$1 \ mol$ of $N_2$ and $2 \ mol$ of $H_2$ are allowed to react in a $1 \ dm^3$ vessel. At equilibrium,$0.8 \ mol$ of $NH_3$ is formed. What is the concentration of $H_2$ at equilibrium (in $M$)?

Consider the following reaction approaching equilibrium at $27^{\circ} C$ and $1 \ atm$ pressure. Given the rate constants for the forward and backward reactions are $K_{f} = 10^{3} \ s^{-1}$ and $K_{b} = 10^{2} \ s^{-1}$ respectively,calculate the standard Gibb's energy change $(\Delta_{r} G^{\circ})$ at $27^{\circ} C$ in $kJ \ mol^{-1}$ (Nearest integer). (Given: $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ and $\ln 10 = 2.3$)

For the equilibrium $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$,the partial pressures of $SO_2$,$O_2$,and $SO_3$ are $0.662 \ atm$,$0.101 \ atm$,and $0.331 \ atm$ respectively. If the equilibrium concentrations of $SO_2$ and $SO_3$ are made equal,the partial pressure of $O_2$ will be ..... $atm$.

At a certain temperature,the equilibrium constant $(K_C)$ is $16$ for the reaction:
$SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$
If we take one mole of each of the four gases in a $1 \ L$ container,what would be the equilibrium concentration of $NO$ and $NO_2$ respectively?

Which of the following is true for the reaction $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$ at $100 \ ^\circ C$ and $1 \ \text{atm}$ pressure?

Consider a reaction that is first order in both directions: $A \underset{K_b}{\stackrel{K_f}{\rightleftharpoons}} B$. Initially only $A$ is present,and its concentration is $A_{0}$. Assume $A_{t}$ and $A_{\text{eq}}$ are the concentrations of $A$ at time $t$ and at equilibrium,respectively. The time $t$ at which $A_{t} = (A_{0} + A_{\text{eq}})/2$ is $....$