In which of the following reactions is $K_p = K_c$?

The equilibrium constant for the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$ is $K.$ Then,the equilibrium constant for the equilibrium $NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2$ is

For the reversible reaction $A + B \rightleftharpoons C + D$,the equilibrium concentrations of $C$ and $D$ are $0.8 \ mol/L$ each. If the initial concentrations of $A$ and $B$ were $1 \ mol/L$ each,calculate the equilibrium constant $K_c$.

Consider the equilibrium,$H_2 + I_2 \rightleftharpoons 2 HI$. Calculate the equilibrium constant of the reverse reaction when the equilibrium concentrations of $H_2$,$I_2$,and $HI$ are $1.14 \times 10^{-2} \ mol \ L^{-1}$,$0.12 \times 10^{-2} \ mol \ L^{-1}$,and $2.52 \times 10^{-2} \ mol \ L^{-1}$,respectively.

For the reactions $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ and $NO_{(g)} \rightleftharpoons 1/2N_{2(g)} + 1/2O_{2(g)}$,the equilibrium constants are $K_1$ and $K_2$ respectively. What is the relationship between $K_1$ and $K_2$?

The value of $\log \ K$ for the reaction $A \rightleftharpoons B$ at $298 \ K$ is (Nearest integer).
Given: $\Delta H^{\circ} = -54.07 \ kJ \ mol^{-1}$
$\Delta S^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$
(Take $2.303 \times 8.314 \times 298 = 5705$)