# A mixture initially contains A, B, and C in the following concentrations: [A]=0.550 M, [B]=1.05 M, [C]=.550 M. The following reaction occurs and equilibrium is established: A+2B <-> C At equilibrium, [A]=.390 M and [C]=0.710 M. What is the value of the equilibrium constant, Kc?

Oct 7, 2014

The equilibrium constant is 3.4.

#### Explanation:

First, write the balanced chemical equation with an ICE table.

$\textcolor{w h i t e}{m m m m m m m l} \text{A"color(white)(l) +color(white)(m) "2B" ⇌ "C}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l} 0.550 \textcolor{w h i t e}{m l l} 1.05 \textcolor{w h i t e}{m} 0.550$
$\text{C/mol·L"^"-1": color(white)(mll)"-"xcolor(white)(mml)"+"xcolor(white)(mll)"+} x$
$\text{E/mol·L"^"-1": color(white)(m)"0.550-} x \textcolor{w h i t e}{m l} x \textcolor{w h i t e}{m m l} x$

At equilibrium, ["A"] = "0.390 mol/L" = ("0.550 -"x) "mol/L"

So $x = \text{0.550 – 0.390 = 0.160 mol/L}$

["B"] = ("1.05 - 2"x)color(white)(l)"mol/L" = "(1.05 – 2×0.160) mol/L =0.73 mol/L"

["C"] = ("0.550 +"x)color(white)(l) "mol/L" = "(0.550 + 0.160) mol/L = 0.710 mol/L"

K_"c" = "[C]"/("[A][B]"^2) = 0.710/( 0.390 × 0.73^2) = 3.4