Question #4f54f
1 Answer
Explanation:
For starters, calculate the number of moles of hydrogen iodide present in the equilibrium mixture by using the compound's molar mass
#1.9 color(red)(cancel(color(black)("g"))) * "1 mole HI"/(127.9color(red)(cancel(color(black)("g")))) = "0.01486 moles HI"#
Next, calculate the initial concentration and the equilibrium concentration of hydrogen iodide by using the volume of the container
#["HI"]_ 0 = "0.0172 moles"/"2 L" = "0.00860 M"#
#["HI"]_ "equi" = "0.01486 moles"/"2 L" = "0.00743 M"#
Now, the equilibrium reaction looks like this
#color(red)(2)"HI"_ ((g)) rightleftharpoons "H"_ (2(g)) + "I"_ (2(g))#
You know that you started with
#["HI"] _ "react" = "0.00860 M" - "0.00743 M"#
#["HI"]_ "react" = "0.00117 M"#
Notice that it takes
#["H"_ 2]_ "equi" = ["HI"]_ "react"/color(red)(2) = "0.00117 M"/color(red)(2) = "0.000585 M"#
#["I"_ 2]_ "equi" = ["HI"]_ "react"/color(red)(2) = "0.00117 M"/color(red)(2) = "0.000585 M"#
By definition, the equilibrium constant for this reaction takes the form
#K_c = (["H"_ 2]_ "equi" * ["I"_ 2]_ "equi")/(["HI"]_"equi"^color(red)(2))#
Plug in your values to find
#K_c = (0.000585color(red)(cancel(color(black)("M"))) * 0.000585color(red)(cancel(color(black)("M"))))/(0.00743^color(red)(2) color(red)(cancel(color(black)("M"^color(red)(2))))) = color(darkgreen)(ul(color(black)(0.0062)))#
I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the volume of the container.
Finally, does the result make sense?
Notice that most of the hydrogen iodide that was initially placed in the container remains unreacted. This tells you that