Consider the following two equilibrium reactions:
$i$. $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$
$ii$. $2ND_{3(g)} \rightleftharpoons N_{2(g)} + 3D_{2(g)}$
What is the difference in their equilibrium constants $(K_c)$?

If the volume of the container for the reaction $2NO + O_2 \rightleftharpoons 2NO_2$ is reduced to half of its initial volume,the rate of the reaction will become .......

$K_{c}$ for the following reaction is $99.0$: $A_{2(g)} \rightleftharpoons B_{2(g)}$. In a $1 \ L$ flask,$2 \ moles$ of $A_{2}$ were heated to $T(K)$ and equilibrium was reached. The concentrations at equilibrium of $A_{2}$ and $B_{2}$ are $C_{1}(A_{2})$ and $C_{2}(B_{2})$ respectively. Now,$1 \ mole$ of $A_{2}$ was added to the flask and heated to $T(K)$ to establish equilibrium again. The concentrations of $A_{2}$ and $B_{2}$ are $C_{3}(A_{2})$ and $C_{4}(B_{2})$ respectively. What is the value of $C_{3}(A_{2})$ in $mol \ L^{-1}$?

$CH_3COCH_{3(g)} \rightleftharpoons C_2H_{6(g)} + CO_{(g)}$. The initial pressure of $CH_3COCH_3$ is $100 \ mm$. When equilibrium is set up,the mole fraction of $CO_{(g)}$ is $\frac{1}{4}$. Hence,the partial pressure of $CO$ is:

For the reaction,$A \rightleftharpoons n B$,the concentration of $A$ decreases from $0.06 \ mol \ L^{-1}$ to $0.03 \ mol \ L^{-1}$ and that of $B$ rises from $0$ to $0.06 \ mol \ L^{-1}$ at equilibrium. The values of $n$ and the equilibrium constant for the reaction,respectively,are

In a $20 \ L$ vessel,if $1 \ mol$ each of $CO$,$H_2O$,and $CO_2$ are initially present,then for the equilibrium $CO + H_2O \rightleftharpoons CO_2 + H_2$,which of the following is true?