Prediction of amount of products and reactants on the base of vale of equilibrium constant of the following reactions at constant temperature.(a)H_(2(g)) hArr 2HI_((g)) , K_c=57.0 (b)N_2O_(4(g)) hArr 2NO_(2(g)), K_c=4.64xx10^(-3) (c)N_(2(g)) + O_(2(g)) hArr 2NO_((g)) , K_c=4.8xx10^(-31) (d) H_(2(g)) + Cl_(2(g)) hArr 2HCl_((g)) , K_c=4.0xx10^31

Prediction of amount of products and reactants on the base of vale of

1.	Prediction of amount of products and reactants on the base of vale of equilibrium constant of the following reactions at constant temperature.(a)H_(2(g)) hArr 2HI_((g)) , K_c=57.0 (b)N_2O_(4(g)) hArr 2NO_(2(g)), K_c=4.64xx10^(-3) (c)N_(2(g)) + O_(2(g)) hArr 2NO_((g)) , K_c=4.8xx10^(-31) (d) H_(2(g)) + Cl_(2(g)) hArr 2HCl_((g)) , K_c=4.0xx10^31
Answer» Solution :Reaction (a) : The value of `K_c` is between `10^(-3)` to `10^3` [Products] `GT` [Reactants] because `K_c gt1`. Reaction (b) : The value of `K_c` is between `10^(-3)` to `10^(+3)` So, [Products] `LT`[REACTANT] because `K_c lt 1`. Reaction (c): The value of `K_c` is very less. So, the amount of PRODUCT at equilibrium is nigligible. Reaction (d) : `K_c` is much HIGH, So, only mainly products present at equilibrium.