Question

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`

Answer 1

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!