The equilibrium constant of the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $64$. If the volume of the container is reduced to one-fourth of its original volume,the value of the equilibrium constant will be

Given the reactions at $500 \ K$:
$(i) \ H_{2(g)} + 1/2 O_{2(g)} \rightleftharpoons H_2O_{(g)}$,$K_c = 2.4 \times 10^{47}$
$(ii) \ H_2O_{(g)} \rightleftharpoons H_{2(g)} + 1/2 O_{2(g)}$,$K_c = 4.1 \times 10^{-48}$
What does the magnitude of $K_c$ indicate about the formation of the product in reaction $(i)$?

For the equilibrium $A \rightleftharpoons B$,the variation of the rate of the forward $(a)$ and reverse $(b)$ reaction with time is given by:

For the equilibrium process $H_2O_{(\ell)} \rightleftharpoons H_2O_{(g)}$ at $1 \, \text{atm}$ and $373 \, K$,which of the following is true?

Explain solid-liquid equilibrium by giving an example.

For the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$,the value of $K_c$ does not depend on: $(a)$ Initial concentration of reactants $(b)$ Pressure $(c)$ Temperature $(d)$ Catalyst