{c} = 1.7 \cdot {10}^{- 3}$

Right from the start, the fact that ${K}_{c} < 1$ should tell you that when equilibrium is reached, the reaction vessel will contain 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

${\text{ ""N"_ (2(g)) " "+" " "O"_ (2(g)) " "rightleftharpoons" " color(red)(2)"NO}}_{\left(g\right)}$

color(purple)("I")color(white)(aaaacolor(black)(2.24)aaaaaaaacolor(black)(0.55)aaaaaaaaaacolor(black)(0)
color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaacolor(black)((-x))aaaaaaacolor(black)((+color(red)(2)x))
color(purple)("E")color(white)(aacolor(black)(2.24-x)aaaaacolor(black)(0.55-x)aaaaaaacolor(black)(color(red)(2)x)

By definition, the equilibrium constant will be equal to

${K}_{c} = \left(\left[{\text{NO"]^color(red)(2))/(["N"_2] * ["O}}_{2}\right]\right)$

In this case, you will have

${K}_{c} = {\left(\textcolor{red}{2} x\right)}^{\textcolor{red}{2}} / \left(\left(2.24 - x\right) \cdot \left(0.55 - x\right)\right) = \frac{4 {x}^{2}}{\left(2.24 - x\right) \left(0.55 - x\right)}$

Now, because the value of ${K}_{c}$ is very small, you can use the approximations

$2.24 - x \approx 2.24 \text{ }$ and $\text{ } 0.55 - x \approx 0.55$

This will get you

$1.7 \cdot {10}^{- 3} = \frac{4 {x}^{2}}{2.24 \cdot 0.55}$

$x = \sqrt{\frac{2.24 \cdot 0.55 \cdot 1.7 \cdot {10}^{- 3}}{4}} = 0.0223$

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

$4 {x}^{2} - 1.7 \cdot {10}^{- 3} \cdot \left(1.232 - 2.79 x + {x}^{2}\right) = 0$

$4 {x}^{2} + 0.004743 - 0.0020944 = 0$

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

$x \approx 0.0022297$

This shows that the approximation holds because

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