The equilibrium constant for the reaction is 4949. What is the equilibrium concentration of "HI"HI if 0.5000.500 mol "H"_2H2 and 0.5000.500 mol "I"_2I2 were mixed in a "1.00-L"1.00-L container initially?
"H"_2(g) + "I"_2(g) rightleftharpoons 2"HI"(g)H2(g)+I2(g)⇌2HI(g)
Select one:
a. 0.39 M
b. 0.78 M
c. 0.22 M
d. 0.11 M
e. 0.89 M
Select one:
a. 0.39 M
b. 0.78 M
c. 0.22 M
d. 0.11 M
e. 0.89 M
1 Answer
Explanation:
You know that at an unspecified temperature, the following equilibrium reaction
"H"_ (2(g)) + "I"_ (2(g)) rightleftharpoons 2"HI"_ ((g))H2(g)+I2(g)⇌2HI(g)
has an equilibrium constant equal to
K_c = 49Kc=49
By definition, the equilibrium constant is equal to
K_c = (["HI"]^2)/(["H"_2] * ["I"_2])Kc=[HI]2[H2]⋅[I2]
Now, the volume of the container is equal to
In other words, you have
["H"_ 2]_ 0 = "0.500 M"[H2]0=0.500 M
["I"_ 2]_ 0 = "0.500 M"[I2]0=0.500 M
If you take
This is the case because the reaction consumes hydrogen gas and iodine gas in a
So at equilibrium, the reaction vessel will contain
["H"_ 2] = ["H"_ 2]_ 0 - x[H2]=[H2]0−x
["H"_ 2] = (0.500 - x) quad "M"
["I"_ 2] = ["I"_ 2]_ 0 - x
["I"_ 2] = (0.500 - x) quad "M"
and
["HI"] = (2x) quad "M"
The equilibrium constant will thus be equal to
K_c = (2x)^2/((0.500 -x)(0.500-x))
49 = (2x)^2/(0.500-x)^2
49 = ((2x)/(0.500-x))^2
Take the square roots of both sides to get--remember, we're looking for concentration here, so you can discard the negative solution.
7 = (2x)/(0.500-x)
Rearrange to solve for
3.500 - 7x = 2x
x = 3.500/9 = 0.389
This means that the equilibrium concentration of hydrogen iodide will be
["HI"] = (2x) quad "M"
["HI"} = (2 * 0.389) quad "M" = "0.778 M"
The answer should be rounded to three sig figs, the number of sig figs you have for your values, but since the options given to you are rounded to two sig figs, you can say that
["HI"] = "0.78 M"
