# If 2.0 mol #NO# and 1.0 mol #Cl_2# are placed in a 1.0-L flask, calculate the equilibrium concentrations of all species?

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(I am aware this likely isn't the proper constant. But I'd like to know if I did everything else correctly...)

At #35°C# , #K_c=1.6xx10^-5# for the reaction

#2NOCl(g)\leftrightarrow2NO(g)+Cl_2(g)#

(I am aware this likely isn't the proper constant. But I'd like to know if I did everything else correctly...)

At

##### 2 Answers

Work is incomplete; can someone check my process?

#### Explanation:

ICE table:

As

As a result, the equilibrium concentrations are

#color(blue)(["NO"]) = 2.0 - 2("0.9752 M") = color(blue)("0.0496 M")#

#color(blue)(["Cl"_2]) = 1.0 - 0.9752 = color(blue)("0.0248 M")#

#color(blue)(["NOCl"]) = 2(0.9752) = color(blue)("1.9504 M")#

It is important to realize that

Since

#["products"]# #"<<"# #["reactants"]#

is expected.

Hence, what we should have done is the following.

Gas-phase equilibrium constants aren't as well-known, so it's fine to use the one you are given. (

Since these gases are in a **ICE table**:

#"2NOCl"(g) rightleftharpoons 2"NO"(g) + "Cl"_2(g)#

#"I"" "0" "" "" "" "" "2.0" "" "" "1.0#

#"C"" "+2x" "" "" "-2x" "" "-x#

#"E"" "2x" "" "" "" "2.0-2x" "1.0-x#

Given **coefficients go into the exponents and change in concentration**,

#1.6 xx 10^(-5) = ((2.0-2x)^2(1.0-x))/(2x)^2#

However, it is unreasonable to use the small

The way this reaction is written, it is clearer to flip everything to get:

#1/(1.6xx10^(-5)) = 62500 = (2x)^2/((2.0-2x)^2(1.0-x))#

And so, instead, we have to say that

#62500(4.0 - 8x + 4x^2)(1.0 - x) = 4x^2#

#= 62500(4.0 - 4x - 8x + 8x^2 + 4x^2 - 4x^3)#

#=> -4x^3 + 12x^2 - 12x + 4.0 = 1/62500 cdot 4x^2#

#=> -x^3 + 3x^2 - 3x + 1.0 = 1/62500 cdot x^2#

#=> x^3 - 3x^2 + 3x - 1.0 ~~ 0#

This cubic has one real solution, which is

#color(blue)(["NO"]) = 2.0 - 2("0.9752 M") = color(blue)("0.0496 M")#

#color(blue)(["Cl"_2]) = 1.0 - 0.9752 = color(blue)("0.0248 M")#

#color(blue)(["NOCl"]) = 2(0.9752) = color(blue)("1.9504 M")#

Verify

#1/62500 = 1.6 xx 10^(-5) = ((2.0-2x)^2(1.0-x))/(2x)^2#

#stackrel(?" ")(=) ((0.0496^2)(0.0248))/(1.9504^2)#

#= 1.60386 xx 10^(-5) ~~ 1.6 xx 10^(-5)# #color(blue)(sqrt"")#