# The following reaction is at equilibrium at a particular temperature:H_2(g) + I_2(g) -> 2HI(g). The [H_2]_(eq) = .012 M, [I_2]_(eq) = .15 M, and [HI]_(eq) = .30 M.What is the magnitude of K_c for the reaction?

Dec 10, 2015

${K}_{c} = 50.$

#### Explanation:

Even without doing any calculation, you should be able to look at the values given for the equilibrium concentrations of the three chemical species that take part in the reaction and predict that ${K}_{c}$ will be greater than one.

As you know, the equilibrium constant tells you what the ratio is between the equilibrium concentrations of the products and the equilibrium concentrations of the reactants, all raised to the power of their respective stoichiometric coefficients.

In this case, notice that the concentration of the product, hydrogen iodide, $\text{HI}$, is bigger than the equilibrium concentrations of the two reactants, hydrogen gas, ${\text{H}}_{2}$, and iodine, ${\text{I}}_{2}$.

Moreover, the reaction produces twice as many moles of hydrogen iodide than moles of hydrogen and iodine that take part in the reaction, you can expect ${K}_{c} > 1$.

${\text{H"_text(2(g]) + "I"_text(2(g]) -> color(red)(2)"HI}}_{\textrm{\left(g\right]}}$

the equilibrium constant takes the form

${K}_{c} = \left(\left[{\text{HI"]^color(red)(2))/(["H"_2] * ["I}}_{2}\right]\right)$

Plug in your values to get

${K}_{c} = \left({\left(0.30\right)}^{\textcolor{red}{2}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{M"^color(red)(2)))))/(0.012 color(red)(cancel(color(black)("M"))) * 0.15color(red)(cancel(color(black)("M}}}}\right) = \textcolor{g r e e n}{50.}$

Indeed, ${K}_{c} > 1$ as initially predicted.