$$Ba{O}_2\left(s\right)\underset{r_2}{\overset{r_1}{\rightleftharpoons }}BaO\left(s\right)+O_2\left(g\right),\Delta H=+ve$$
According to law of mass action, the rate of forward reaction $$ = {r_1}$$
$$\eqalign{ & {r_1} \propto \left[ {Ba{O_2}} \right] \cr & {\text{or}}\,\,\,{r_1} = {k_1}\left[ {Ba{O_2}} \right] \cr} $$
$$Ba{O_2}$$  is solid substance in pure state concentration $$ = 1\,m$$
then, $${r_1} = {k_1}$$
Similarly the rate of backward reaction $$ = {r_2}$$
$$\eqalign{ & {r_2} \propto \left[ {BaO} \right]\left[ {{O_2}} \right] \cr & {\text{or}}\,\,\,{r_2} = {k_2}\left[ {BaO} \right]\left[ {{O_2}} \right] \cr} $$
$$\because $$ Concentration of solid $$\left[ {BaO} \right] = 1\,\,\,\,\left[ {{O_2}\left( g \right)} \right]$$
$$\therefore \,\,\,{r_2} = {k_2}\left[ {{O_2}} \right]$$
$$\eqalign{ & {\text{At equilibrium,}} \cr & {r_1} = {r_2} \cr & {K_1} = {K_2}\left[ {{O_2}} \right] \cr & {\text{or}}\,\,\,{K_1} = {K_2} \cdot \,{p_{{O_2}}} \cr & {\text{where,}}\,\,{p_{{O_2}}} = {\text{partial pressure of}}\,{O_2} \cr & {\text{or}}\,\,\,\,\,\frac{{{K_1}}}{{{K_2}}} = {p_{{O_2}}}\,{\text{(equilibrium constant)}} \cr & \because \,\,\frac{{{K_1}}}{{{K_2}}} = K\,\,\,{\text{or}}\,\,K = {p_{{O_2}}} \cr} $$
So, from the above it is clear that pressure of $${O_2}$$  does not depend upon the concentration of reactants. The given equation is an endothermic reaction. If the temperature of such reaction is increased, then dissociation of $$Ba{O_2}$$  would increase and more $${O_2}$$  is produced.

\[\begin{align} & \text{Given,} \\ & \begin{matrix} {} & A & + \\ \text{No}\text{. of moles initially} & 1 & {} \\ \text{At equilibrium} & 1-a & {} \\ \end{matrix}\,\,\,\begin{matrix} B & \rightleftharpoons & C \\ 1 & {} & 1 \\ 1-a & {} & 1+a \\ \end{matrix}\,\,\,\,\,\,\begin{matrix} + \\ {} \\ {} \\ \end{matrix}\,\,\,\,\,\,\begin{matrix} D \\ 1 \\ 1+a \\ \end{matrix} \\ \end{align}\]
$$\eqalign{ & \therefore \,\,{K_c} = {\left( {\frac{{1 + a}}{{1 - a}}} \right)^2} = 100 \cr & \therefore \,\,\frac{{1 + a}}{{1 - a}} = 10 \cr & {\text{On solving}} \cr & a = 0.81 \cr & {\left[ D \right]_{{\text{At}}\,{\text{eq}}}} = 1 + a = 1 + 0.81 = 1.81 \cr} $$

$$\eqalign{ & HI\left( g \right) \rightleftharpoons \frac{1}{2}{H_2}\left( g \right) + \frac{1}{2}{I_2}\left( g \right) \cr & K = \frac{{{{\left[ {{I_2}} \right]}^{\frac{1}{2}}}{{\left[ {{H_2}} \right]}^{\frac{1}{2}}}}}{{\left[ {HI} \right]}}\,\,\,...{\text{(i)}} \cr & {H_2}\left( g \right) + {I_2}\left( g \right) \rightleftharpoons 2HI\left( g \right) \cr & K' = \frac{{{{\left[ {HI} \right]}^2}}}{{\left[ {{H_2}} \right]\left[ {{I_2}} \right]}}\,\,\,...{\text{(ii)}} \cr & {\text{From Eqs}}{\text{. (i) and (ii)}} \cr & K \times \sqrt {K'} = 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,K' = \frac{1}{{{K^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{{{\left( 8 \right)}^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{64}} \cr} $$

$$2Cu{\left( {N{O_3}} \right)_{2\left( s \right)}} \rightleftharpoons $$    $$2Cu{O_{\left( s \right)}} + 4N{O_{2\left( g \right)}} + {O_{2\left( g \right)}}$$
Since conc. of solids is taken as 1, the expression for $${K_c}$$  becomes $${K_c} = {\left[ {N{O_{2\left( g \right)}}} \right]^4}\left[ {{O_{2\left( g \right)}}} \right]$$

\[\begin{matrix} {} \\ \text{Initial}\,\,\text{conc}\text{.} \\ \text{At}\,\,\text{equilibrium} \\ \end{matrix}\begin{matrix} PC{{l}_{5\left( g \right)}} \\ \frac{5}{5}=1 \\ 1-0.4 \\ \end{matrix}\begin{matrix} \rightleftharpoons \\ {} \\ {} \\ \end{matrix}\begin{matrix} PC{{l}_{3\left( g \right)}} \\ 0 \\ 0.4 \\ \end{matrix}\begin{matrix} + \\ {} \\ {} \\ \end{matrix}\begin{matrix} C{{l}_{2\left( g \right)}} \\ 0 \\ 0.4 \\ \end{matrix}\]

$$\eqalign{ & {K_c} = \frac{{\left[ {PC{l_3}} \right]\left[ {C{l_2}} \right]}}{{\left[ {PC{l_5}} \right]}} \cr & \,\,\,\,\,\,\,\, = \frac{{0.4\,M \times 0.4\,M}}{{0.6\,M}} \cr & \,\,\,\,\,\,\,\, = 0.266\,M \cr} $$

For the reaction
$$2Ba{O_2}\left( s \right) \rightleftharpoons 2BaO\left( s \right) + {O_2}\left( g \right);\Delta H = + ve.$$
At equilibrium $${K_p} = {P_{{O_2}}}$$
Hence, the value of equilibrium constant depends only upon partial pressure of $${O_2}.$$ Further on increasing temperature formation of $${O_2}$$  increases as this is an endothermic reaction. Hence, pressure of $${O_2}$$  is dependent on temperature.

$$N_{2\left(g\right)}+3H_{2\left(g\right)}\underset{\text{endo}}{\overset{\text{exo}}{\rightleftharpoons }}2N{H}_{3\left(g\right)};$$      $$\Delta H < 0$$
Initially, with increase in temperature $$\left( {{T_2} > {T_1}} \right)\% $$    yield increases. Afterwards, equilibrium is reached and if the temperature is increased, i.e., heat is supplied to the system, then according to Le Chateliers principle, the equilibrium will shift in the backward direction, where the heat is absorbed. Hence, the % yield decreases.

Le-Chatelier's and Braun French chemists made certain generalisations to explain the effect of changes in concentrations, temperature or pressure on the state of system in equilibrium. When a system is subjected to a change in one of these factors, the equilibrium gets disturbed and the system re-adjusts itself until it return to equilibrium.

$$\eqalign{ & {K_c} = \frac{{{{\left[ {N{O_2}} \right]}^2}}}{{\left[ {{N_2}{O_4}} \right]}} \cr & = \frac{{{{\left[ {1.2 \times {{10}^{ - 2}}} \right]}^2}}}{{\left[ {4.8 \times {{10}^{ - 2}}} \right]}} \cr & = 3 \times {10^{ - 3}}mol/L \cr} $$

$$\eqalign{ & {\text{For the reaction : }}2N{O_2}\left( g \right) \rightleftharpoons 2NO\left( g \right) + {O_2}\left( g \right) \cr & {\text{Given}}\,\,\,{K_c} = 1.8 \times {10^{ - 6}}\,{\text{at}}\,{184^ \circ }C \cr & R = 0.0831\,kj/mol.k \cr & {K_p} = 1.8 \times {10^{ - 6}} \times 0.0831 \times 457 = 6.836 \times {10^{ - 6}} \cr & \left[ {\because \,\;{{184}^ \circ }C = \left( {273 + 184} \right) = 457k,\,\,\Delta n = \left( {2 + 1 - 2} \right) = 1} \right] \cr & {\text{Hence it is clear that}}\,\,{K_p} > {K_c} \cr} $$