2HI <==> H2 + I2 Keq = 81.0
A 2.00L container is initially filled with 400 mol HI. Calculate the [HI] at equilibrium.
This is what I did so far...
HI = 4.00/2 = 2M
___2HI___<==>___H2___I2_
I 2.0M_________----__----
C -2X___________X_____X__
E 2-2X_________X______X__
81 = x^2/2-2x^2
Then I square root both sides
9 = x/2-2x
9(2-2x) = X
18-18x = X
But I'm having trouble of what to do next, or if I'm doing it right at all.... please teach me
You're ok so far. You've done the hard part. What's left? move the x to one side and numbers to the other and solve for x
18 = 19x
x = ?
i know this is kind of a stupid question... but I get divide it by 19 on both sides to get X on its own right?
so x = 0.947
with sig figs. 0.95?
Yes, that's exactly what you do.
19x = 18
19x/19 = 18/19
x = 18/19 = 0.947
But I wouldn't round to 0.95. You're allowed 3 s.f. You have 4.00 mols, 2.00 L, 81.0 for K and all of those have 3 s.f. You're multiplying and/or dividing so that allows 3 s.f. I would keep the 0.947 as is.
A couple of other points I didn't mention my last post and they aren't biggies; I let them go because I assumed they were just typos. In your post you have 400 mols and I assumed that was 4.00 since you used that in later calculations. Then in your calculation you show 81 = x^2/2-2x^2 and it should be 81 = x^2/(2-2x)^2. Again, you corrected that in the next step and I just assumed you omitted the parentheses in your haste.
I'm not grading papers today so you get by with these little typos but on an exam I would have marked some points off. That's the nice part about being retired; I don't have to be so picky about some things. ;-)
Ohh yeah I think I was typing too fast haha, thank you very much for taking so much time to help me. One final question though, because I am calculating the [HI] which 2-2x, would my answer be, 2-2(0.947) = 1.89M ?
Yes, 2-2x is right but you didn't finish.
2x is 2*0.947 = 1.89
Then 2-1.89 = 0.11 = (HI)
To solve this equilibrium problem, you are on the right track. Let's go through the steps together:
1. Start with the balanced equation: 2HI ⇌ H2 + I2
2. Assign variables to the changes in concentration. Let's use "x" for the change in concentration of H2 and I2, and "2x" for the change in concentration of HI. This is because the stoichiometry of the reaction is 2:1:1. So, if "x" H2 and I2 are formed, "2x" HI must have been consumed.
3. Write the equilibrium expression using the concentrations (or molarities) of the substances at equilibrium: Keq = [H2][I2]/[HI]^2 = 81.0
4. Calculate the concentrations of H2, I2, and HI at equilibrium using the initial concentration and the values found in step 2. Let's refer to the initial concentration of HI as [HI]0.
The final concentration of HI will be [HI] = [HI]0 - 2x.
The final concentration of H2 will be [H2] = x.
The final concentration of I2 will also be [I2] = x.
5. Substitute the equilibrium concentrations into the equilibrium expression:
Keq = [H2][I2]/[HI]^2
Substitute the values from step 4:
81.0 = (x)(x)/([HI]0 - 2x)^2
6. Solve the equation for x:
81.0([HI]0 - 2x)^2 = x^2
You can start by expanding the squared term:
81.0([HI]0 - 2x)([HI]0 - 2x) = x^2
81.0([HI]0)^2 - 162.0([HI]0)x + 324.0x^2 = x^2
Combine like terms:
81.0([HI]0)^2 - 162.0([HI]0)x + (324.0 - 1)x^2 = 0
Simplify:
(324.0 - 1)x^2 - 162.0([HI]0)x + 81.0([HI]0)^2 = 0
This is a quadratic equation. We can solve it using the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / (2a)
Substitute the values:
x = [-(-162.0([HI]0)) ± sqrt((-162.0([HI]0))^2 - 4(324.0 - 1)(81.0([HI]0)^2))] / (2(324.0 - 1))
7. After calculating the value of x, substitute it back into the equilibrium concentrations to find the equilibrium concentration of HI, [HI] = [HI]0 - 2x.
Remember to use the given values, such as the initial concentration of HI (400 mol in a 2.00 L container, so [HI]0 = 200 mol/L) and work through the calculations step by step.