At 500 K , equilibrium constant, $K_c$, for the following reaction is 5 .
$
\frac{1}{2} \mathrm{H}_{

Question

At 500 K , equilibrium constant, $K_c$, for the following reaction is 5 .
$
\frac{1}{2} \mathrm{H}_{2(g)}+\frac{1}{2} \mathrm{I}_{2(g)} \rightleftharpoons \mathrm{HI}_{(g)}
$
What would be the equilibrium constant $K_c$ for the reaction : $2 \mathrm{HI}_{(g)} \rightleftharpoons \mathrm{H}_{2(g)}+\mathrm{I}_{2(g)}$ ?

Professor PC Thomas Classes · Answer

(a) : $\frac{1}{2} \mathrm{H}_{2(g)}+\frac{1}{2} \mathrm{I}_{2(g)} ightleftharpoons \mathrm{HI}_{(g)} ; K_c=5$ \ Multiply eqn (i) by 2 , \ $ \mathrm{H}_{2(g)}+\mathrm{I}_{2(g)} ightleftharpoons 2 \mathrm{HI}_{(g)} ; K_c=(5)^2  $ \ Now, reverse the reaction $ 2 \mathrm{HI}_{(g)} ightleftharpoons \mathrm{H}_{2(g)}+\mathrm{I}_{2(g)} ; K_c=\frac{1}{(5)^2}=\frac{1}{25}=0.04 $