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 = ?
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
5 answers
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?
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. ;-)
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)
2x is 2*0.947 = 1.89
Then 2-1.89 = 0.11 = (HI)