At $T(K)$,$K_c$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $99.0$. Two moles of $A_{2(g)}$ were heated to $T(K)$ in a $1 \ L$ closed flask to reach the above equilibrium. What are the concentrations (in $mol \ L^{-1}$) of $A_{2(g)}$ and $B_{2(g)}$ respectively at equilibrium?

For the chemical equilibrium $A + B \rightleftharpoons C + D$,when one mole of each reactant is mixed,$0.4 \ mol$ of each product is formed. The equilibrium constant $K_c$ is:

$K_{a_1}, K_{a_2}$ and $K_{a_3}$ are the respective ionization constants for the following reactions $(a), (b),$ and $(c)$.
$(a)$ $H_2C_2O_4 \rightleftharpoons H^{+} + HC_2O_4^-$
$(b)$ $HC_2O_4^- \rightleftharpoons H^{+} + C_2O_4^{2-}$
$(c)$ $H_2C_2O_4 \rightleftharpoons 2H^{+} + C_2O_4^{2-}$
The relationship between $K_{a_1}, K_{a_2}$ and $K_{a_3}$ is given as

Write the equilibrium constant (equilibrium law) for the following reactions:
$(i)$ $4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_{2}O_{(g)}$
$(ii)$ $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$
$(iii)$ $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

The reaction,$2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$ is begun with the concentrations of $A$ and $B$ both at an initial value of $1.00 \ M$. When equilibrium is reached,the concentration of $D$ is measured and found to be $0.25 \ M$. The value for the equilibrium constant for this reaction is given by the expression:

Write the expressions for $K_{P}$ and $K_{C}$ for the reaction: $3Fe_{(s)} + 4H_2O_{(g)} \rightleftharpoons Fe_3O_{4(s)} + 4H_{2(g)}$.