# 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?

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#"H"_2(g) + "I"_2(g) rightleftharpoons 2"HI"(g)#

##### 1 Answer

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

#K_ (c \ "reverse") = 1/0.543 = color(darkgreen)(ul(color(black)(1.84)))#

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

#"H"_ (2(g)) + "I"_ (2(g)) rightleftharpoons 2"HI"_ ((g))#

You will have

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

Now do the same for the **reverse reaction**

#2"HI"_ ((g)) rightleftharpoons "H"_ (2(g)) + "I"_ (2(g))#

This time, you will have

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

As you can see, you have

#K_ (c \ "forward") = (["HI"]^2)/(["H"_ 2] * ["I"_ 2]) = 1/((["H"_ 2] * ["I"_ 2])/(["HI"]^2)) = 1/K_ (c \ "reverse")#

and so

#K_ (c \ "reverse") = 1/0.543 = color(darkgreen)(ul(color(black)(1.84)))#

The answer is rounded to three **sig figs**.