There is 0.150 moles of H2 and I2, and are brought to equilibrium in a 3.30 L flask. The equation is H2 + I2 = 2HI. Find the equilibrium amounts of H2, I2, and HI.
I'm not getting the correct answer. I set up the formula as x^2/(.04545-x) (.04545-x). To get moles I multiplied the resulting molar concentration by the 3.30 liters. what am i doing something wrong
and KC = 50.2
If you are letting x stand for the amount of H2 and I2 that combine, then HI = 2x; otherwise what you have looks ok to me.
so the equation would be 2x/(.04545-x)(.04545-x)?
i mean 2x^2?
I would use 2x^2. yes.
To solve this equilibrium problem, we will use the stoichiometry of the balanced reaction equation and the concept of the equilibrium constant (Kc).
Given:
- Initial moles of H2 = I2 = 0.150 moles
- Volume of the flask = 3.30 L
- Reaction equation: H2 + I2 ⇌ 2HI
Let's denote the amount of H2 and I2 that react and form HI as 'x' (in moles).
The initial moles of H2 and I2 are both 0.150 moles, and since they react in a 1:1 molar ratio, the amount of HI formed will also be 'x' moles.
At equilibrium, the moles of H2 and I2 will be (0.150 - x) moles, and the moles of HI will be (2x) moles.
Now, we need to set up the expression for the equilibrium constant (Kc) using the concentrations of the species at equilibrium.
Kc = [HI]^2 / ([H2] * [I2])
Since the volume of the flask is 3.30 L and we're dealing with moles, we need to divide the moles by the volume to get the molar concentrations.
Kc = [(2x / 3.30)^2] / [((0.150 - x) / 3.30) * ((0.150 - x) / 3.30)]
Simplifying the above equation gives:
Kc = (4x^2 / 10.89) / ((0.150 - x)^2 / 10.89)
To find the equilibrium amounts of H2, I2, and HI, we need to solve for 'x' in the equation for Kc.
However, the equation you mentioned, x^2 / (0.04545 - x)(0.04545 - x), seems incorrect. It appears to be using an incorrect value of 0.04545 instead of 1/3.30, which is the molar concentration of reactants and products in the flask.
So, you can solve the equation for 'x' using the correct equation for Kc and then calculate the equilibrium amounts of H2, I2, and HI using the formulas:
[H2] = (0.150 - x) / 3.30
[I2] = (0.150 - x) / 3.30
[HI] = (2x) / 3.30
Note: Make sure to calculate 'x' accurately to obtain the correct equilibrium amounts.