For the balanced reaction $A + B \rightleftharpoons 2C$,if the equilibrium concentrations of both $A$ and $B$ are $0.20 \ mol/L$,and the concentration of $C$ is $0.60 \ mol/L$,then the equilibrium constant for this reaction will be:

The equilibrium constant $(K_c)$ for the reaction,$CaSO_4 \cdot 5H_2O_{(s)} \rightleftharpoons CaSO_4 \cdot 3H_2O_{(s)} + 2H_2O_{(g)}$ is equal to

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

For the reaction $A + 2B \rightleftharpoons C$,the expression for the equilibrium constant is.........

In a reaction $A + B \rightleftharpoons C + D$,the concentrations of $A$,$B$,$C$ and $D$ (in $mol/L$) are $0.5$,$0.8$,$0.4$ and $1.0$ respectively. The equilibrium constant is

Consider the following equilibrium reaction in gaseous state at $T(K)$.
$A(g) + 2B(g) \rightleftharpoons 2C(g) + D(g)$
The initial concentration of $B$ is $1.5$ times that of $A$. At equilibrium,the concentrations of $A$ and $B$ are equal. The equilibrium constant for the reaction is