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 andheated to a particular temperature, the equilibrium concentration of CO2(g) is measured to be 0.039 mol/L. Use an ICE table to determine the equuilibrium concentrations of CO(g) and O2(g)
It's difficult to do spacing on these boards but if I turn the equation down instead of sideways, we may be able to do it. First will be I, followed by C and E horizontally. Let me know if this is too confusing. (First, (CO2) = 2.0 mols/5 L = 0.4 mols/L.
2CO2(g)  0.4   -2x   0.039
    |
    |
    v
2CO(g)     0   +2x
    +
O2(g)     0   +x
If equilibrium for CO2 is 0.039 and it was 0.4 initially, that means it must have changed by 0.4 - 0.039 = 0.361 and that is 2x. Then CO must be the same and O2 must be 1/2 that.
Calculate the solubility of zinc hydroxide at 25 °C. The Ksp of Zn(OH)2(s) is 4.5 x 10^-17 at 25 °C?
What did I do wrong?
Zn(OH)2 <----> Zn2+ + 2 OH-
let x = moles/L of Zn(OH)2 that dissolve. This gives us x moles/L of Zn2+ and 2x moles / L of OH-
[Zn2+] = x
[OH-] = 2x
Ksp = [Zn2+][OH-]^2 = x ( 2x)^2 = 4x^3
4.5 x 10^-17 = 4x^3
x = molar solubility = 2.2 x 10^-6 M
2.2 x 10^-6 M looks ok to me. Perhaps they want g/L instead of mols/L.
To determine the equilibrium concentrations of CO(g) and O2(g), we can use an ICE table.
Firstly, let's write down the balanced equation and express it in terms of moles, taking into account the initial concentration and the changes that occur during the reaction:
2CO2(g) ---> 2CO(g) + O2(g)
Initial: 2.0 mol 0 mol 0 mol
Change: -2x 2x x
Equilibrium: 2.0 - 2x 2x x
Here, 'x' represents the change in concentration at equilibrium.
According to the problem, the equilibrium concentration of CO2(g) is measured to be 0.039 mol/L. We can use this information to set up an equation:
0.039 mol/L = 2x
Solving for 'x,' we find:
x = 0.039 mol/L / 2
x = 0.0195 mol/L
Now we can substitute this value back into the ICE table to find the equilibrium concentrations:
Equilibrium: CO2 = 2.0 - (2 * 0.0195) = 2.0 - 0.039 = 1.961 mol/L
CO = 2 * 0.0195 = 0.039 mol/L
O2 = 0.0195 mol/L
Therefore, the equilibrium concentrations of CO(g) and O2(g) are approximately 0.039 mol/L and 0.0195 mol/L, respectively.