In a 10 litre evacuated chamber, 0.5 mole of H2 and 0.5 mole I2 gases are reacted to produce HI gas at 445°.the equilibrium constant at this temperature is 50. How many moles of iodine remain unreacted at equilibrium?
1 Answer
Here's what I got.
Explanation:
Before doing any calculations, try to predict what you expect to see once the reaction reaches equilibrium.
Take a look at the value of the equilibrium constant,
The fact that you have
The bigger the value of
Now, use the volume of the chamber and the number of moles of each reactant to calculate their initial concentrations
#color(blue)(c = n/V)#
#["H"_2] = "0.5 moles"/"10 L" = "0.05 M"#
#["I"_2] = "0.5 moles"/"10 L" = "0.05 M"#
You can now use an ICE table to help you find the equilibrium concentrations of the three species
#" " "H"_text(2(g]) " "+" " " " "I"_text(2(g]) " "rightleftharpoons" " color(red)(2)"HI"_text((g])#
By definition, the equilibrium constant for this reaction will be equal to
#K_c = ["HI"]^color(red)(2)/( ["H"_2] * ["I"_2]) = (color(red)(2)x)^color(red)(2)/( (0.050 -x )(0.050 -x))#
This is equivalent to
#K_c = (4x^2)/(0.050 - x)^2 = 50#
Take the square root of both sides to get
#sqrt((4x^2)/(0.050 - x)^2) = sqrt(50)#
#(2x)/(0.050 -x ) = sqrt(50)#
Solve this for
#2x = 0.050 * sqrt(50) - x * sqrt(50)#
#x * (2 + sqrt(50)) = 0.050 * sqrt(50)#
#x = (0.050 * sqrt(50))/(2 + sqrt(50)) = 0.0390#
This means that the equilibrium concentrations for the species involved in this reaction will be
#["H"_2] = "0.050 M" - "0.0390 M" = "0.011 M"#
#["I"_2] = "0.050 M" - "0.0390 M" = "0.011 M"#
#["HI"] = 2 * "0.0390 M" = "0.078 M"#
As predicted, the equilibrium concentration of hydrogen iodide,
Finally, to get the number of moles of iodine,
#color(blue)(c = n/V implies n = c * V)#
This will get you
#n = 0.011 "moles"/color(red)(cancel(color(black)("L"))) * 10 color(red)(cancel(color(black)("L"))) = color(green)("0.11 moles I"_2)#
I'll leave the answer rounded to two sig figs.