Consider the followingStatement-I: Kolbe's electrolysis of sodium propionate gives n-hexane as product. Statement-II: In Kolbe's process, CO$_2$ is liberated at anode and H$_2$ is liberated at cathode.
Match the following polymers with their applications:
At 700 K, the equilibrium constant $K_e$ for the reaction $ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) $ is 0.2 mol L$^{-2}$. What is the value of $K$ for the reverse reaction?