# Question 5fb4f

Sep 27, 2015

["N"_2] = "0.0180 M"
["O"_2] = "0.0180 M"
["NO"] = "0.00400 M"

#### Explanation:

So, you know that the equilibrium constant, ${K}_{c}$, for this reaction, is equal to $0.050$.

Right from the start, this tells you that the equilibrium will favor the reactants, since ${K}_{c} < 1$. This means that you can expect the equilibrium concentrations of the reactants to be 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 $\text{NO}$ will be smaller than what you started with, i.e. smaller than $\text{0.0400 M}$.

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

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

color(purple)("I")" " " " " "0" " " " " " " " 0" " " " " " " " "0.0400
color(purple)("C")" " " "(+x)" " " "(+x)" " " " "(0.0400-color(red)(2)x)
color(purple)("E")" " " " " "(x)" " " " " "(x)" " " " " " "(0.0400-color(red)(2)x)

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

$3.95 {x}^{2} - 0.16 x + 0.0016 = 0$

This quadratic equation will produce two positive solutions for $x$, but only one will be valid in this context because you need $\left(0.0400 - x\right)$, which is the equilibrium concentration of $\text{NO}$, to be positive as well.

${x}_{1} = 0.01799 \text{ }$ and $\text{ } \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}_{2} = 0.02252}}}$

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

$\left[\text{N"_2] = x = "0.01799 M" = color(green)("0.0180 M}\right)$

$\left[\text{O"_2] = x = color(green)("0.0180 M}\right)$

$\left[\text{NO"] = 0.0400 - 2x = 0.0400 - 2 * 0.0180 = color(green)("0.00400 M}\right)$

Sep 27, 2015

$\left[{N}_{2}\right] = 0.0180 M$
$\left[{O}_{2}\right] = 0.0180 M$
$\left[N O\right] = 0.0040 M$

#### Explanation:

The forward reaction is:

N_2(g)+O_2(g) â‡Œ 2NO(g) " " " " K_c=([NO]^2)/([N_2]*[O_2])=0.050

You should know that the ${K}_{c}$ is given for the forward reaction, however, since we only have $N O$ the reaction will proceed to the left, which means the reverse direction.

" " " " " " " " " "N_2(g)+O_2(g) â‡Œ 2NO(g)#
$I n i t i a l \text{ " " " " "0M " " 0M" " " " } 0.0400 M$
$\text{Change" " " " "+x/2M "" +x/2M " } - x M$
$\text{Equilibrium" " " x/2M " "x/2M " } 0.04 - x M$

Therefore, the equilibrium concentrations are:
$\left[{N}_{2}\right] = \frac{x}{2} M$
$\left[{O}_{2}\right] = \frac{x}{2} M$
$\left[N O\right] = 0.0400 - x M$

Replacing the concentrations in the expression of ${K}_{c}$ we get:
${K}_{c} = \frac{{\left(0.0400 - x\right)}^{2}}{\left(\frac{x}{2}\right) \cdot \left(\frac{x}{2}\right)} = 0.050$

Solve for $x \implies x = 0.0360 M$

Therefore, the equilibrium concentrations are:
$\left[{N}_{2}\right] = 0.0180 M$
$\left[{O}_{2}\right] = 0.0180 M$
$\left[N O\right] = 0.0040 M$