Question #b584a

1 Answer
Jun 5, 2017

Answer:

  • #"H"_2#: #0.322# #M#

  • #"I"_2#: #0.00749# #M#

  • #"HI"#: #0.347# #M#

Explanation:

We're asked to find the molar concentrations of the substances at equilibrium from the given data.

From the chemical equation

#"H"_2 (g) + "I"_2 (g) rightleftharpoons 2"HI" (g)#

the equilibrium-constant expression is

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

The given volume is #1# #"dm"^3#, which is equivalent to #1# #"L"#, so the given mole values in the equation are the same as their molar concentrations.

Let's create a makeshift I.C.E. chart, using bullet points:

Initial Concentration (#M#):

  • #"H"_2#: #0.496#

  • #"I"_2#: #0.181#

  • #"HI"#: #0#

Change in Concentration (#M#):

  • #"H"_2#: #-x#

  • #"I"_2#: #-x#

  • #"HI"#: #+2x#

Final Concentration (#M#):

  • #"H"_2#: #0.496-x#

  • #"I"_2#: #0.00749#

  • #"HI"#: #2x#

Since we're given both the initial and final concentrations for #"I"_2#, we can calculate the change in concentration, and thus #x#:

#x = overbrace(0.181)^("initial I"_2)-overbrace(0.00749)^("final I"_2) = color(red)(0.17351#

Now that we know #x#, we can use this to calculate the remaining equilibrium concentrations:

Final Concentrations (#M#):

  • #"H"_2#: #0.496-color(red)(0.17351) = 0.322#

  • #"I"_2#: #0.00749#

  • #"HI"#: #2(color(red)(0.17351)) = 0.347#

The equilibrium concentrations are thus

  • #"H"_2#: #0.322# #M#

  • #"I"_2#: #0.00749# #M#

  • #"HI"#: #0.347# #M#