For the reaction H2+I2----2HI Kc is 49. Calculate the concentration of HI at equilibrium when initially one mole of H2 is mixed with one mole of I2 in 2 litre flask.?

1 Answer
Jun 25, 2017

Answer:

#["HI"] = 0.778M#

Explanation:

We're asked to find the equilibrium concentration of #"HI"# with an initial concentration of #1# #"mol H"_2# and #1# #"mol I"_2#.

The equilibrium constant expression for this reaction is

#K_c = (["HI"]^2)/(["H"_2]["I"_2]) = 49#

The initial concentrations of both #"H"_2# and #"I"_2# are

#(1color(white)(l)"mol")/(2color(white)(l)"L") = 0.5M#

So our initial concentrations for each species are

Initial:

  • #"H"_2#: #0.5M#

  • #"I"_2#: #0.5M#

  • #"HI"#: #0#

From the coefficients of the chemical equation, we can predict the changes in concentration with the quantity #x#:

Change:

  • #"H"_2#: #-x#

  • #"I"_2#: #-x#

  • #"HI"#: #+2x#

and the final concentrations are the sum of the initial and change:

Final:

  • #"H"_2#: #0.5M-x#

  • #"I"_2#: #0.5M-x#

  • #"HI"#: #2x#

It's algebra time! Let's plug these into our equilibrium constant expression to start solving for #x#:

#K_c = ((2x)^2)/((0.50-x)(0.50-x)) = 49#

#4x^2 = 49(0.25-x+x^2)#

#49x^2 - 49x + 12.25 = 4x^2#

#45x^2 - 49x + 12.25 = 0#

#x = (49 +-sqrt((-49)^2-4(45)(12.25)))/(2(45)) = 0.7# or #0.389#

We can neglect the solution that is greater than #0.5#, because then the equilibrium concentrations of hydrogen and iodine would be negative! So the one to use is #color(blue)(0.389#.

Therefore, the final equilibrium concentration of #"HI"# is

#["HI"] = 2x = 2(color(blue)(0.389)) = color(red)(0.778M#