The plot of $log_{10} K$ vs $\frac{1}{T}$ gives a straight line. The intercept and slope respectively are (where $K$ is equilibrium constant).

$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$
The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).

Consider the following equilibrium,$CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_3OH_{(g)}$. $0.1 \ mol$ of $CO$ along with a catalyst is present in a $2 \ dm^3$ flask maintained at $500 \ K$. Hydrogen is introduced into the flask until the pressure is $5 \ bar$ and $0.04 \ mol$ of $CH_3OH$ is formed. The $K_{p}^0$ is $......... \times 10^{-3}$ (nearest integer). Given: $R = 0.08 \ dm^3 \ bar \ K^{-1} \ mol^{-1}$. Assume only methanol is formed as the product and the system follows ideal gas behaviour.

For which of the following reactions is the relation $\log \frac{K_P}{K_C} + \log RT = 0$ correct?

The equilibrium constant for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ at temperature $T$ is $4 \times 10^{-4}$. The value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}$ at the same temperature is

$2N_2O_{4(g)} \rightleftharpoons 4NO_{2(g)}$ has $K_p = 0.15 \, atm$ at $298 \, K$. Calculate $K_p$ in $torr$ and $K_c$ in $mol \, L^{-1}$. ($1 \, atm = 760 \, torr$; $R = 0.082 \, L \, atm \, mol^{-1} \, K^{-1}$).