The value of $K_{C}$ for the equilibrium reaction: $CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ at $T \ K$ is $0.036$. If the equilibrium concentration of $CO_{2(g)}$ is $0.004 \ M$,the concentration of $CO_{(g)}$ in $mol \ L^{-1}$ is:

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).

One mole $H_2O_{(g)}$ and one mole $CO_{(g)}$ are taken in a $1 \ L$ flask and heated to $725 \ K$. At equilibrium,$40 \%$ of water reacted with $CO_{(g)}$ as follows:
$H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$
Its $K_c$ value is:

The equilibrium concentrations of $N_2, H_2$ and $NH_3$ in the formation of $NH_3$ at $500 \ K$ are $1.25 \times 10^{-2} \ M, 4.0 \times 10^{-2} \ M$ and $1.6 \times 10^{-2} \ M$ respectively. The equilibrium constant $K_{p}$ at the same temperature is

One mole $H_2O_{(g)}$ and one mole $CO_{(g)}$ are taken in a $1 \ L$ flask and heated to $725 \ K$. At equilibrium,$40 \%$ (by mass) of water reacted with $CO_{(g)}$ as follows: $H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2_{(g)}} + CO_{2_{(g)}}$. The value of $K_p$ is:

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.