Question #7bc48

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.