Question

CO (g) + ½ O2(g) ↔ CO2(g In an experiment, 4.95 mol of CO2, 0.050 mol of CO and 0.050 mol of O2 are placed in a 5.0 L reaction vessel at 1400 °K. If Kc = 1.05 x 10-5, what will the equilibrium concentrations be for each reactant and product?

Answer 1

The new concentrations are `["CO"] = "1.00 mol/L"`, `["O"_2] = "0.50 mol/L"`, and `["CO"_2] = 7.5 × 10^-6 "mol/L"`

Explanation:

The initial concentration are.

`["CO"]_0 = "0.050 mol"/"5.0 L" = "0.010 mol/L"`

`["O"_2]_0 = "0.050 mol"/"5.0 L" = "0.010 mol/L"`

`["CO"_2]_0 = "4.95 mol"/"5.0 L" = "0.99 mol/L"`

The equation for the reaction is

`"CO" + 1/2"O"_2 ⇌ "CO"_2`

`K_"c" = (["CO"_2]) /(["CO"]["O"_2]^(1/2)) = 1.05×10^-5`

If we solve in the usual way, we get

`K_"c" = (0.99-x)/((0.010+x)(0.010+x)^(1/2)) = 1.05 × 10^-5`

If `x≪0.010`,

`K_"c" = 0.99/(0.010×0.010^(1/2)) = 1.05 × 10^-5`

There is no `x` for which to solve.

The way around this is to recognize that the system will reach equilibrium no matter what the starting point is.

So, let's decompose all the `"CO"_2` to starting materials and calculate the new equilibrium from that starting point.

The new initial concentrations are

`["CO"]_0 =("0.010 +0.99) mol/L" = "1.00 mol/L"`

`["O"_2]_0 = ("0.010+"0.99/2)color(white)(1)"mol/L" = "(0.010+0.495) mol/L" = "0.505 mol/L"`

`["CO"_2]_0 = ("0.99-0.99) mol/L" = "0 mol/L"`

If we set up an ICE table to calculate the new concentrations, we get

`" "" "" "" "" "" "color(white)(1)"CO"" "+" "color(white)(1)1/2"O"_2" "⇌" ""CO"_2`
`"I/mol·L"^-1":"" "" "1.00" "" "" "color(white)(1)0.505" "" "" "" "0`
#"C/mol·L"^-1":"" " -x" "" "" "color(white)(1)-1/2x " "" "" "color(white)(1)+x#
`"E/mol·L"^-1":"" "1.00-x" "color(white)(1)0.505-1/2x" "" "color(white)(1)x`

`K_"c" = (["CO"_2]) /(["CO"]["O"_2]^(1/2)) = x/((1.00-x)(0.505-1/2x)^(1/2)) = 1.05×10^-5`

Now we can assume that `x≪0.505`, and the equation becomes

`x/(1.00×sqrt0.505) = 1.05×10^-5`

`x = sqrt0.505 × 1.05 × 10^-5 = 7.46 × 10^-6`

The new concentrations are

`["CO"] = (1.00-x) "mol/L" = (1.00-7.46 × 10^-6)color(white)(1)"mol/L" = "1.00 mol/L"`

`["O"_2] = (0.505-x) "mol/L" = (0.505-7.46 × 10^-6)color(white)(1)"mol/L" = "0.50 mol/L"`

`["CO"_2] = x color(white)(1)"mol/L" = 7.5 × 10^-6color(white)(1)"mol/L"`

Check:

`(7.5 × 10^-6)/(1.00×sqrt0.50) = 1.1×10^-5`.

Close enough!