# A reaction mixture initially contains 2.8 M #H_2O# and 2.6 M #SO_2#. How do you determine the equilibrium concentration of #H_2S# if Kc for the reaction at this temperature is #1.3 xx 10^-6#?

##### 1 Answer

#### Explanation:

The first thing to do here is write the equilibrium reaction

#2"SO"_ (2(g)) + 2"H"_ 2"O"_ ((g)) rightleftharpoons 2"H"_ 2"S"_ ((g)) + 3"O"_ (2(g))#

Now, you know that at a certain temperature, the equilibrium constant for this reaction is equal to

#K_c = 1.3 * 10^(-6)#

Right from the start, you can tell just by looking at the value of **lower** than the equilibrium concentration of the two reactants.

This is the case because you have **more reactants** than products.

The next thing to do here is use an **ICE table** to find the equilibrium concentration of hydrogen sulfide

#" "2"SO"_ (2(g)) " "+" " 2"H"_ 2"O"_ ((g)) rightleftharpoons 2"H"_ 2"S"_ ((g)) " "+" " 3"O"_ (2(g))#

By definition, the equilibrium constant for the reaction will be

#K_c = (["H"_2"S"]^2 * ["O"_2]^3)/(["SO"_2]^2 * ["H"_2"O"]^2)#

In your case, this expression is equivalent to

#K_c = ( (2x)^2 * (3x)^3)/( (2.6 - 2x)^2 * (2.8 - 2x)^2)#

#K_c = (4x^2 * 27x^3)/( (2.6 - 2x)^2 * (2.8 - 2x)^2)#

#K_c = (108x^5)/( (2.6 - 2x)^2 * (2.8 - 2x)^2) = 1.3 * 10^(-6)#

Now, because the value of

#2.8 - 2x ~~ 2.8" "# and#" "2.6 - 2x ~~ 2.6#

This will give you

#1.3 * 10^(-6) = (108x^5)/(2.6^2 * 2.8^2)#

which allows you to calculate

#x = root(5)( (1.3 * 2.6^2 * 2.8^2 * 10^(-3))/108) = 0.05767#

Keep in mind that because the equilibrium concentration of hydrogen sulfide is

#["H"_2"S"] = 2 xx "0.05767 M" = "0.11534 M"#

Rounded to two **sig figs**, the number of sig figs you have for the initial concentrations of sulfur dioxide and water vapor, the answer will be

#["H"_2"S"] = color(green)(|bar(ul(color(white)(a/a)color(black)("0.12 M")color(white)(a/a)|)))#

As predicted, the equilibrium concentration of hydrogen sulfide is **lower** than the equilibrium concentrations of the two reactants, which are

#["SO"_2] = 2.6 - 2 * "0.05767 M" = "2.5 M"#

#["H"_2"O"] = 2.8 - 2 * "0.05767 M" = "2.7 M"#