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?
(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
"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
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"")