When $1 \ mol$ of $H_2$ and $1 \ mol$ of $N_2$ are enclosed in a $5 \ L$ vessel and the reaction is allowed to attain equilibrium,it is found that at equilibrium there is $x \ mol$ of $H_2$. The number of moles of $NH_3$ would be

For a reaction,$A \rightleftharpoons P$,the plots of $[A]$ and $[P]$ with time at temperatures $T_1$ and $T_2$ are given below. If $T_2 > T_1$,the correct statement$(s)$ is (are) (Assume $\Delta H^{\ominus}$ and $\Delta S^{\ominus}$ are independent of temperature and ratio of $\ln K$ at $T_1$ to $\ln K$ at $T_2$ is greater than $T_2 / T_1$. Here $H, S, G$ and $K$ are enthalpy,entropy,Gibbs energy and equilibrium constant,respectively.)
$(A)$ $\Delta H^{\ominus} < 0, \Delta S^{\ominus} < 0$
$(B)$ $\Delta G^{\ominus} < 0, \Delta H^{\ominus} > 0$
$(C)$ $\Delta G^{\ominus} < 0, \Delta S^{\ominus} < 0$
$(D)$ $\Delta G^{\ominus} < 0, \Delta S^{\ominus} > 0$

The two substances $A$ and $B$ are in equilibrium with $C$ and $D$ as $2A + B \rightleftharpoons 3C + 2D$. If the initial pressure of $A$ and $B$ are in the ratio of $4:1$ and at equilibrium,the partial pressures of $A$ and $D$ are equal,find the correct relation.

The equilibrium composition for the reaction $PCl_3 + Cl_2 \rightleftharpoons PCl_5$ at $298 \, K$ is given below.
$[PCl_3]_{eq} = 0.2 \, mol \, L^{-1}$
$[Cl_2]_{eq} = 0.1 \, mol \, L^{-1}$
$[PCl_5]_{eq} = 0.40 \, mol \, L^{-1}$
If $0.2 \, mol$ of $Cl_2$ is added at the same temperature,the equilibrium concentration of $PCl_5$ is $.... \times 10^{-2} \, mol \, L^{-1}$. Given: $K_c$ for the reaction at $298 \, K$ is $20$.

$1 \ mol$ of $N_2$ and $2 \ mol$ of $H_2$ are allowed to react in a $1 \ dm^3$ vessel. At equilibrium,$0.8 \ mol$ of $NH_3$ is formed. What is the concentration of $H_2$ at equilibrium (in $M$)?

$CH_3COCH_{3(g)} \rightleftharpoons C_2H_{6(g)} + CO_{(g)}$. The initial pressure of $CH_3COCH_3$ is $100 \ mm$. When equilibrium is set up,the mole fraction of $CO_{(g)}$ is $\frac{1}{4}$. Hence,the partial pressure of $CO$ is: