CO(g) + CL2(g) <--> Cl2CO(g)
Set up the ICE chart and substitute the equilibrium values into the Kdq you have writte. Calculate Keq from that. Can you set up the ICE chart?
The initial concentration for the compounds involved in the reaction shown were determine to be [CO]=1.80, [CL2]=9.70; [Cl2CO]=0. Calculate the value of the equilibrium constant (Kc) at 527C, if the equilibrium concentration of CO was 1.05E-3.
CO(g) + CL2(g) <--> Cl2CO(g)
so what i am thinking is i first set it as:
Kc= [Cl2CO]/[CO][CL2]
but after that i am lost and don't know what to do...
9 answers
What is an ICE chart?
By the way, you need to specify if these concns are partial pressures, molarity, or mols in a particular volume.
it is molarity
Is the ICE chart is the same as a list of initial, change and equilibrium values?
CO(g) + Cl2(g) <--> Cl2CO(g)
K= what you wrote.
I = initial concns (before any reaction):
(CO) = 1.80 M from the problem.
(Cl2) = 9.70 M from the problem.
(Cl2CO) = 0
C = change in concn:
(Cl2CO) = +x
(Cl2) = -x
(CO) = -x
E = equilibrium concns:
(Cl2CO) = 0+x = x
(Cl2) = 9.70-x
(CO) = 0.00105 (from the problem)
If we started with 1.80 M for (CO) and we ended up with 0.0105, then how much reacted? That MUST be 1.80-0.0105 (that makes x = 0.0105 doesn't it) = 1.7995 which you might keep as 1.799 or round to 1.80--not much difference.
Now go through and assign equilibrium values to CL2 (that is 9.70-x) and to Cl2CO (that is +x)
Plug those values into Keq that you have set up and solve for K. I don't know how picky your teacher is about significant figures, but if so you need to watch them at the end. You may have 3 s.f.
Any questions?
K= what you wrote.
I = initial concns (before any reaction):
(CO) = 1.80 M from the problem.
(Cl2) = 9.70 M from the problem.
(Cl2CO) = 0
C = change in concn:
(Cl2CO) = +x
(Cl2) = -x
(CO) = -x
E = equilibrium concns:
(Cl2CO) = 0+x = x
(Cl2) = 9.70-x
(CO) = 0.00105 (from the problem)
If we started with 1.80 M for (CO) and we ended up with 0.0105, then how much reacted? That MUST be 1.80-0.0105 (that makes x = 0.0105 doesn't it) = 1.7995 which you might keep as 1.799 or round to 1.80--not much difference.
Now go through and assign equilibrium values to CL2 (that is 9.70-x) and to Cl2CO (that is +x)
Plug those values into Keq that you have set up and solve for K. I don't know how picky your teacher is about significant figures, but if so you need to watch them at the end. You may have 3 s.f.
Any questions?
i set up the equation as:
Keq= x/(17.36-1.7895x)
where would i go from here?
Keq= x/(17.36-1.7895x)
where would i go from here?
I made two typos in my response. I typed If we started with 1.80 M for (CO) and we ended up with 0.0105, then how much reacted? That MUST be 1.80-0.0105 (that makes x = 0.0105 doesn't it) = 1.7995 which you might keep as 1.799 or round to 1.80--not much difference. but it should have read, If we started with 1.80 M for (CO) and we ended up with 0.00105, then how much reacted? That MUST be 1.80 - 0.00105 (that makes x = 1.80 - 0.00105 = 1.80 for all practical purposes.
Keq = (Cl2CO)/(Cl2)(CO)
You KNOW (Cl2CO). It's x. Didn't we put a number on x? Put the number in, not x. You know what it is.
Now redo your E part of the ICE table and redo the Keq. I don't have the foggiest where you found (17.36-1.7895x).
(Cl2O) = 1.80 M in round numbers but you may want to make it closer than that.
(CO) = 0.00105 M
(Cl2) = 9.70-1.80=7.90
Check my work. It's late and I'm bleary eyed.
Keq = (Cl2CO)/(Cl2)(CO)
You KNOW (Cl2CO). It's x. Didn't we put a number on x? Put the number in, not x. You know what it is.
Now redo your E part of the ICE table and redo the Keq. I don't have the foggiest where you found (17.36-1.7895x).
(Cl2O) = 1.80 M in round numbers but you may want to make it closer than that.
(CO) = 0.00105 M
(Cl2) = 9.70-1.80=7.90
Check my work. It's late and I'm bleary eyed.
thanks for your help
1 answer