Question #072cf
1 Answer
Here's what I got.
Explanation:
All you have to do here is make sure that you write the correct expression for the equilibrium constant,
The first thing to do here is write the balanced chemical equation
#"H"_ (2(g)) + "I"_ (2(g)) rightleftharpoons color(red)(2)"HI"_((g))#
Notice that one mole of hydrogen gas,
By definition, the equilibrium constant for this reaction, which uses equilibrium concentrations raised to the power of their associated stoichiometric coefficients, looks like this
#K_c = (["HI"]^color(red)(2))/(["H"_2] * ["I"_2])#
The problem tells you that at
#K_c > 1#
you can say that the equilibrium will lie mostly to the right, i.e. the forward reaction will be favored.
You can thus expect the equilibrium concentration of hydrogen iodide to be significantly higher than the equilibrium concentrations of the two reactants.
Now, the problem also provides initial concentrations for the three chemical species. In order to determine in which direction will the equilibrium shift once the reaction starts, you need to calculate the reaction quotient,
The reaction quotient takes the same form as the equilibrium constant
#Q_c = (["HI"]_0^color(red)(2))/(["H"_2]_0 * ["I"_2]_0)#
with the important difference being that
In your case, this moment will be right before the reaction starts and
So, plug in the values given to you for the initial concentrations of the three chemical species to get
#Q_c = (0.0424^color(red)(2)color(red)(cancel(color(black)("M"^color(red)(2)))))/(0.00623 color(red)(cancel(color(black)("M"))) * 0.00414color(red)(cancel(color(black)("M")))) = 69.7#
Since you have
Use an ICE table to find the equilibrium concentrations - remember, the equilibrium shifts to the left, which implies that hydrogen iodide will be consumed and hydrogen gas and iodine gas will be produced
#" ""H"_ (2(g)) " "+" " "I"_ (2(g)) " "rightleftharpoons" " color(red)(2)"HI"_((g))#
The equilibrium constant will be equal to
#K_c = ((0.0424 - color(red)(2)x)^color(red)(2))/(( 0.00623+x) * (0.00414 + x))#
SIDE NOTE This is equivalent to saying that
#2"HI"_ ((g)) rightleftharpoons "H"_ (2(g)) + "I"_(2(g))#
with the equilibrium constant being
#K_"c reverse" = 1/K_c#
To keep things simple, I used the forward reaction as the base for the ICE table.
END OF SIDE NOTE
This will be equivalent to
#54.3 = (4x^2 - 0.0848x + 0.001798)/((x^2 + 0.01037x + 0.0000258)#
Rearrange to get
#50.3x^2 + 0.6479x - 0.000397 = 0#
This quadratic equation will produce two values for
#x = 0.000586#
The equilibrium concentrations for the three chemical species will be
#["HI"] = 0.0424 - color(red)(2) * 0.00568 = color(green)(|bar(ul(color(white)(a/a)"0.0412 M"color(white)(a/a)|)))#
#["H"_2] = 0.00623 + 0.000586 = color(green)(|bar(ul(color(white)(a/a)"0.00682 M"color(white)(a/a)|)))#
#["I"_2] = 0.00414 + 0.000586 = color(green)(|bar(ul(color(white)(a/a)"0.00473 M"color(white)(a/a)|)))#
I'll leave the answers rounded to three sig figs.
Notice that despite the fact that the concentration of hydrogen iodide decreased and the concentrations of hydrogen gas and iodine gas increased, the value of