Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached?

The reversible chemical reaction
"A"+"B"rightleftharpoons"C"+"D"A+B⇌C+D

has the following equilibrium constant:
K_c=([C][D])/([A][B])=1.8Kc=[C][D][A][B]=1.8

1 Answer
Ernest Z.
Mar 3, 2018

The equilibrium concentration of "A"A is 0.85 mol/L.

Explanation:

We can solve this problem by setting up an ICE table.

color(white)(mmmmmmml)"A" +color(white)(m) "B" ⇋ color(white)(ll)"C" + "D"
"I/mol·L"^"-1": color(white)(mll)2.0color(white)(mm)2.0color(white)(mmll)0color(white)(mll)0
"C/mol·L"^"-1": color(white)(mll)"-"xcolor(white)(mml)"-"xcolor(white)(mm)"+"xcolor(white)(m)"+"x
"E/mol·L"^"-1": color(white)(m)"2.0-"xcolor(white)(m)"2.0-"xcolor(white)(mm)xcolor(white)(mll)x

K_text(c) = (["C"]["D"])/(["A"]["B"]) = (x × x)/(("2.0-"x)("2.0-"x)) =x^2/("2.0-"x)^2 = 1.8

x/("2.0-"x) = sqrt(1.8) = 1.34

x = 1.34(2.0 - x) = 2.68 - 1.34x

2.34x = 2.68

"x = 2.68/2.34 = 1.15

∴ ["A"] = "(2.0 - 1.15) mol/L = 0.85 mol/L"

Check:

1.15^2/0.85^2 = 1.8

It checks!

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