# Question 7bc48

Dec 3, 2017

Here's what I got.

#### Explanation:

Start by calculating the initial concentration of nitrosyl chloride

["NOCl"] = "0.80 moles"/"2.0 L" = "0.40 mol L"^(-1)

Now, the balanced chemical equation that describes this equilibrium looks like this

$2 {\text{NOCl"_ ((g)) rightleftharpoons 2"NO"_ ((g)) + "Cl}}_{2 \left(g\right)}$

As you can see, for every $2$ moles of nitrosyl chloride that undergo decomposition, you get $2$ moles of nitric oxide and $1$ mole of chlorine gas.

This means that if you take $x$ ${\text{mol L}}^{- 1}$ to be the concentration of nitrosyl gas that dissociates, you can say that, at equilibrium, you will have

• ["NO"] = xcolor(white)(.)"mol L"^(-1)
• ["Cl"_2] = (1/2 * x) color(white)(.)"mol L"^(-1)
• ["NOCl"] = (0.40 - x)color(white)(.)"mol L"^(-1)

This basically means that in order for the reaction to produce $x$ ${\text{mol L}}^{- 1}$ of nitric oxide and $\frac{1}{2} x$ ${\text{mol L}}^{- 1}$ of chlorine gas, the concentration of nitrosyl chloride must decrease by $x$ ${\text{mol L}}^{- 1}$.

By definition, the equilibrium constant for this reaction looks like this

${K}_{c} = \left({\left[\text{NO"]^2 * ["Cl"_2])/(["NOCl}\right]}^{2}\right)$

$1.60 \cdot {10}^{- 5} = \frac{{x}^{2} \cdot \left(\frac{1}{2} x\right)}{{\left(0.40 - x\right)}^{2}}$

$1.60 \cdot {10}^{- 5} = \frac{\frac{1}{2} {x}^{3}}{{\left(0.40 - x\right)}^{2}}$

This will be equivalent to

${x}^{3} + 3.2 \cdot {10}^{- 5} \cdot {x}^{2} + 1.024 \cdot {10}^{- 5} \cdot x - 0.512 \cdot {10}^{- 5} = 0$

This cubic equation will produce three solutions, one positive and two negative. Since $x$ represents concentration, you can discard the negative solutions to get

$x = 0.017$

This means that, at equilibrium, the reaction vessel will contain

["NOCl"] = (0.40 - 0.017)color(white)(.)"mol L"^(-1) = "0.38 mol L"^(-1)

["NO"] = "0.017 mol L"^(-1)

["Cl"_2] = (1/2 * 0.017)color(white)(.)"mol L"^(-1) = "0.0090 mol L"^(-1)

All the values are rounded to two sig figs.

$\textcolor{w h i t e}{a}$

SIDE NOTE: Notice that you can't use the approximation

$0.40 - x \approx 0.40$

because, in that case, you would get

$1.60 \cdot {10}^{- 5} = \frac{\frac{1}{2} {x}^{3}}{0.40} ^ 2$

Rearrange to solve for $x$

$x = \sqrt[3]{\frac{1.60 \cdot {10}^{- 5} \cdot 0.40}{\frac{1}{2}}} = 0.023$

In order for that approximation to hold, you need to have

(["NO"])/(["NOCl"]_0) xx 100% < color(red)(5%)

However, this is not the case because

(0.023 color(red)(cancel(color(black)("mol L"^(-1)))))/(0.40color(red)(cancel(color(black)("mol L"^(-1))))) xx 100% = 5.75% > color(red)(5%)#

which implies that the approximation is not valid.