Question

Question #7bc48

Answer 1

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"NOCl"_ ((g)) rightleftharpoons 2"NO"_ ((g)) + "Cl"_ (2(g))`

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` `"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` `"mol L"^(-1)` of nitric oxide and `1/2x` `"mol L"^(-1)` of chlorine gas, the concentration of nitrosyl chloride must decrease by `x` `"mol L"^(-1)`.

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

`K_c = (["NO"]^2 * ["Cl"_2])/(["NOCl"]^2)`

In your case, you have--I won't add the units here

`1.60 * 10^(-5) = (x^2 * (1/2x))/((0.40 - x)^2)`

`1.60 * 10^(-5) = (1/2x^3)/((0.40 - x)^2)`

This will be equivalent to

`x^3 + 3.2 * 10^(-5) * x^2 + 1.024 * 10^(-5) * x - 0.512 * 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.

`color(white)(a)`

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

`0.40 -x ~~ 0.40`

because, in that case, you would get

`1.60 * 10^(-5) = (1/2x^3)/0.40^2`

Rearrange to solve for `x`

`x = root(3)( (1.60 * 10^(-5) * 0.40)/(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.