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]?
If 5.0M HI is initially placed into a container, what will be the equilibrium [HI]?
2 Answers
Your numbers given are wrong. To calculate the conc. of HI we need to know the initial concentration of
Explanation:
[HI] = 1M
I got
Given that
#"H"_2(g) + "I"_2(g) rightleftharpoons 2"HI"(g)#
we have
#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
#color(blue)(["HI"]_(eq)) = "5.0 M" - 2("0.553 M")#
#=# #color(blue)("3.9 M")#