Question #9a69f

1 Answer
Sep 14, 2017

Answer:

#K_c = 1#

Explanation:

Start by calculating the initial concentrations of the two reactants and the equilibrium concentration of the product.

#["H"_ 2]_ 0 = "4.5 moles"/"10 L" = "0.45 mol L"^(-1)#

#["I"_ 2]_ 0 = "4.5 moles"/"10 L" = "0.45 mol L"^(-1)#

#["HI"] = "3 moles"/"10 L" = "0.30 mol L"^(-1)#

Now, the equilibrium reaction given to you looks like this

#"H"_ (2(g)) + "I"_ (2(g)) rightleftharpoons color(red)(2)"HI"_ ((g))#

By definition, the equilibrium constant that describes this reaction is equal to

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

In order to find the equilibrium concentrations of the two reactants, use the fact that the reaction consumes equal numbers of moles of hydrogen gas and iodine gas and that you get #color(red)(2)# moles of hydrogen iodide for every #1# mole of hydrogen gas and iodine gas that takes part in the reaction.

You can say that in order for the equilibrium mixture to contain #"0.3 mol L"^(-1)# of hydrogen iodide, the reaction must consume #"1.15 mol L"^(-1)# of hydrogen gas and #"0.15 mol L"^(-1)# of iodine gas.

This is the case because

#3color(red)(cancel(color(black)("moles HI"))) * "1 mole H"_2/(color(red)(2)color(red)(cancel(color(black)("moles HI")))) = "1.5 moles H"_2#

which is equivalent to

#["H"_ 2]_ "consumed" = "1.5 moles"/"10 L" = "0.15 mol L"^(-1)#

#["I"_ 2]_ "consumed" = "1.5 moles"/"10 L" = "0.15 mol L"^(-1)#

The equilibrium concentrations of the two reactants will be

#["H"_ 2] = ["H"_ 2]_ 0 - ["H"_ 2]_ "consumed"#

#["H"_ 2] = "0.45 mol L"^(-1) - "0.15 mol L"^(-1)#

#["H"_ 2] = "0.30 mol L"^(-1)#

and

#["I"_ 2] = "0.30 mol L"^(-1)#

This means that the equilibrium constant for this reaction at a specific temperature is equal to--I'll show the calculations without added units!

#K_c = (0.3)^color(red)(2)/(0.3 * 0.3) = 1#

The fact that #K_c# is equal to #1# tells you that at equilibrium, the mixture will contain approximately equal--equal in your case--of reactants and products, i.e. the reaction reaches equilibrium as an intermediate mixture.