$X_{2(g)} + Y_{2(g)} \rightleftharpoons 2Z_{(g)}$
$X_{2(g)}$ and $Y_{2(g)}$ are added to a $1 \ L$ flask and it is found that the system attains the above equilibrium at $T \ K$ with the number of moles of $X_{2(g)}$, $Y_{2(g)}$ and $Z_{(g)}$ being $3$, $3$ and $9 \ mol$ respectively (equilibrium moles). Under these conditions of equilibrium, $10 \ mol$ of $Z_{(g)}$ is added to the flask and the temperature is maintained at $T \ K$. Then the number of moles of $Z_{(g)}$ in the flask when the new equilibrium is established is . . . . . . . (Nearest integer).

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

At $1 \ bar$ pressure and $310 \ K$ temperature,$25\%$ of $N_2O_4$ decomposes. The reaction is: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$.
$(i)$ Find $K_p$.
$(ii)$ At $0.1 \ bar$ pressure and $310 \ K$,what is the percentage of $N_2O_4$ decomposed?

$A$ reaction mixture containing $H_2, N_2$ and $NH_3$ has partial pressures of $2 \ atm, 1 \ atm$ and $3 \ atm$ respectively at $725 \ K.$ If the value of $K_P$ for the reaction,$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is $4.28 \times 10^{-5} \ atm^{-2}$ at $725 \ K,$ in which direction will the net reaction proceed?

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)

From the given data of equilibrium constants for the following reactions:
$(1) \ CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)} \ ; \ K_1$
$(2) \ CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)} \ ; \ K_2$
Wait,the provided question text has a typo in the reaction equations. Assuming the standard problem format where we relate equilibrium constants for reverse or combined reactions,if the target reaction is the same as reaction $(1)$,the answer is $K_1$. However,based on the options provided,this is likely a question asking for the relationship between $K_1$ and $K_2$ where reaction $(2)$ is the reverse of reaction $(1)$. If reaction $(2)$ is the reverse of reaction $(1)$,then $K_2 = \frac{1}{K_1}$. Given the options,please re-verify the input. Assuming the question asks for the equilibrium constant of a reaction derived from these,if the target reaction is $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$,the answer is $K_1^{-1}$. Given the options,if we assume the target reaction is the reverse of reaction $(1)$,then $K = \frac{1}{K_1}$.