Finding initial concentration by using Kc and equilibrium concentration?!

At a certain temp the Kc for the Equillibrium reaction #H_2+ I_2 rightleftharpoons 2HI# is 51,50.
An unknown amount of HI was placed in a 1 dm^3 container. At equilibrium 0,218 mol of Hydrogen gas (H2) was present.

i)How much HI was originally placed in the container?
ii) How much I2 and HI were present at equilibrium?

1 Answer
Mar 26, 2017

Answer:

Here's what I get.

Explanation:

This looks like the appropriate time to set up an ICE table.

#color(white)(mmmmmmm)"H"_2 + "I"_2 ⇌ "2HI"#
#"I/mol·dm"^3: color(white)(mll)0color(white)(mll)0color(white)(mmm)y#
#"C/mol·dm"^3: color(white)(ll)"+"xcolor(white)(m)"+"xcolor(white)(mm)"-2"x#
#"E/mol·dm"^3: color(white)(l)"0,218"color(white)(ll)xcolor(white)(mml)y"-"2x#

The equilibrium concentration of #"H"_2# is #"0,218 mol/dm"^3#, so #x" = 0,218"#.

Our ICE table now becomes

#color(white)(mmmmmmml)"H"_2color(white)(m) +color(white)(m) "I"_2color(white)(mm) ⇌color(white)(ml) "2HI"#
#"I/mol·dm"^3: color(white)(mm)0color(white)(mmmm)0color(white)(mmmmmm)y#
#"C/mol·dm"^3: color(white)(ll)"+0,218"color(white)(m)"+0,218"color(white)(mmm)"-0,436"#
#"E/mol·dm"^3: color(white)(mll)"0,218"color(white)(mm)"0,218"color(white)(mml)y-"0,436"#

We can now insert these numbers into the equilibrium constant expression.

#K_text(eq) = "[HI]"^2/(["H"_2]["I"_2]) = (y" - 0,436")^2/("0,218" × "0,218") = "51,50"#

#(y" - 0,436")^2= "0,218"^2 × "51,50" ="2,447"#

#y" - 0,436" = 1.564"#

#y = "1,564+ 0,436" = "2,000"#

#["I"_2] = "0,218 mol·dm"^"-3"#

#["HI"] =(y" - 0,436") color(white)(l)"mol·dm"^"-3" = ("2,000 - 0,436")color(white)(l)"mol·dm"^"-3" = "1,56 mol·dm"^"-3"#

The volume of the container was #"1 dm"^3#, so

#"i)"color(white)(m)"the initial amount of HI was 2,00 mol."#
#"ii)"color(white)(ll) "the equilibrium amount of I"_2 color(white)(l)"was 0,218 mol".#
#color(white)(mll)"the equilibrium amount of HI was 1,56 mol."#