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"#

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

1 Answer
Mar 3, 2018

The equilibrium concentration of #"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!