For a chemical reaction $A \rightleftharpoons B, K_1 = 2.0$ and $B \rightleftharpoons C, K_2 = 0.01$. The equilibrium constant for the reaction $2C \rightleftharpoons 2A$ is $:-$

The equilibrium constant for the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$ is $5 \times 10^{-2} \ atm^{-1/2}$. The equilibrium constant of the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$ would be

If the equilibrium constants for the given reactions are $K_1$ and $K_2$ respectively,find the relationship between $K_2$ and $K_1$.
$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$

The equilibrium constant,$K_c$ for $3 C_2H_{2(g)} \rightleftharpoons C_6H_{6(g)}$ is $4 \, L^2 \, mol^{-2}$. If the equilibrium concentration of benzene is $0.5 \, mol \, L^{-1}$,the equilibrium concentration of acetylene in $mol \, L^{-1}$ is:

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

Observe the following equilibrium in a $1 \text{ L}$ flask. $A_{(g)} \rightleftharpoons B_{(g)}$. At $T \text{ K}$, the equilibrium concentrations of $A$ and $B$ are $0.5 \text{ M}$ and $0.375 \text{ M}$ respectively. $0.1 \text{ moles}$ of $A$ is added into the flask and heated to $T \text{ K}$ to establish the equilibrium again. The new equilibrium concentrations (in $\text{M}$) of $A$ and $B$ are respectively.