# Question 9fff9

Aug 7, 2015

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.

${\text{CO" + "Cl"_2 ⇌ "COCl}}_{2}$

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.

${\text{ "" "" "" "" "color(white)(1)"CO"" "+" "color(white)(1)"Cl"_2" "⇌color(white)(1)"COCl}}_{2}$
$\text{I/mol·L"^-1" "0.400" "" "" "0.500" "" "" } 0.400$
"C/mol·L"^-1" " -x" "" "" "color(white)(1)-x " "" "" "+x
$\text{E/mol·L"^-1" "0.400-x" "0.500-x" } 0.400 + x$

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.

$0.400 + x = 8.00 \left(0.400 + x\right) \left(0.500 - x\right) = 8.00 \left(0.200 - 0.900 x + {x}^{2}\right) = 1.60 - 7.20 + 8 {x}^{2}$

$8 {x}^{2} - 8.20 x + 1.20 = 0$

${x}^{2} - 1.025 x + 0.150 = 0$

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!