# The K_c for the equation shown below is 0.543 at 425^@"C". What is the value of K_c for the reverse equation?

## $\text{H"_2(g) + "I"_2(g) rightleftharpoons 2"HI} \left(g\right)$

Mar 15, 2018

$1.84$

#### Explanation:

The thing to remember here is that the equilibrium constant of the forward reaction is always equal to the inverse of the equilibrium constant of the reverse reaction or vice versa.

K_ (c \ "forward") = 1/K_ (c \ "reverse"

So in your case, you would have

${K}_{c \setminus \text{reverse") = 1/K_ (c \ "forward}}$

${K}_{c \setminus \text{reverse}} = \frac{1}{0.543} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.84}}}$

To prove that this is the case, simply write the expression of the equilibrium constant for the forward reaction

${\text{H"_ (2(g)) + "I"_ (2(g)) rightleftharpoons 2"HI}}_{\left(g\right)}$

You will have

K_ (c\ "forward") = (["HI"]^2)/(["H"_2] * ["I"_2])

Now do the same for the reverse reaction

$2 {\text{HI"_ ((g)) rightleftharpoons "H"_ (2(g)) + "I}}_{2 \left(g\right)}$

This time, you will have

K_ (c \ "reverse") = (["H"_2] * ["I"_2])/(["HI"]^2)

As you can see, you have

${K}_{c \setminus \text{forward") = (["HI"]^2)/(["H"_ 2] * ["I"_ 2]) = 1/((["H"_ 2] * ["I"_ 2])/(["HI"]^2)) = 1/K_ (c \ "reverse}}$

and so

${K}_{c \setminus \text{reverse}} = \frac{1}{0.543} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.84}}}$

The answer is rounded to three sig figs.