The rate of forward reaction is two times that of reverse reaction at a given temperature and identical concentration. $K_{equilibrium}$ is

If the equilibrium constant for $A \rightleftharpoons B + C$ is $K_{eq}^{(1)}$ and that of $B + C \rightleftharpoons P$ is $K_{eq}^{(2)}$,the equilibrium constant for $A \rightleftharpoons P$ is :-

Consider the following gaseous equilibrium reactions $(I)$,$(II)$ and $(III)$ with equilibrium constants $K_1$,$K_2$ and $K_3$ respectively:
$I$) $\frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3$
$II$) $2 NO \rightleftharpoons N_2 + O_2$
$III$) $H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2 O$
The correct expression for the equilibrium constant for the gaseous equilibrium reaction $2 NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2 NO + 3 H_2 O$ is

The dissociation of $SO_{3(g)}$ into $SO_{2(g)}$ and $O_{2(g)}$ is carried out in a closed container at a constant temperature $T$. The equilibrium constant for the reaction is $K_P = x \ atm$. The partial pressure variation for the three gases is as shown in the graph. The value of $x$ is

Explain the calculation of equilibrium concentrations given the value of the equilibrium constant.

In the equilibrium $AB \rightleftharpoons A + B$; if the equilibrium concentration of $A$ is doubled,the equilibrium concentration of $B$ would become: