For the reaction $H_2O_{(s)} \rightleftharpoons H_2O_{(l)}$ at $0\,^\circ C$ and normal pressure:

Explain the equilibrium when sublimation of solid takes place in a closed vessel.

For the process $H_2O_{(s)} \rightleftharpoons H_2O_{(l)}$ at $0^{\circ}C$ and $1 \ atm$ pressure,which of the following is true?

In a reversible reaction $A \overset{k_1}{\underset{k_2}{\longleftrightarrow}} B$,the initial concentrations of $A$ and $B$ are $a$ and $b$,and the equilibrium concentrations are $(a - x)$ and $(b + x)$ respectively. Express $x$ in terms of $k_1, k_2, a$,and $b$.

In a reversible chemical reaction at equilibrium, if the concentration of any one of the reactants is doubled, then the equilibrium constant will

In the equilibrium $CH_3COOH + H_2O \rightleftharpoons CH_3COO^{-} + H_3O^{+}$,the equilibrium constant may change when: