Calculate the concentration at equilibrium?

#H_2+I_2 rightleftharpoons 2HI#
#Keq = 49.5#

If 5.0M HI is initially placed into a container, what will be the equilibrium [HI]?

2 Answers
SupernovaT
Feb 14, 2018

Your numbers given are wrong. To calculate the conc. of HI we need to know the initial concentration of #H_2# and #I_2#. Assuming we have 5.0M of #H_2# and #I_2#; conc. of HI = 1M

Explanation:

#Keq = [HI]^2/([H_2][I_2])#

#49.5 = [HI]^2/(5.00*5.00)#

#49.5*25 = [HI]^2#

#sqrt1237.5 = [HI]#

[HI] = 1M

Truong-Son N.
Feb 14, 2018

I got #"3.9 M"#.

Given that #K = 49.5# for this reaction

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

we have #K' = 1/K = 0.0202# for this reaction:

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

#"I"" "5.0" "" "" "0" "" "" "0#
#"C"" "-2x" "" "+x" "+x#
#"E"" "5.0-2x" "x" "" "x#

So the mass action expression for this reaction is:

#K' = (["H"_2]_(eq)["I"_2]_(eq))/(["HI"]_(eq)^2)#

#= 0.0202 = x^2/(5.0 - 2x)^2#

This is readily solvable:

#sqrt(0.0202) = x/(5.0 - 2x)#

#5.0sqrt(0.0202) - 2sqrt(0.0202)x = x#

#5.0sqrt(0.0202) = (2sqrt(0.0202) + 1)x#

#x = (5.0sqrt(0.0202))/(2sqrt(0.0202) + 1)#

#=# #"0.553 M"#

That means the equilibrium #["HI"]# is:

#color(blue)(["HI"]_(eq)) = "5.0 M" - 2("0.553 M")#

#=# #color(blue)("3.9 M")#

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