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!
