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?
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.
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.
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]
You are OK to here.
54=2x^2/(0.2-x)(0.2-x)
This should be
54 = (2x)^2/(0.2-x)(0.2-x)
54 = 4x^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?
yes, 7.348 = 2x/(0.2-x)
Then I get stuck at
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).
7.348= 4x/0.2-x
I don't know where else to go from here and if I did it correct.
To find the equilibrium concentration of HI, we need to use the given equilibrium constant (Kc) and the initial concentrations of H2 and I2.
The balanced equation for the reaction is:
H2(g) + I2(g) ⇌ 2HI(g)
The equilibrium constant expression (Kc) is defined as the ratio of the products' concentrations to the reactants' concentrations, each raised to the power of their stoichiometric coefficients. In this case, it is:
Kc = [HI]^2 / ([H2] * [I2])
Given that Kc = 54, we can rearrange the equation to solve for [HI]:
[HI]^2 = Kc * ([H2] * [I2])
Since we are given the initial concentrations of H2 and I2 as 1.0 mole each, and the volume of the container as 5.0 L, we can calculate the initial concentration of HI:
[H2] = 1.0 mole / 5.0 L = 0.2 M
[I2] = 1.0 mole / 5.0 L = 0.2 M
Substituting these values into our equation, we get:
[HI]^2 = 54 * (0.2 * 0.2)
Solving for [HI], we take the square root of both sides:
[HI] = √(54 * 0.2 * 0.2) = 0.6 M
Therefore, the equilibrium concentration of HI is 0.6 M.