When carbon dioxide is heated in a closed container, it decomposes into carbon monoxide and oxygen according to the following equilibrium equation: 2CO2(g) 2CO(g) + O2(g)
When 2.0 mol of CO2(g) is placed in a 5.0-L closed container and heated to a particular temperature, the equilibrium concentration of CO2(g) is measured to be 0.39 mol/L. Use an ICE table to determine the equilibrium concentrations of CO(g) and O2(g).
I worked this problem for Thomas last night. Here is a link.
I worked this problem for Thomas last night. Here is a link.
http://www.jiskha.com/display.cgi?id=1206928168
To solve the problem using an ICE table, we need to determine the initial concentration (I), the change in concentration (C), and the equilibrium concentration (E) for each substance involved in the reaction.
Given:
Initial concentration of CO2(g) (I) = 2.0 mol / 5.0 L = 0.40 mol/L (since the volume and moles are directly proportional)
Equilibrium concentration of CO2(g) (E) = 0.39 mol/L
Using the balanced equation:
2CO2(g) ⇌ 2CO(g) + O2(g)
Let x be the change in concentration for both CO(g) and O2(g). Therefore, the change in concentration for CO2(g) would be 2x (since 2 moles of CO2 produce 2 moles of CO).
The ICE table is as follows:
2CO2(g) ⇌ 2CO(g) + O2(g)
I: 0.40 0 0
C: -2x +2x +x
E: 0.40 - 2x 2x x
According to the problem, the equilibrium concentration of CO2(g) is measured to be 0.39 mol/L. Thus:
0.40 - 2x = 0.39
2x = 0.40 - 0.39
2x = 0.01
x = 0.01 / 2
x = 0.005
So, the equilibrium concentrations are:
[CO2(g)] = 0.40 - 2x = 0.40 - 2(0.005) = 0.39 mol/L
[CO(g)] = 2x = 2(0.005) = 0.01 mol/L
[O2(g)] = x = 0.005 mol/L
Therefore, at equilibrium, the concentration of CO(g) is 0.01 mol/L, and the concentration of O2(g) is 0.005 mol/L.
To determine the equilibrium concentrations of CO(g) and O2(g), we will use an ICE table.
An ICE table stands for Initial, Change, and Equilibrium. It helps us organize and calculate the changes in concentration during a reaction.
1. Write down the balanced equation:
2CO2(g) ⇌ 2CO(g) + O2(g)
2. Establish the initial concentrations:
- CO2(g) starts with 2.0 mol in a 5.0-L container, so the initial concentration is 2.0 mol/5.0 L = 0.4 mol/L.
- CO(g) and O2(g) start with 0 mol, so their initial concentrations are both 0 mol/L.
3. Define the changes:
- Two moles of CO2 decompose to form two moles of CO and one mole of O2.
- Therefore, the change in CO2 is -2x, the change in CO is +2x, and the change in O2 is +x.
4. Express the equilibrium concentrations in terms of x:
The equilibrium concentration of CO2 is given as 0.39 mol/L. Therefore, the final concentration of CO2 is (0.4 - 2x) mol/L.
The equilibrium concentrations of CO and O2 are both 2x and x mol/L, respectively.
5. Set up the equilibrium expression:
Kc = [CO]^2[O2]/[CO2]^2
Where [CO], [O2], and [CO2] represent the equilibrium concentrations.
6. Substitute the equilibrium concentrations into the equilibrium expression:
Kc = ([CO]^2[O2])/([CO2]^2)
Kc = (2x)^2(x)/(0.4 - 2x)^2
7. Solve for x:
We can use the given value of 0.39 mol/L to solve for x. Plug in 0.39 for [CO2] and solve for x:
Kc = (2x)^2(x)/(0.4 - 2x)^2
0.39^2 = (2x)^2(x)/(0.4 - 2x)^2
Solving this equation will give us the value of x, which represents the equilibrium concentrations of CO(g) and O2(g).