Asked by Anonymous
Gaseous CO2 decomposes to form the gaes CO and O2. At 20.0 degrees Celsius, =2.0*10^-8.
2.0 mol carbon dioxide is placed in a 5.0 L flask. What are the equilibrium concentrations of each species.
My attempt to get the answer:
Balanced equation: 2CO2--> 2CO+O2
To get the equilibrium for CO and O2, I know i have to find the initial-change=equil.
I think the Initial for CO2 is 2.0 mol and change is zero, so the equil. for CO2 is 2.0. How do I find the equil. for CO and O2?
2.0 mol carbon dioxide is placed in a 5.0 L flask. What are the equilibrium concentrations of each species.
My attempt to get the answer:
Balanced equation: 2CO2--> 2CO+O2
To get the equilibrium for CO and O2, I know i have to find the initial-change=equil.
I think the Initial for CO2 is 2.0 mol and change is zero, so the equil. for CO2 is 2.0. How do I find the equil. for CO and O2?
Answers
Answered by
DrBob222
You don't say if 2.0 x 10^-8 is Kp or Kc.
Your balanced equation is ok but thereafter is not.
2CO2 ==> 2CO + O2
If the K you omitted is Kc, then,
initial:
(CO2) = 2.0
(CO)=0
(O2)=0
change:
(CO) = +x
(O2) = +x
(CO2) = -2x
equilibrium:
(CO) = +x
(O2) = +x
(CO2) = 2.0 - 2x
Now plug these equilibrium values into Kc expression and solve.
If K is Kp, you need to change Kp to Kc or solve it with partial pressures.
Your balanced equation is ok but thereafter is not.
2CO2 ==> 2CO + O2
If the K you omitted is Kc, then,
initial:
(CO2) = 2.0
(CO)=0
(O2)=0
change:
(CO) = +x
(O2) = +x
(CO2) = -2x
equilibrium:
(CO) = +x
(O2) = +x
(CO2) = 2.0 - 2x
Now plug these equilibrium values into Kc expression and solve.
If K is Kp, you need to change Kp to Kc or solve it with partial pressures.
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