$56 \, g$ of nitrogen and $8 \, g$ of hydrogen gas are heated in a closed vessel. At equilibrium,$34 \, g$ of ammonia are present. The equilibrium number of moles of nitrogen,hydrogen,and ammonia are respectively:

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 the reaction $X(s) \rightleftharpoons Y(s) + Z(g)$,the plot of $\ln \frac{p_z}{p^\ominus}$ versus $\frac{10^4}{T}$ is given below,where $p_z$ is the pressure (in bar) of the gas $Z$ at temperature $T$ and $p^\ominus = 1 \ bar$.
(Given,$\frac{d(\ln K)}{d(\frac{1}{T})} = -\frac{\Delta H^\ominus}{R}$,where the equilibrium constant,$K = \frac{p_z}{p^\ominus}$ and the gas constant,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$)
$(1)$ The value of standard enthalpy,$\Delta H^\ominus$ (in $kJ \ mol^{-1}$) for the reaction is. . . . . . .
$(2)$ The value of $\Delta S^\ominus$ (in $J \ K^{-1} \ mol^{-1}$) for the given reaction,at $1000 \ K$ is. . . . . .
Give the answer for $(1)$ and $(2)$

Consider the reaction,$P(aq) \rightleftharpoons Q(aq)$ with an equilibrium constant $K=1.5$. The reaction is started in a vessel with a concentration of $[P]$ of $2 \ M$ and concentration of $[Q]=0$. When the equilibrium is established,half the amount of $P$ is removed,and the reaction is allowed to re-equilibrate. The concentration of $Q$ in the vessel (in $M$) is closest to

$A$ plot of $\ln \ K$ against $\frac{1}{T}$ (abscissa) is expected to be a straight line with an intercept on the ordinate axis equal to:

At a certain temperature,the equilibrium constant $(K_C)$ is $16$ for the reaction:
$SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$
If we take one mole of each of the four gases in a $1 \ L$ container,what would be the equilibrium concentration of $NO$ and $NO_2$ respectively?