In a $1 \, L$ container,the reaction of $2 \, mol$ $N_2$ and $5 \, mol$ $H_2$ occurs. If the equilibrium concentration of $NH_3$ is half the equilibrium concentration of $N_2$,what is the equilibrium constant $(K_c)$?

The reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2_{(g)}} + S_{2_{(g)}}$ is in equilibrium. If $0.5 \ mol$ of $H_2S$,$0.10 \ mol$ of $H_2$,and $0.4 \ mol$ of $S_2$ are taken in a $1 \ L$ vessel,the value of the equilibrium constant $(K)$ is .... $mol \ L^{-1}$.

$A$ plot of $\ln K$ against $\frac{1}{T}$ ($x$-axis) is expected to be a straight line,with intercept on $Y$-axis equal to

In the reaction $2P_{(g)} + Q_{(g)} \rightleftharpoons 3R_{(g)} + S_{(g)}$,if $2 \text{ moles}$ of each $P$ and $Q$ are taken initially in a $1 \text{ L}$ flask,which of the following is true at equilibrium?

Given three reactions and their equilibrium constants:
$N_2 + 3H_2 \rightleftharpoons 2NH_3 ; k_1$
$N_2 + O_2 \rightleftharpoons 2NO ; k_2$
$H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O ; k_3$
The equilibrium constant for the reaction $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $k_1, k_2,$ and $k_3$ is:

One mole each of $He$ and $A(g)$ are taken in a $10 \text{ L}$ closed flask and heated to $400 \text{ K}$ to establish the following equilibrium: $A(g) \rightleftharpoons B(g)$. $K_{c}$ for this reaction at $400 \text{ K}$ is $4.0$. The partial pressures (in $\text{atm}$) of $He$ and $B(g)$ are respectively (at equilibrium) (Assume $He$,$A(g)$ and $B(g)$ behave as ideal gases) (Given: $R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1}$)