Could someone please correct my answer to the Equilibrium reaction question?
Consider the equilibrium reaction.
H2(g) + I2(g) ⇌ 2 HI(g)
In this case, 1.000 M H2 reacts with 2.000 M of I2 at a temperature of 441°C. The value of Kc = 67.
Determine the equilibrium concentrations of H2, I2, and HI.
These were MY answers, but, they are wrong.
[H2] = 1.896 M
[I2] = 0.052
[HI] = 1.052 M
Any teaching is Much Appreciated. Thank you.
Consider the equilibrium reaction.
H2(g) + I2(g) ⇌ 2 HI(g)
In this case, 1.000 M H2 reacts with 2.000 M of I2 at a temperature of 441°C. The value of Kc = 67.
Determine the equilibrium concentrations of H2, I2, and HI.
These were MY answers, but, they are wrong.
[H2] = 1.896 M
[I2] = 0.052
[HI] = 1.052 M
Any teaching is Much Appreciated. Thank you.
1 Answer
Maybe your process was correct? Check which concentration is which.
["H"_2] = "0.0511 M"
["I"_2] = "1.0511 M"
["HI"] = "1.898 M"
Construct the ICE table:
"H"_2(g) + "I"_2(g) rightleftharpoons 2"HI"(g)
"I"" "1.000" "2.000" "" "0
"C"" "-x" "" "-x" "+2x
"E"" "1-x" "2-x" "" "2x
This gives:
K_c = 67 = (["HI"]^2)/(["H"_2]["I"_2])
= (2x)^2/((1.000 - x)(2.000 - x))
This
67(1.000 - x)(2.000 - x) = 4x^2
67(2.000 - 3.000x + x^2) = 4x^2
134.000 - 201.000x + 67x^2 = 4x^2
63x^2 - 201.000x + 134.000 = 0
This quadratic formula is set up as:
x = (-(-201.000) pm sqrt(201.000^2 - 4(63)(134.000)))/(2(63))
Once you solve this, you should get
x = "2.242 M", "0.9489 M"
The physically reasonable solution is
color(blue)(["H"_2]) = 1.000 - 0.9489 = color(blue)("0.0511 M")
color(blue)(["I"_2]) = 2.000 - 0.9489 = color(blue)("1.0511 M")
color(blue)(["HI"]) = 2(0.9489) = color(blue)("1.898 M")
Keep your answers in order, is all.
