# Question #2c380

Apr 30, 2015

The equilibrium constant for this reaction will be ${K}_{\text{c}} = 3.13$.

So, you've got a generic equilibrium reaction and you're asked to determine the equilibrium constant, ${K}_{c}$. In order to do that, you need to know the equilibrium concentration of all the species involved in the reaction.

Since you only know the equilibrium concentrations of $\text{A}$ and $\text{C}$, you can use the initial concentrations to determine what you're missing - the equilibrium concentration of $\text{C}$.

Before doing any actual calculations, you can predict that ${K}_{c}$ will be greater than 1, since the equilibrium concentration of the product increases and the equilibrium concentration of the $A$ decreases - compared with their initial cocnentrations.

Use an ICE table (more here: https://en.wikipedia.org/wiki/RICE_chart) to find a relationship between the initial and the equilibrium concentrations of all the species

$\text{ "A + " " " "color(red)(2)B " "rightleftharpoons " } C$
I....0.650..........1.20...............0.700
C...(-x).............(-$\textcolor{red}{2}$x)................(+x)
E.0.650-x........1.20-2x.............0.700+x

By definition, the equilibrium constant will be

${K}_{c} = \frac{0.700 + x}{\left(0.650 - x\right) \cdot {\left(1.20 - 2 x\right)}^{2}}$

You know that

$0.650 - x = 0.450 \implies x = 0.200$

(you get the same result if you use 0.700 + x = 0.900)

Replace the value of $x$ into the equation and you'll get

${K}_{c} = \frac{0.700 + 0.200}{\left(0.650 - 0.200\right) \cdot {\left(1.20 - 2 \cdot 0.200\right)}^{2}}$

${K}_{c} = \frac{0.900}{0.450 \cdot {0.800}^{2}} = 3.125$

Rounded to three sig figs, the answer will be

${K}_{c} = \textcolor{g r e e n}{3.13}$