At 500 K, equilibrium constant, Kc, for the following reaction is 5.

$\frac{1}{2}{\mathrm{H}}_{2}\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{I}}_{2}\left(\mathrm{g}\right)⇌\mathrm{HI}\left(\mathrm{g}\right)$

What would be the equilibrium constant Kc for the reaction?

$2\mathrm{HI}\left(\mathrm{g}\right)⇌{\mathrm{H}}_{2}\left(\mathrm{g}\right)+{\mathrm{I}}_{2}\left(\mathrm{g}\right)$

1.  0.04

2.  0.4

3.  25

4.  2.5

Hint: ${\mathrm{K}}_{\mathrm{c}}=\frac{\left[\mathrm{HI}\right]}{{\left[{\mathrm{H}}_{2}\right]}^{1/2}{\left[{\mathrm{I}}_{2}\right]}^{1/2}}$

Step 1:

The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward

direction
For the reaction, $\frac{1}{2}{\mathrm{H}}_{2}\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{I}}_{2}\left(\mathrm{g}\right)⇌\mathrm{HI}\left(\mathrm{g}\right)$
${\mathrm{K}}_{\mathrm{c}}=\frac{\left[\mathrm{HI}\right]}{{\left[{\mathrm{H}}_{2}\right]}^{1/2}{\left[{\mathrm{I}}_{2}\right]}^{1/2}}=5$
Step 2:
Thus, for the reaction, $2\mathrm{HI}\left(\mathrm{g}\right)⇌{\mathrm{H}}_{2}\left(\mathrm{g}\right)+{\mathrm{I}}_{2}\left(\mathrm{g}\right)$

${\mathrm{K}}_{{\mathrm{c}}_{1}}=\frac{\left[{\mathrm{H}}_{2}\right]\left[{\mathrm{I}}_{2}\right]}{{\left[\mathrm{HI}\right]}^{2}}={\left(\frac{1}{{\mathrm{K}}_{\mathrm{c}}}\right)}^{2}={\left(\frac{1}{5}\right)}^{2}=\frac{1}{25}=0.04$