The reaction quotient $Q$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is given by $Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$. The reaction will proceed from right to left when:

The following equilibrium constants are given: $N_2 + 3 H_2 \rightleftharpoons 2 NH_3$ $(k_1)$,$N_2 + O_2 \rightleftharpoons 2 NO$ $(k_2)$,$H_2 + 1/2 O_2 \rightleftharpoons H_2 O$ $(k_3)$. The equilibrium constant for the oxidation of $1 \text{ mole } NH_3$ by oxygen to give $NO$ according to the reaction $NH_3 + 5/4 O_2 \rightleftharpoons NO + 3/2 H_2 O$ is:

For the reaction $2 H_{2(g)} + 2 NO_{(g)} \rightarrow N_{2(g)} + 2 H_2O_{(g)}$,the observed rate expression is $rate = k_f [NO]^2 [H_2]$. The rate expression of the reverse reaction is:

(i) $H_3PO_{4\text{(aq)}} \rightleftharpoons H^+{_{\text{(aq)}}} + H_2PO_4^-{_{\text{(aq)}}}$
(ii) $H_2PO_4^-{_{\text{(aq)}}} \rightleftharpoons H^+{_{\text{(aq)}}} + HPO_4^{2-}{_{\text{(aq)}}}$
(iii) $HPO_4^{2-}{_{\text{(aq)}}} \rightleftharpoons H^+{_{\text{(aq)}}} + PO_4^{3-}{_{\text{(aq)}}}$
The equilibrium constants for the above reactions at a certain temperature are $K_1$,$K_2$,and $K_3$ respectively. The equilibrium constant for the reaction $H_3PO_{4\text{(aq)}} \rightleftharpoons 3H^{+}{_{\text{(aq)}}} + PO_4^{3-}{_{\text{(aq)}}}$
$K = K_1 \times K_2 \times K_3$ is

On a given condition,the equilibrium concentrations of $HI$,$H_2$,and $I_2$ are $0.80 \ mol/L$,$0.10 \ mol/L$,and $0.10 \ mol/L$ respectively. The equilibrium constant for the reaction $H_2 + I_2 \rightleftharpoons 2HI$ will be:

The dissociation of $SO_{3(g)}$ into $SO_{2(g)}$ and $O_{2(g)}$ is carried out in a closed container at a constant temperature $T$. The equilibrium constant for the reaction is $K_P = x \ atm$. The partial pressure variation for the three gases is as shown in the graph. The value of $x$ is