$A_{(g)} + 3B_{(g)} \rightleftharpoons 4C_{(g)}$. The starting concentrations of $A$ and $B$ are equal. At equilibrium,the concentrations of $A$ and $C$ are the same. The value of $K_c$ is:

For the reversible reaction in equilibrium
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
If the rate constant for the forward reaction is $k_1 = 2.1 \times 10^{-3} \ s^{-1}$ and for the backward reaction is $k_2 = 4.2 \times 10^{-4} \ s^{-1}$,then the equilibrium constant $K_c$ for the above reaction is:

The gas phase reaction $2A_{(g)} \rightleftharpoons A_{2(g)}$ at $400 \ K$ has $\Delta G^{\circ} = +25.2 \ kJ \ mol^{-1}$. The equilibrium constant $K_{C}$ for this reaction is $...... \times 10^{-2}$. (Round off to the nearest integer) $[$Use: $R = 8.3 \ J \ mol^{-1} \ K^{-1}$,$\ln 10 = 2.3$,$\log_{10} 2 = 0.30$,$1 \ atm = 1 \ bar]$ $[$antilog $(-3.3) = 5.01 \times 10^{-4}]$

For the reaction: $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$,if the initial concentration of $[H_2] = [CO_2] = 1 \ M$ and $x \ mol/L$ of hydrogen is consumed at equilibrium,the correct expression for $K_c$ is:

For the formation of $NH_3$ from $N_2$ and $H_2$ at $500 \ K$,the concentrations of $N_2, H_2$ and $NH_3$ at equilibrium are $1.5 \times 10^{-2} \ M, 3.0 \times 10^{-2} \ M$ and $1.2 \times 10^{-2} \ M$,respectively. The equilibrium constant for the reverse reaction is

$NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$
The reaction was started with some amount of $NH_4HS$. The equilibrium pressure at $25^{\circ}C$ is $0.5 \ atm$. What is $K_p$ for the reaction (in $atm^2$)?