Question

At a particular temperature, Kc = 54 for the reaction

If 1.0 mole of H2 and 1.0 mole of I2 are placed in a 5.0 L container, what would be the equilibrium concentration of HI?

Answer 1

Write the equation and balance it. Set up an ICE chart, substitute into the expression for Kc and solve. Post your work if you get stuck.

Answer 2

H2 + I2 = 2HI
I 1.0/5.0 1.0/5.0 0


C -x -x 2x


E 0.2-x 0.2-x 2x


54= [HI]^2/[H2]*[I]


54=2x^2/(0.2-x)(0.2-x)


need to solve for x
then would I take square root of both sides to get rid of the squares?

Then I get stuck at

7.348= 4x/0.2-x

I don't know where else to go from here and if I did it correct.

Answer 3

H2 + I2 = 2HI
I 1.0/5.0 1.0/5.0 0


C -x -x 2x


E 0.2-x 0.2-x 2x


54= [HI]^2/[H2]*[I]
<b>You are OK to here.</b>

54=2x^2/(0.2-x)(0.2-x)
<b>This should be
54 = (2x)^2/(0.2-x)(0.2-x)
54 = 4x^2(0.2-x)(0.2-x).</b>

need to solve for x
then would I take square root of both sides to get rid of the squares?
<b>yes, 7.348 = 2x/(0.2-x)</b>
Then I get stuck at
<b>7.348*(0.2-x) = 2x
1.4696-7.348x = 2x
1.4696 = 2x+7.348x
1.46960 = 9.348x
x = 0.1572 but check my arithmetic.
So HI is 2x and H2 and I2 are 0.2-x. Then I would round everything to two places (the 1.0 and 5.0 gives us two significant figures).</b>
7.348= 4x/0.2-x

I don't know where else to go from here and if I did it correct.