In a $0.25 \ L$ tube,$4 \ mol$ of $NO$ undergoes dissociation. If the degree of dissociation is $10\%$,then the value of $K_c$ for the reaction $2NO \rightleftharpoons N_2 + O_2$ will be:

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 $\ln 10 = 2.3$)

$K_{c}$ 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 $mol \ L^{-1}$?

$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are taken in a $4 \ L$ closed vessel at a constant temperature. The reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$. If $0.25\%$ of $N_2$ is converted into ammonia,calculate $K_c$ for this reaction and the $K_c'$ for the reaction $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$.

When $1 \ mol$ of $H_2$ and $1 \ mol$ of $N_2$ are enclosed in a $5 \ L$ vessel and the reaction is allowed to attain equilibrium,it is found that at equilibrium there is $x \ mol$ of $H_2$. The number of moles of $NH_3$ would be

At high temperature,$2 \, \text{mol}$ of $NH_3$ is placed in a $500 \, \text{mL}$ vessel. For the decomposition reaction $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$,if $1 \, \text{mol}$ of $NH_3$ remains at equilibrium,then $K_c$ is equal to: