the term i accidently ommited was I205(S). THANKS FOR YOUR HELP!!
6 answers
I'm still confused. I understand about the I2O5 but that won't give the Keq expression you listed. What about doing us both a favor and typing the entire question in one place; i.e., here. And is there an initial concentration of I2O5?
Since the I2O5 is a solid, it doesn't need a concentration. But if
I2O5 + 5CO ==> 5CO2 + I2, then it is balanced and
Keq = (CO2)^5(I2)/(CO)^5
which is not the same as you initially wrote Keq. The I2 was squared initially. I will assume (I2) without the squared number is the correct way to write it.
If (CO) = 1 initially
then I2 is 0 initially and
CO2 is 0 initally.
If (CO) is 0.25 at equilibrium, that means CO changed by -0.75.
(CO2) changed by + 0.75
(I2) changed by 0.75/5 = +0.15
Add them together to obtain equilibrium conditions.
(CO) = 1 - 0.75 = 0.25
(I2) = 0 + 0.15 = 0.15
(CO2) = 0 + 0.75 0.75
Plug those values into Keq expression and solve.
I2O5 + 5CO ==> 5CO2 + I2, then it is balanced and
Keq = (CO2)^5(I2)/(CO)^5
which is not the same as you initially wrote Keq. The I2 was squared initially. I will assume (I2) without the squared number is the correct way to write it.
If (CO) = 1 initially
then I2 is 0 initially and
CO2 is 0 initally.
If (CO) is 0.25 at equilibrium, that means CO changed by -0.75.
(CO2) changed by + 0.75
(I2) changed by 0.75/5 = +0.15
Add them together to obtain equilibrium conditions.
(CO) = 1 - 0.75 = 0.25
(I2) = 0 + 0.15 = 0.15
(CO2) = 0 + 0.75 0.75
Plug those values into Keq expression and solve.
i got 36 is that right?
I ran through it quickly and obtained 36.4 which rounds to 36.
thanks:) i still don't get the logic behind it... so can you please explain to me why your subtract 1.0 -.25, and then why you got .15. thanks:)
I2O5(s) + 5CO ==> 5CO2 + I2
Have you used an ICE chart? I tried to do that in the first post but I had to write them vertically as initial, change, and equilibrium. These boards can't handle more than one space at a time, SO, let me put periods for spacing. Just ignore the periods. For example
...1..........2..............3........4
would be four columes labeled 1, 2, 3, and 4, and I could put things under those columns like this
...1...........2..............3........4
..54.6........18.4...........16.4.....10
etc.
In addition, I will add a number in parentheses which will be a sequence number. That will tell you the sequence in which you are to read the numbers.
I = Initial
C = Change
E = Equilibrium
The equation again:
I2O5(s) + 5CO ==> 5CO2 + I2
I.......(1)1M...(1)0M..(1).0M
C.. (3)-0.75 ...(4)0.75..(4)0.15
E..(2)..0.25....(5)0.75..(5)0.15
Here is the way the ICE chart is formed. Sequence 1: The problem states that the CO is 1 M initially. Of course CO2 and I2 are 0 since the reaction hasn't started. Next, you go to sequence 2; the problem states that at equilibrium the CO is 0.25 so that goes at E conditions and we immediately know that (sequence 3) that the concn was 1 to begin, 0.25 at the end and it must have changed by -0.75.
That allows us to go to sequence 4 and fill in the CO2 and I2. The coefficients of CO and CO2 are 5; therefore, if CO changed by -0.75, we know CO2 must have changed by +0.75. And we know I2 changed by 0.75 x (1 mol I2/5 mol CO) = 0.15. The final sequence is (5) which just adds the Initial value to the change value to obtain the equilibrium value. I hope this helps.
Have you used an ICE chart? I tried to do that in the first post but I had to write them vertically as initial, change, and equilibrium. These boards can't handle more than one space at a time, SO, let me put periods for spacing. Just ignore the periods. For example
...1..........2..............3........4
would be four columes labeled 1, 2, 3, and 4, and I could put things under those columns like this
...1...........2..............3........4
..54.6........18.4...........16.4.....10
etc.
In addition, I will add a number in parentheses which will be a sequence number. That will tell you the sequence in which you are to read the numbers.
I = Initial
C = Change
E = Equilibrium
The equation again:
I2O5(s) + 5CO ==> 5CO2 + I2
I.......(1)1M...(1)0M..(1).0M
C.. (3)-0.75 ...(4)0.75..(4)0.15
E..(2)..0.25....(5)0.75..(5)0.15
Here is the way the ICE chart is formed. Sequence 1: The problem states that the CO is 1 M initially. Of course CO2 and I2 are 0 since the reaction hasn't started. Next, you go to sequence 2; the problem states that at equilibrium the CO is 0.25 so that goes at E conditions and we immediately know that (sequence 3) that the concn was 1 to begin, 0.25 at the end and it must have changed by -0.75.
That allows us to go to sequence 4 and fill in the CO2 and I2. The coefficients of CO and CO2 are 5; therefore, if CO changed by -0.75, we know CO2 must have changed by +0.75. And we know I2 changed by 0.75 x (1 mol I2/5 mol CO) = 0.15. The final sequence is (5) which just adds the Initial value to the change value to obtain the equilibrium value. I hope this helps.