When $5 \ mol$ of $SO_2$ and $5 \ mol$ of $O_2$ are reacted,$60\%$ of $SO_2$ is consumed at equilibrium. If the total pressure of the equilibrium mixture is $1 \ atm$,then the partial pressure of $O_2$ will be ...... $atm$.

Nitric oxide reacts with $Br_{2}$ and gives nitrosyl bromide as per the reaction given below:
$2 NO_{(g)} + Br_{2(g)} \longleftrightarrow 2 NOBr_{(g)}$
When $0.087 \ mol$ of $NO$ and $0.0437 \ mol$ of $Br_{2}$ are mixed in a closed container at constant temperature,$0.0518 \ mol$ of $NOBr$ is obtained at equilibrium. Calculate the equilibrium amount of $NO$ and $Br_{2}$.

For the reaction,$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,if dinitrogen tetroxide is $50\%$ dissociated at $60^\circ C$,the standard free energy change at this temperature and $1 \ atm$ pressure is:

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 $K = \frac{1}{K_1}$.

For the reaction,$0.5 C_{(s)} + 0.5 CO_{2(g)} \rightleftharpoons CO_{(g)}$,the equilibrium pressure is $12 \ atm$. If $CO_2$ conversion is $50 \%$,the value of $K_p$,in $atm$,is:

$A$ mixture of $1.57 \ mol$ of $N_2$,$1.92 \ mol$ of $H_2$ and $8.13 \ mol$ of $NH_3$ is introduced into a $20 \ L$ reaction vessel at $500 \ K$. At this temperature,the equilibrium constant,$K_c$ for the reaction $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$ is $1.7 \times 10^2$. Is the reaction mixture at equilibrium? If not,what is the direction of the net reaction?