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?

Md Ariful Islam

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Md Ariful Islam

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

$color(white)(mmmmmmml)"A"color(white)(l) +color(white)(m) "2B" ⇌ "C"$
$"I/mol·L"^"-1": color(white)(ml)0.550 color(white)(mll)1.05color(white)(m)0.550$
$"C/mol·L"^"-1": color(white)(mll)"-"xcolor(white)(mml)"+"xcolor(white)(mll)"+"x$
$"E/mol·L"^"-1": color(white)(m)"0.550-"xcolor(white)(ml)xcolor(white)(mml)x$

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

So $x = "0.550 – 0.390 = 0.160 mol/L"$

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

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

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

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