If for a reaction $\Delta G^{o} > 0$,then:

For the reaction: $HI_{(g)} \rightleftharpoons \frac{1}{2}H_{2(g)} + \frac{1}{2}I_{2(g)}$,the equilibrium constant is $8$. What is the equilibrium constant for the reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$?

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.

One mole of $A_{(g)}$ is heated to $T(K)$ until the following equilibrium is obtained:
$A_{(g)} \rightleftharpoons B_{(g)}$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium,$0.5 \ mol$ of $A_{(g)}$ is added and heated. The equilibrium is again established. The value of $\frac{[A]}{[B]}$ is:

For the reversible reaction in equilibrium
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
If the rate constant for the forward reaction is $k_1 = 2.1 \times 10^{-3} \ s^{-1}$ and for the backward reaction is $k_2 = 4.2 \times 10^{-4} \ s^{-1}$,then the equilibrium constant $K_c$ for the above reaction is:

At $298 \ K$,the value of $K_c$ for the following reaction is $x \ mol \ L^{-1}$. What is the approximate $K_p$ value for this reaction? $(R=0.082 \ L \ atm \ mol^{-1} \ K^{-1})$ $A_2O_{4(g)} \rightleftharpoons 2AO_{2(g)}$