9. Hydricarbon — Chemistry STD 11 Science — Question

${R_2}SiC{l_2} + {H_2}O \to (1)\xrightarrow{{polymerisation}}(2)$ 

Compound $(2)$ in above reaction is

${N_2}\left( g \right) + {O_2}\left( g \right)\underset{{{k_2}}}{\overset{{{k_1}}}{\longleftrightarrow}}2NO\left( g \right)$

$C_0 = Ce^{-2.1×10^{-3}\ t}$ for the forward reaction and

$C_0'= C'e^{-4.2×10^{-4}\ t}$ for the backward reaction, hence $K_c$ for the above equilibrium is