Keq for the reaction 2HI <----> H2 +I2 has a value of 1.85x 10^-2 at 425 degrees celsius. If 0.18 mol of HI is placed in a 2.0 L flask and allowed to come to equilibrium at this temperature. What will the equilibrium of [I2} be?
To find the equilibrium concentration of I2, we will first need to set up an ICE (Initial, Change, Equilibrium) table and use the given information.
The balanced chemical equation for the reaction is:
2HI <----> H2 + I2
Let's start by determining the initial concentrations in moles per liter (M) for each substance. We are given that 0.18 mol of HI is placed in a 2.0 L flask, so the initial concentration of HI is calculated as follows:
[HI]initial = moles of HI / volume of flask
[HI]initial = 0.18 mol / 2.0 L
[HI]initial = 0.09 M
Since we have a value for Keq, we can use it to determine the equilibrium concentrations using the equation:
Keq = [H2][I2] / [HI]^2
We know the Keq value is 1.85 x 10^-2, but we don't have the equilibrium concentrations yet. Let's set up the ICE table and assume x mol/L of I2 is formed at equilibrium:
2HI <----> H2 + I2
Initial: 0.09 0 0
Change: -2x +x +x
Equilibrium: 0.09 - 2x x x
Plugging these values into the equation for Keq:
1.85 x 10^-2 = (x)(x) / (0.09 - 2x)^2
Simplifying this equation:
1.85 x 10^-2 = x^2 / (0.09 - 2x)^2
Now, we can solve this equation for x. Rearranging it:
1.85 x 10^-2 * (0.09 - 2x)^2 = x^2
Expanding and rearranging further:
0.03075 - 2.22x + 0.049x^2 = x^2
Combining like terms:
0.049x^2 - x^2 + 2.22x - 0.03075 = 0
Solving this quadratic equation using any suitable method will give you the value of x, which represents the equilibrium concentration of I2 in mol/L.