The equilibrium constants $(K_C)$ of two reactions $H_2 + I_2 \rightleftharpoons 2 HI$ and $N_2 + 3 H_2 \rightleftharpoons 2 NH_3$ are $50$ and $1000$,respectively. The equilibrium constant of the reaction $N_2 + 6 HI \rightleftharpoons 2 NH_3 + 3 I_2$ is closest to

The equilibrium constants $K_{p1}$ and $K_{p2}$ for the reactions $X \rightleftharpoons 2Y$ and $Z \rightleftharpoons P + Q$ respectively are in the ratio of $1 : 9$. If the degree of dissociation of $X$ and $Z$ be equal,then the ratio of total pressures at these equilibria is:

Using the data provided,find the value of the equilibrium constant for the following reaction at $298 \ K$ and $1 \ atm$ pressure: $NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
$\Delta_{f} H^0(NO_{(g)}) = 90.4 \ kJ \cdot mol^{-1}$
$\Delta_{f} H^0(NO_{2(g)}) = 32.48 \ kJ \cdot mol^{-1}$
$\Delta S^{\circ} = -70.8 \ J \cdot K^{-1} \cdot mol^{-1}$
$\text{antilog}(6.4) = 2.51 \times 10^6$ (Note: Calculation based on standard thermodynamic relations)

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

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 reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $B$ was $1.5$ times that of $[A]$,but at equilibrium,the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is: