$A$ vessel at $1000 \ K$ contains $CO_2$ with a pressure of $0.5 \ atm$. Some of $CO_2$ is converted into $CO$ on addition of graphite. If total pressure at equilibrium is $0.8 \ atm$,then $K_P$ is : (in $atm$)

At $1000 \ K$ in a $0.654 \ L$ vessel,$CaCO_{3(s)}$ is taken. For the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the equilibrium constant $K_p$ is $3.9 \times 10^{-2} \ bar$. Find the weight of $CaO$ produced at equilibrium. $(Ca=40, C=12, O=16)$

$(i) X(g) \rightleftharpoons Y(g) + Z(g), K_{p1} = 3$
$(ii) A(g) \rightleftharpoons 2B(g), K_{p2} = 1$
If the degree of dissociation and initial concentration of both the reactants $X(g)$ and $A(g)$ are equal,then the ratio of the total pressure at equilibrium $\left( \frac{p_1}{p_2} \right)$ is equal to $x : 1$. The value of $x$ is $......$ (Nearest integer)

The equilibrium constant at $298 \ K$ for a reaction $A + B \rightleftharpoons C + D$ is $100$. If the initial concentration of all the four species were $1 \ M$ each,then the equilibrium concentration of $D$ (in $mol \ L^{-1}$) will be:

The surface of copper gets tarnished by the formation of copper oxide. $N_2$ gas was passed to prevent the oxide formation during heating of copper at $1250 \ K$. However,the $N_2$ gas contains $1 \ \text{mole}\%$ of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:
$2 Cu_{(s)} + H_2O_{(g)} \longrightarrow Cu_2O_{(s)} + H_{2(g)}$
$p_{H_2}$ is the minimum partial pressure of $H_2$ (in $\text{bar}$) needed to prevent the oxidation at $1250 \ K$. The value of $\ln(p_{H_2})$ is . . . . .
(Given: total pressure $= 1 \ \text{bar}$,$R = 8 \ J \ K^{-1} \ mol^{-1}$,$\ln(10) = 2.3$. $Cu_{(s)}$ and $Cu_2O_{(s)}$ are mutually immiscible.
At $1250 \ K$: $2 Cu_{(s)} + 1/2 O_{2(g)} \longrightarrow Cu_2O_{(s)}; \Delta G^\theta = -78,000 \ J \ mol^{-1}$
$H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(g)}; \Delta G^\theta = -1,78,000 \ J \ mol^{-1}$)

The amount of $PCl_5$ (in moles) that needs to be added to a $1\,L$ vessel at $250\,^oC$ in order to obtain $0.1\,mol$ of $Cl_2$ for the given reaction is:
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$; $K_C = 0.0414\,mol\,L^{-1}$