Question #9fff9

1 Answer

Answer:

The new concentrations are #["COCl"""_2] = "0.577 mol/L"#, #["CO"] = "0.223 mol/L"#, and #["Cl"_2] = "0.323 mol/L"#

Explanation:

The first step is to determine the equilibrium constant for the reaction.

#"CO" + "Cl"_2 ⇌ "COCl"""_2#

#K_"eq" = (["COCl"""_2]) /(["CO"]["Cl"_2]#

# K_"eq" = 0.400/(0.100 × 0.500) = 8.00#

Now we can set up an ICE table to calculate the new concentrations.

#" "" "" "" "" "color(white)(1)"CO"" "+" "color(white)(1)"Cl"_2" "⇌color(white)(1)"COCl"""_2#
#"I/mol·L"^-1" "0.400" "" "" "0.500" "" "" "0.400#
#"C/mol·L"^-1" " -x" "" "" "color(white)(1)-x " "" "" "+x#
#"E/mol·L"^-1" "0.400-x" "0.500-x" "0.400+x#

#K_"eq" = (["COCl"""_2]) /(["CO"]["Cl"_2]) = (0.400+x)/(( 0.400-x)(0.500-x)) = 8.00#

We can't assume that #x≪0.400#, so we must solve a quadratic equation.

#0.400+x = 8.00(0.400+x)(0.500-x) = 8.00(0.200-0.900x+x^2) = 1.60-7.20+8x^2#

#8x^2-8.20x+1.20=0#

#x^2-1.025x+0.150 = 0#

#x= (-b±sqrt(b^2-4ac))/(2a) = (1.025±sqrt(1.025^2-4×1×0.150))/(2×1) = (1.025-sqrt(1.051-0.600))/2 = (1.025±sqrt0.451)/2 = (1.025±0.671)/2#

#x = 0.177# or #x = 0.848#

Since #x# cannot be greater than 0.400, #x=0.177#

The new concentrations are

#["COCl"""_2] = (0.400+x) "mol/L" = (0.400+0.177) "mol/L" = "0.577 mol/L"#

#["CO"] = (0.400-x) "mol/L" = (0.400-0.177) "mol/L" = "0.223 mol/L"#

#["Cl"_2] = (0.500-x) "mol/L" = (0.500-0.177) "mol/L" = "0.323 mol/L"#

Check:

#0.577/(0.223×0.323) = 8.01#.

Close enough!