Question #9bbc5

Question

Question #9bbc5

1 Answer

Impact of this question

2016 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-01T00:55:54

Here's what I got.

Explanation:

Start by writing the balanced chemical equation for this equilibrium reaction

$2 {NO}_{2 (g)} ⇌ 2 {NO}_{(g)} + O_{2 (g)}$ $color(red)(2)"NO"_ (2(g)) rightleftharpoons color(blue)(2)"NO"_ ((g)) + "O"_(2(g))$

The first thing to notice here is that the equilibrium constant for this reaction

$K_{c} = 1.2 \cdot 10^{- 5}$ $K_c = 1.2 * 10^(-5)$

is smaller than $1$ $1$ . This tells you that the equilibrium lies to the left, i.e. the reverse reaction is favored.

In other words, you can expect the equilibrium concentrations of the two reactants, nitric oxide, $NO$ $"NO"$ , and oxygen gas, $O_{2}$ $"O"_2$ , to be significantly smaller than the equilibrium concentration of the reactant, nitrogen dioxide, ${NO}_{2}$ $"NO"_2$ .

Use the number of moles of nitrogen dioxide and the volume of the reaction vessel to determine the initial concentration of the reactant

$[{NO}_{2}] = \frac{0.50 moles}{2.0 L} = 0.25 M$ $["NO"_2] = "0.50 moles"/"2.0 L" = "0.25 M"$

Use an ICE table to help you determine the equilibrium concentrations of the three chemical species

$2 {NO}_{2 (g)} ⇌ 2 {NO}_{(g)} + O_{2 (g)}$ $" "color(red)(2)"NO"_ (2(g)) " "rightleftharpoons" " color(blue)(2)"NO"_ ((g)) " "+" " "O"_(2(g))$

$I a a a a a 0.25 a a a a a a a a a a a a 0 a a a a a a a a a a a 0$ $color(purple)("I")color(white)(aaaaacolor(black)("0.25)aaaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0))$
$C a a a (- 2 x) a a a a a a a a (+ 2 x) a a a a a a a (+ x)$ $color(purple)("C")color(white)(aaacolor(black)((-color(red)(2)x))aaaaaaaacolor(black)((+color(blue)(2)x))aaaaaaacolor(black)((+x))$
$E a a a 0.25 - 2 x a a a a a a a a a 2 x a a a a a a a a a a x$ $color(purple)("E")color(white)(aaacolor(black)(0.25-color(red)(2)x)aaaaaaaaacolor(black)(color(blue)(2)x)aaaaaaaaaacolor(black)(x)$

By definition, the equilibrium constant for this reaction will look like this

$K_{c} = \frac{[O_{2}] \cdot {[NO]}^{2}}{{[{NO}_{2}]}_{2}^{2}}$ $K_c = (["O"_2] * ["NO"]^color(blue)(2))/(["NO"_2]^color(red)(2))$

In this case, you will have

$K_{c} = \frac{x \cdot {(2 x)}^{2}}{{(0.25 - 2 x)}^{2}}$ $K_c =(x * (color(blue)(2)x)^color(blue)(2))/((0.25 - color(red)(2)x)^color(red)(2)$

$K_{c} = \frac{4 x^{3}}{{(0.25 - 2 x)}^{2}}$ $K_c = (4x^3)/(0.25 - 2x)^2$

Now, because $K_{c}$ $K_c$ is so small, you can use the following approximation

$0.25 - 2 x \approx 0.25$ $0.25 - 2x ~~ 0.25$

This means that you'll have

$K_{c} = \frac{4 x^{3}}{{0.25}^{2}}$ $K_c = (4x^3)/0.25^2$

This will get you

$x = \sqrt[3]{\frac{K_{c} \cdot {0.25}^{2}}{4}} = \sqrt[3]{\frac{1.2 \cdot 10^{- 5} \cdot 0.0625}{4}}$ $x = root(3)( (K_c * 0.25^2)/4) = root(3)((1.2 * 10^(-5) * 0.0625)/4)$

$x = 0.005724$ $x = 0.005724$

The equilibrium concentrations of the three species will thus be

$[{NO}_{2}] = 0.25 M - 2 \cdot 0.005724 M = 0.2386 M$ $["NO"_2] = "0.25 M" - color(red)(2) * "0.005724 M" = "0.2386 M"$

$[NO] = 2 \cdot 0.005724 M = 0.01145 M$ $["NO"] = color(blue)(2) * "0.005724 M" = "0.01145 M"$

$[O_{2}] = 0.005724 M$ $["O"_2] = "0.005724 M"$

Rounded to two sig figs, the answers will be

$["NO"_2] = color(green)(|bar(ul(color(white)(a/a)"0.24 M"color(white)(a/a)|)))$

$["NO"] = color(green)(|bar(ul(color(white)(a/a)"0.011 M"color(white)(a/a)|)))$

$["O"_2] = color(green)(|bar(ul(color(white)(a/a)"0.0057 M"color(white)(a/a)|)))$

As predicted, the equilibrium concentrations of the products are significantly smaller than the equilibrium concentration of the reactant.

Science

Math

Humanities

... and beyond

Question #9bbc5

1 Answer

Explanation:

Related questions

Impact of this question