# Question #c8765

May 23, 2015

The equilibrium concentration of $D$ is 0.890 M.

The first part of the problem was already answered here, check it out:

http://socratic.org/questions/initially-only-a-and-b-are-present-each-at-2-00-m-what-is-the-final-concentratio?source=search

Now for the second part. This time, you start with 1.00 M of $A$ and 2.00 M of $B$. Once again, use an ICE table to determine the equilibrium concentration of $D$.

$\text{ "A " "+ " "B " "rightleftharpoons " "C " "+" } D$
I....1.00.............2.00...............0...................0
C....(-x)...............(-x)...............(+x)...............(+x)
E...1.00-x..........2.00-x...........x....................x

${K}_{c} = \frac{\left[C\right] \cdot \left[D\right]}{\left[A\right] \cdot \left[B\right]} = \frac{x \cdot x}{\left(1.00 - x\right) \cdot \left(2.00 - x\right)}$

${K}_{c} = {x}^{2} / \left(\left(1.00 - x\right) \left(2.00 - x\right)\right) = 6.5$

$5.5 {x}^{2} - 19.5 x + 13 = 0$
This equation will produce two solutions for $x$, 2.66 and 0.890. The first solution will be eliminated, since it implies that the equilibrium concentrations of $A$ and $B$ are negative.
This means that the equilibrium concentration of $D$ will be
$\left[D\right] = \textcolor{g r e e n}{\text{0.890 M}}$