Law of equilibrium and Equilibrium constant Questions in English

(D) For reaction $(i)$: $N_2 + 2O_2 \rightleftharpoons 2NO_2$ with equilibrium constant $K_1$.
For reaction $(ii)$: $2NO_2 \rightleftharpoons N_2 + 2O_2$,which is the reverse of $(i)$,so $K_2 = 1/K_1$.
For reaction $(iii)$: $NO_2 \rightleftharpoons 1/2 N_2 + O_2$,which is half of reaction $(ii)$,so $K_3 = (K_2)^{1/2} = \sqrt{K_2}$.
From $K_2 = 1/K_1$,we get $K_1 = 1/K_2$.
From $K_3 = \sqrt{K_2}$,we get $K_3^2 = K_2$,so $K_2 = K_3^2$.
Substituting $K_2$ in $K_1 = 1/K_2$,we get $K_1 = 1/K_3^2$.

(C) The reaction is $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$.
Initial moles: $AB_3 = 8, A_2 = 0, B_2 = 0$.
At equilibrium,$x$ moles of $A_2$ are formed. Given $x = 2 \, \text{mol}$.
Change in moles: $AB_3 = -2x = -4, A_2 = +x = +2, B_2 = +3x = +6$.
Equilibrium moles: $AB_3 = 8 - 4 = 4, A_2 = 2, B_2 = 6$.
Since the volume is $1.0 \, \text{dm}^3$,concentrations are equal to the number of moles.
$K_c = \frac{[A_2][B_2]^3}{[AB_3]^2} = \frac{(2)(6)^3}{(4)^2} = \frac{2 \times 216}{16} = \frac{432}{16} = 27$.

(A) The reaction is: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$
Initial moles: $H_2 = 4.5$,$I_2 = 4.5$,$HI = 0$
At equilibrium,$2x = 3 \ mol$ of $HI$ is formed,so $x = 1.5 \ mol$.
Equilibrium moles: $H_2 = 4.5 - 1.5 = 3 \ mol$,$I_2 = 4.5 - 1.5 = 3 \ mol$,$HI = 3 \ mol$.
Since the volume is $10 \ L$,the concentrations are:
$[H_2] = 3/10 = 0.3 \ M$
$[I_2] = 3/10 = 0.3 \ M$
$[HI] = 3/10 = 0.3 \ M$
The equilibrium constant $K_c$ is given by:
$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.3)^2}{(0.3)(0.3)} = \frac{0.09}{0.09} = 1$.

(D) For a heterogeneous equilibrium,the concentration of pure solids is taken as unity $(1)$.
Given reaction: $P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$
Applying the law of mass action: $K_c = \frac{[P_4O_{10}]}{[P_4][O_2]^5}$
Since $P_{4(s)}$ and $P_4O_{10(s)}$ are solids,their concentrations are considered as $1$.
Therefore,$K_c = \frac{1}{1 \times [O_2]^5} = \frac{1}{[O_2]^5}$.

(A) The equilibrium reaction is: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
The expression for the equilibrium constant $K_c$ is:
$K_c = \frac{[NO_2]^2}{[N_2O_4]}$
Substituting the given equilibrium concentrations:
$K_c = \frac{(1.2 \times 10^{-2})^2}{4.8 \times 10^{-2}}$
$K_c = \frac{1.44 \times 10^{-4}}{4.8 \times 10^{-2}}$
$K_c = 0.3 \times 10^{-2} = 3 \times 10^{-3}$

(C) For a reversible reaction,the reaction quotient $Q$ is compared with the equilibrium constant $K_c$ to predict the direction of the reaction.
If $Q < K_c$,the ratio of product concentrations to reactant concentrations is less than the equilibrium value,so the reaction proceeds in the forward direction to reach equilibrium.
If $Q > K_c$,the reaction proceeds in the backward direction.
If $Q = K_c$,the reaction is at equilibrium.

(C) The given reactions are:
$(1) N_2 + 3H_2 \rightleftharpoons 2NH_3$ with constant $K_1$
$(2) N_2 + O_2 \rightleftharpoons 2NO$ with constant $K_2$
$(3) H_2 + 1/2 O_2 \rightleftharpoons H_2O$ with constant $K_3$
We want to obtain the reaction: $2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O$
To get this,we reverse reaction $(1)$,add reaction $(2)$,and add $3$ times reaction $(3)$:
Reverse $(1): 2NH_3 \rightleftharpoons N_2 + 3H_2$ (Constant $= 1/K_1$)
$(2): N_2 + O_2 \rightleftharpoons 2NO$ (Constant $= K_2$)
$3 \times (3): 3H_2 + 3/2 O_2 \rightleftharpoons 3H_2O$ (Constant $= K_3^3$)
Adding these: $(2NH_3 + N_2 + 3H_2 + O_2 + 3/2 O_2) \rightleftharpoons (N_2 + 3H_2 + 2NO + 3H_2O)$
Simplifying: $2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O$
The equilibrium constant $K$ is: $K = (1/K_1) \times K_2 \times K_3^3 = \frac{K_2 K_3^3}{K_1}$.

(C) For the first reaction: $K_1 = \frac{[SO_3]}{[SO_2][O_2]^{1/2}}$
For the second reaction: $K_2 = \frac{[SO_2]^2[O_2]}{[SO_3]^2}$
Comparing the two expressions,we can see that $K_2 = \left( \frac{1}{K_1} \right)^2 = \frac{1}{K_1^2}$.

(B) The given reaction is $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$ with $K_{c1} = 1.8 \times 10^{-6}$.
For the target reaction $NO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons NO_{2(g)}$,we observe that the reaction is reversed and the stoichiometric coefficients are halved.
Therefore,the new equilibrium constant $K_{c2} = \sqrt{\frac{1}{K_{c1}}}$.
$K_{c2} = \sqrt{\frac{1}{1.8 \times 10^{-6}}} = \sqrt{0.555 \times 10^6} = 0.745 \times 10^3 \approx 7.5 \times 10^2$.

(D) The equilibrium constant for a reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.
Given the reaction $2HI \rightleftharpoons H_2 + I_2$ has $K_1 = 0.25$.
The reaction $H_2 + I_2 \rightleftharpoons 2HI$ is the reverse of the given reaction.
Therefore,$K_2 = \frac{1}{K_1} = \frac{1}{0.25} = 4$.

(D) Given the equilibrium constants for the individual steps:
$1) A \rightleftharpoons B; K_{c1} = 2$
$2) B \rightleftharpoons C; K_{c2} = 4$
$3) C \rightleftharpoons D; K_{c3} = 6$
When reactions are added,their equilibrium constants are multiplied to obtain the equilibrium constant of the net reaction.
For the net reaction $A \rightleftharpoons D$,which is the sum of the three given reactions:
$K_{c(net)} = K_{c1} \times K_{c2} \times K_{c3}$
$K_{c(net)} = 2 \times 4 \times 6 = 48$

(A) The equilibrium constant expression is: $K_c = \frac{[H_2]^2 [S_2]}{[H_2S]^2}$
First,calculate the molar concentrations (moles / volume in $L$):
$[H_2S] = \frac{1 \ mol}{2 \ L} = 0.5 \ M$
$[H_2] = \frac{0.2 \ mol}{2 \ L} = 0.1 \ M$
$[S_2] = \frac{0.8 \ mol}{2 \ L} = 0.4 \ M$
Substitute these values into the $K_c$ expression:
$K_c = \frac{(0.1)^2 \times (0.4)}{(0.5)^2}$
$K_c = \frac{0.01 \times 0.4}{0.25} = \frac{0.004}{0.25} = 0.016$

Species	$2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$
Equilibrium Moles	$1, 0.2, 0.8$
Concentration $(M)$	$0.5, 0.1, 0.4$

(B) Step $1$: Calculate the molar masses of the substances:
$M(SO_2) = 32 + 2 \times 16 = 64 \ g/mol$
$M(O_2) = 2 \times 16 = 32 \ g/mol$
$M(SO_3) = 32 + 3 \times 16 = 80 \ g/mol$
Step $2$: Calculate the number of moles at equilibrium:
$n(SO_2) = \frac{12.8 \ g}{64 \ g/mol} = 0.2 \ mol$
$n(O_2) = \frac{9.6 \ g}{32 \ g/mol} = 0.3 \ mol$
$n(SO_3) = \frac{48 \ g}{80 \ g/mol} = 0.6 \ mol$
Step $3$: Since the volume is $1 \ L$,the concentrations are equal to the number of moles:
$[SO_2] = 0.2 \ M, [O_2] = 0.3 \ M, [SO_3] = 0.6 \ M$
Step $4$: Calculate $K_c$ using the equilibrium expression:
$K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]} = \frac{(0.6)^2}{(0.2)^2 \times 0.3} = \frac{0.36}{0.04 \times 0.3} = \frac{0.36}{0.012} = 30$

(B) The equilibrium constant $K$ for a general reversible reaction $aA + bB \rightleftharpoons cC + dD$ is defined as the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants,each raised to the power of their stoichiometric coefficients.
For the reaction $CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}$,the expression is:
$K = \frac{[H_3O^{+}][CH_3COO^{-}]}{[CH_3COOH][H_2O]}$
Comparing this with the given options,option $B$ matches this expression.

(B) For the first reaction: $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
$K_1 = \frac{[SO_3]^2}{[SO_2]^2 [O_2]}$
For the second reaction: $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$
$K_2 = \frac{[SO_3]}{[SO_2] [O_2]^{1/2}}$
Comparing the two expressions,we can see that:
$K_2 = \sqrt{\frac{[SO_3]^2}{[SO_2]^2 [O_2]}} = \sqrt{K_1}$
Therefore,$K_2 = \sqrt{K_1}$.

(A) The equilibrium reaction is: $Ag^{+} + 2NH_3 \rightleftharpoons Ag(NH_3)_2^{+}$
The expression for the equilibrium constant $K_c$ is:
$K_c = \frac{[Ag(NH_3)_2^{+}]}{[Ag^{+}][NH_3]^2}$
Substituting the given values:
$K_c = \frac{10^{-1}}{(10^{-1}) \times (10^3)^2}$
$K_c = \frac{10^{-1}}{10^{-1} \times 10^6}$
$K_c = \frac{1}{10^6} = 10^{-6}$

(A) For the reaction $A + 3B \rightleftharpoons 2C$,the equilibrium constant is $K_1 = \frac{[C]^2}{[A][B]^3}$.
For the reverse reaction $2C \rightleftharpoons A + 3B$,the equilibrium constant is $K_2 = \frac{[A][B]^3}{[C]^2}$.
Comparing the two expressions,we see that $K_2 = \frac{1}{K_1}$.

(D) For the reaction $N_2O_2 \rightleftharpoons 2NO$:

Condition	Reaction: $N_2O_2 \rightleftharpoons 2NO$
Initial moles	$a$ $\quad$ $0$
Moles at equilibrium	$a - x$ $\quad$ $2x$

Concentration at equilibrium:
$[N_2O_2] = \frac{a - x}{V}$
$[NO] = \frac{2x}{V}$
The equilibrium constant $K_c$ is given by:
$K_c = \frac{[NO]^2}{[N_2O_2]}$
$K_c = \frac{(\frac{2x}{V})^2}{(\frac{a - x}{V})}$
$K_c = \frac{4x^2}{V^2} \times \frac{V}{a - x}$
$K_c = \frac{4x^2}{(a - x)V}$

(B) The given reactions are:
$(I) \, CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2_{(g)}} + H_{2_{(g)}} ; \, k_1$
$(II) \, CH_{4_{(g)}} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + 3H_{2_{(g)}} ; \, k_2$
$(III) \, CH_{4_{(g)}} + 2H_2O_{(g)} \rightleftharpoons CO_{2_{(g)}} + 4H_{2_{(g)}} ; \, k_3$
Observe that adding reaction $(I)$ and reaction $(II)$ gives reaction $(III)$:
$(CO_{(g)} + H_2O_{(g)}) + (CH_{4_{(g)}} + H_2O_{(g)}) \rightleftharpoons (CO_{2_{(g)}} + H_{2_{(g)}}) + (CO_{(g)} + 3H_{2_{(g)}})$
Canceling common terms $(CO_{(g)})$ on both sides:
$CH_{4_{(g)}} + 2H_2O_{(g)} \rightleftharpoons CO_{2_{(g)}} + 4H_{2_{(g)}}$
Since reaction $(III) = (I) + (II)$,the equilibrium constant $k_3$ is the product of $k_1$ and $k_2$:
$k_3 = k_1 \times k_2$

(B) The reaction is $A + B \rightleftharpoons C + D$.
Initial moles: $A = 4, B = 4, C = 0, D = 0$.
At equilibrium,$2 \ mol$ of $C$ is formed. Since the stoichiometry is $1:1$,$2 \ mol$ of $A$ and $2 \ mol$ of $B$ are consumed,and $2 \ mol$ of $D$ are produced.
Equilibrium moles: $A = 4 - 2 = 2, B = 4 - 2 = 2, C = 2, D = 2$.
Equilibrium constant $K_c = \frac{[C][D]}{[A][B]} = \frac{(2/V) \times (2/V)}{(2/V) \times (2/V)} = 1$.

(C) The equilibrium constant $K_p$ (or $K_c$) depends only on the temperature of the reaction.
Since the temperature remains constant,the value of the equilibrium constant $K_p$ will remain unchanged regardless of changes in volume,pressure,or concentration.
Therefore,the value of $K_p$ remains $16$.

(D) The balanced chemical equation is $A + B \rightleftharpoons 2C$.
The expression for the equilibrium constant $K_c$ is given by:
$K_c = \frac{[C]^2}{[A][B]}$
Given equilibrium concentrations:
$[A] = 0.20 \ mol/L$
$[B] = 0.20 \ mol/L$
$[C] = 0.60 \ mol/L$
Substituting these values into the expression:
$K_c = \frac{(0.60)^2}{(0.20)(0.20)} = \frac{0.36}{0.04} = 9$
Therefore,the equilibrium constant is $9$.

(B) For a reaction,$K_c = \frac{[\text{Product}]}{[\text{Reactant}]}$.
If $K_c > 1$,then $[\text{Product}] > [\text{Reactant}]$.
Among the given options,only option $B$ has $k = 10$,which is greater than $1$.

(C) For a general reversible reaction $aA + bB \rightleftharpoons cC + dD$,the equilibrium constant $K_c$ is given by the ratio of the product of concentrations of products to the product of concentrations of reactants,each raised to the power of their stoichiometric coefficients.
For the reaction $A + 2B \rightleftharpoons C$,the equilibrium constant $K_c$ is expressed as:
$K_c = \frac{[C]}{[A][B]^2}$

(C) The unit of the equilibrium constant $K_c$ is given by the formula: $\text{Unit of } K_c = (\text{unit of concentration})^{\Delta n}$.
Here,$\Delta n$ is the change in the number of moles of gaseous products and reactants: $\Delta n = (4 + 6) - (4 + 5) = 10 - 9 = 1$.
Substituting the value of $\Delta n$ into the formula: $\text{Unit of } K_c = (\text{mol} \cdot \text{L}^{-1})^1 = \text{mol} \cdot \text{L}^{-1}$.
Therefore,the correct unit is $Mole \ litre^{-1}$.

(D) The direction of a reaction can be predicted by comparing the reaction quotient $Q$ with the equilibrium constant $K_c$.
If $Q > K_c$,the concentration of products is higher than at equilibrium,so the reaction proceeds in the reverse direction (right to left) to reach equilibrium.
If $Q < K_c$,the reaction proceeds in the forward direction (left to right).
If $Q = K_c$,the reaction is at equilibrium.
Therefore,the correct condition for the reaction to proceed from right to left is $Q > K_c$.

(C) The dissociation reaction is: $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$
Let the initial concentration of $HI$ be $2 \, \text{moles}$.
At equilibrium,the amount of $HI$ dissociated is $22\%$ of $2 = 0.44 \, \text{moles}$.
Remaining $HI = 2 - 0.44 = 1.56 \, \text{moles}$.
According to the stoichiometry,$2 \, \text{moles}$ of $HI$ produce $1 \, \text{mole}$ of $H_2$ and $1 \, \text{mole}$ of $I_2$.
Therefore,$0.44 \, \text{moles}$ of $HI$ produce $0.22 \, \text{moles}$ of $H_2$ and $0.22 \, \text{moles}$ of $I_2$.
The equilibrium constant $K_c$ is given by:
$K_c = \frac{[H_2][I_2]}{[HI]^2} = \frac{0.22 \times 0.22}{(1.56)^2} = \frac{0.0484}{2.4336} \approx 0.0199$.

(C) The equilibrium constant $K$ for a general reaction $aA + bB \rightleftharpoons cC + dD$ is given by the expression $K = \frac{[C]^c [D]^d}{[A]^a [B]^b}$.
For the given reaction $2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)}$,the equilibrium constant $K$ is expressed as the ratio of the product of the concentrations of the products raised to their stoichiometric coefficients to the product of the concentrations of the reactants raised to their stoichiometric coefficients.
Thus,$K = \frac{[C]^3}{[A]^2 [B]}$.

(D) For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)},$ the equilibrium constant is $K = \frac{[NO]^2}{[N_2][O_2]}.$
For the reaction $\frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{(g)},$ the equilibrium constant $K'$ is given by $K' = \frac{[NO]}{[N_2]^{1/2}[O_2]^{1/2}}.$
Comparing the two expressions,we see that $K' = \sqrt{\frac{[NO]^2}{[N_2][O_2]}} = \sqrt{K} = K^{1/2}.$

(C) For the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$,the equilibrium constant is $K = 278$.
Reversing the reaction gives $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$,for which the equilibrium constant is $K' = \frac{1}{K} = \frac{1}{278}$.
Dividing the stoichiometric coefficients of the reversed reaction by $2$ gives the target reaction: $SO_{3(g)} \rightleftharpoons SO_{2(g)} + \frac{1}{2} O_{2(g)}$.
The equilibrium constant for this reaction is $K'' = \sqrt{K'} = \sqrt{\frac{1}{278}} \approx 0.0599 \approx 6.0 \times 10^{-2}$.

(C) The equilibrium reaction is: $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$
The expression for the equilibrium constant $K_c$ is:
$K_c = \frac{[AB]^2}{[A_2][B_2]}$
Substituting the given equilibrium concentrations:
$K_c = \frac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}$
$K_c = \frac{2.8 \times 2.8}{3.0 \times 4.2} = \frac{7.84}{12.6} \approx 0.622$
Rounding to two decimal places,the value of $K_c$ is $0.62$.

(C) Given reactions:
$1) N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)} ; K_1$
$2) 2NO_{(g)} + O_{2(g)} \rightleftharpoons 2NO_{2(g)} ; K_2$
Adding equations $(1)$ and $(2)$ gives:
$N_{2(g)} + 2O_{2(g)} \rightleftharpoons 2NO_{2(g)}$
The equilibrium constant for this combined reaction is $K_{eq} = K_1 \times K_2$.
Now,to obtain the target reaction $NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$,we reverse the combined reaction and multiply the coefficients by $\frac{1}{2}$.
Reversing the reaction gives $K' = \frac{1}{K_1 K_2}$.
Multiplying coefficients by $\frac{1}{2}$ gives $K = (K')^{1/2} = \left[ \frac{1}{K_1 K_2} \right]^{1/2}$.

(B) The reaction is $2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$.
Initial concentrations: $[A] = 1.00 \ M$,$[B] = 1.00 \ M$,$[C] = 0 \ M$,$[D] = 0 \ M$.
At equilibrium,$[D] = 0.25 \ M$. Since the stoichiometry of $D$ is $1$,the change in concentration for $D$ is $+0.25 \ M$.
Using stoichiometry:
$[A]_{eq} = 1.00 - 2(0.25) = 0.50 \ M$
$[B]_{eq} = 1.00 - 0.25 = 0.75 \ M$
$[C]_{eq} = 3(0.25) = 0.75 \ M$
$[D]_{eq} = 0.25 \ M$
The equilibrium constant expression is $K_c = \frac{[C]^3[D]}{[A]^2[B]}$.
Substituting the values: $K_c = \frac{(0.75)^3(0.25)}{(0.50)^2(0.75)}$.

(C) The given reaction is $:$
$CH_{4(g)} + 2O_{2(g)} \rightleftharpoons CO_{2(g)} + 2H_2O_{(l)}$
The equilibrium constant expression for this reaction is $K_c = \frac{[CO_2]}{[CH_4][O_2]^2}$ because $H_2O$ is a pure liquid and its activity is taken as $1$.
Option $A$ is true because $\Delta H_r$ is negative,indicating an exothermic reaction.
Option $B$ is true because there is no requirement for the concentrations of products to be equal at equilibrium.
Option $C$ is incorrect because $K_p$ is defined in terms of partial pressures,not concentrations. The expression given is for $K_c$,and it is also missing the pressure terms for the gaseous species.
Option $D$ is true according to Le Chatelier's principle; adding reactants shifts the equilibrium to the right.

(C) Given reactions are:
$(i)$ $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$ ; $K_1$
$(ii)$ $CH_{4(g)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + 3H_{2(g)}$ ; $K_2$
Adding reaction $(i)$ and $(ii)$:
$(CO_{(g)} + H_2O_{(g)}) + (CH_{4(g)} + H_2O_{(g)}) \rightleftharpoons (CO_{2(g)} + H_{2(g)}) + (CO_{(g)} + 3H_{2(g)})$
Canceling $CO_{(g)}$ from both sides:
$CH_{4(g)} + 2H_2O_{(g)} \rightleftharpoons CO_{2(g)} + 4H_{2(g)}$
This is reaction $(iii)$.
When reactions are added,their equilibrium constants are multiplied.
Therefore,$K_3 = K_1 \cdot K_2$.

(D) For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$,the equilibrium constant is $K_c = 4 \times 10^{-4}$.
The given reaction is $NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)}$.
This reaction is the reverse of the original reaction multiplied by a factor of $\frac{1}{2}$.
Therefore,the new equilibrium constant $K_c'$ is given by $K_c' = \frac{1}{\sqrt{K_c}}$.
$K_c' = \frac{1}{\sqrt{4 \times 10^{-4}}} = \frac{1}{2 \times 10^{-2}} = \frac{1}{0.02} = 50$.

(A) The balanced chemical equation for the dissociation of $SO_{3(g)}$ is:
$2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$
From the graph,at equilibrium:
$P_{SO_3} = 2 \ atm$
$P_{SO_2} = 3 \ atm$
$P_{O_2} = 1.5 \ atm$ (The curve for $O_2$ levels off at $1.5$)
The expression for the equilibrium constant $K_P$ is:
$K_P = \frac{(P_{SO_2})^2 \times (P_{O_2})}{(P_{SO_3})^2}$
Substituting the equilibrium partial pressures:
$K_P = \frac{(3)^2 \times (1.5)}{(2)^2} = \frac{9 \times 1.5}{4} = \frac{13.5}{4} = 3.375$
Checking the options,$3.375 = \frac{27}{8} = (\frac{3}{2})^3 = \frac{3^3}{2^3} = \frac{3^{1.5}}{2^{1.5}}$.
Thus,$x = \frac{3^{1.5}}{2^{1.5}}$.

(D) For the reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$

$Reaction$	$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$
$Initial \ moles$	$5, 5, 0$
$Change \ in \ moles$	$-x, -x, +2x$
$Equilibrium \ moles$	$5-x, 5-x, 2x$

Given that at equilibrium,$n_{HI} = 2 \ moles$.
Therefore,$2x = 2 \Rightarrow x = 1$.
Equilibrium moles: $n_{H_2} = 5 - 1 = 4 \ moles$,$n_{I_2} = 5 - 1 = 4 \ moles$,$n_{HI} = 2 \ moles$.
Volume $V = 10 \ L$.
Equilibrium concentrations: $[H_2] = 4/10 = 0.4 \ M$,$[I_2] = 4/10 = 0.4 \ M$,$[HI] = 2/10 = 0.2 \ M$.
$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.2)^2}{(0.4) \times (0.4)} = \frac{0.04}{0.16} = 0.25$.

(C) Given reactions:
$(1) \ 2PQ \rightleftharpoons P_2 + Q_2\,;\,K_1 = 2.5 \times 10^5$
$(2) \ PQ + \frac{1}{2}R_2 \rightleftharpoons PQR\,;\,K_2 = 5 \times 10^{-3}$
To obtain the target reaction $\frac{1}{2}P_2 + \frac{1}{2}Q_2 + \frac{1}{2}R_2 \rightleftharpoons PQR$,we manipulate the given reactions:
Reverse reaction $(1)$ and divide by $2$:
$\frac{1}{2}P_2 + \frac{1}{2}Q_2 \rightleftharpoons PQ\,;\,K_3 = \sqrt{\frac{1}{K_1}} = \sqrt{\frac{1}{2.5 \times 10^5}} = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3}$
Add this new reaction to reaction $(2)$:
$(\frac{1}{2}P_2 + \frac{1}{2}Q_2 \rightleftharpoons PQ) + (PQ + \frac{1}{2}R_2 \rightleftharpoons PQR) \implies \frac{1}{2}P_2 + \frac{1}{2}Q_2 + \frac{1}{2}R_2 \rightleftharpoons PQR$
The equilibrium constant $K_{eq}$ for the target reaction is:
$K_{eq} = K_3 \times K_2 = (2 \times 10^{-3}) \times (5 \times 10^{-3}) = 10 \times 10^{-6} = 1 \times 10^{-5}$

(D) The stability of a compound is inversely proportional to its tendency to decompose into its elements.
Greater the value of the equilibrium constant $(K_{eq})$ for the decomposition reaction,the more unstable the compound is.
Comparing the given $K_{eq}$ values:
$6.7 \times 10^{16} < 1.2 \times 10^{24} < 2.2 \times 10^{30} < 3.5 \times 10^{33}$.
Since $2N_2O_{(g)} \rightleftharpoons 2N_{2(g)} + O_{2(g)}$ has the largest $K_{eq}$ value $(3.5 \times 10^{33})$,$N_2O$ is the least stable oxide of nitrogen.

(D) The overall reaction is the sum of the two given steps:
$Ag^{+} + NH_3 \rightleftharpoons [Ag(NH_3)]^+$,$K_1 = 1.6 \times 10^3$
$[Ag(NH_3)]^+ + NH_3 \rightleftharpoons [Ag(NH_3)_2]^+$,$K_2 = 6.8 \times 10^3$
Adding these reactions gives:
$Ag^{+} + 2NH_3 \rightleftharpoons [Ag(NH_3)_2]^+$
The formation constant $K_f$ (or $K_3$) for the overall reaction is the product of the equilibrium constants of the individual steps:
$K_f = K_1 \times K_2$
$K_f = (1.6 \times 10^3) \times (6.8 \times 10^3)$
$K_f = 10.88 \times 10^6 = 1.088 \times 10^7 \approx 1.08 \times 10^7$

(D) Given reactions:
$(i) \ XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, K_1$
$(ii) \ XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2, K_2$
To obtain the target reaction $XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$,we perform the operation $(ii) - (i)$:
$(XeO_4 + XeF_6) - (XeF_6 + H_2O) \rightleftharpoons (XeOF_4 + XeO_3F_2) - (XeOF_4 + 2HF)$
$XeO_4 + XeF_6 - XeF_6 - H_2O \rightleftharpoons XeOF_4 + XeO_3F_2 - XeOF_4 - 2HF$
$XeO_4 - H_2O \rightleftharpoons XeO_3F_2 - 2HF$
Rearranging gives: $XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$
Since we subtracted reaction $(i)$ from reaction $(ii)$,the new equilibrium constant $K$ is given by $\frac{K_2}{K_1}$.

(D) The equilibrium constant expression for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is given by:
$K_c = \frac{[HI]^2}{[H_2][I_2]}$
Given:
$K_c = 32$
$[HI] = 8 \times 10^{-3} \ M$
$[I_2] = 0.5 \times 10^{-3} \ M$
Substituting the values into the expression:
$32 = \frac{(8 \times 10^{-3})^2}{[H_2] \times (0.5 \times 10^{-3})}$
$32 = \frac{64 \times 10^{-6}}{[H_2] \times 0.5 \times 10^{-3}}$
$[H_2] = \frac{64 \times 10^{-6}}{32 \times 0.5 \times 10^{-3}}$
$[H_2] = \frac{64 \times 10^{-6}}{16 \times 10^{-3}}$
$[H_2] = 4 \times 10^{-3} \ M$

(A) From the graph,at equilibrium,the concentration of species $A$ is $[A] = 0.1 \ M$ and the concentration of species $B$ is $[B] = 0.4 \ M$.
For the reaction $2A_{(g)} \rightleftharpoons B_{(g)}$,the equilibrium constant $K_C$ is given by:
$K_C = \frac{[B]}{[A]^2}$
Substituting the equilibrium concentrations:
$K_C = \frac{0.4}{(0.1)^2} = \frac{0.4}{0.01} = 40$
Since $40 > 1$,the correct option is $K_C > 1$.

(A) The reaction is $A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)}$.
The reaction quotient $Q_c$ is calculated as $Q_c = \frac{[C]^2}{[A][B]^2}$.
Given concentrations in a $1 \, L$ vessel: $[A] = 2 \, M$,$[B] = 1 \, M$,$[C] = 1 \, M$.
Substituting these values: $Q_c = \frac{1^2}{2 \times 1^2} = \frac{1}{2} = 0.5$.
Since $Q_c < K_c$ $(0.5 < 2)$,the reaction will proceed in the forward direction.

(C) The equilibrium reaction is $2HX_{(g)} \rightleftharpoons H_{2(g)} + X_{2(g)}$.
The expression for the equilibrium constant $K$ is $K = \frac{[H_2][X_2]}{[HX]^2}$.
Given $K = 1 \times 10^{-5}$,$[H_2] = 1.2 \times 10^{-3} \ M$,and $[X_2] = 1.2 \times 10^{-4} \ M$.
Substituting the values: $10^{-5} = \frac{(1.2 \times 10^{-3}) \times (1.2 \times 10^{-4})}{[HX]^2}$.
$[HX]^2 = \frac{1.44 \times 10^{-7}}{10^{-5}} = 1.44 \times 10^{-2}$.
$[HX] = \sqrt{1.44 \times 10^{-2}} = 1.2 \times 10^{-1} \ M$.
Converting to the format of the options: $1.2 \times 10^{-1} \ M = 12 \times 10^{-2} \ M$.

(B) For reaction $(I): N_2 + 2O_2 \rightleftharpoons 2NO_2$,the equilibrium constant is $K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2} \dots (i)$
For reaction $(II): 2NO_2 \rightleftharpoons N_2 + 2O_2$,which is the reverse of $(I)$,the equilibrium constant is $K_2 = \frac{[N_2][O_2]^2}{[NO_2]^2} = \frac{1}{K_1} \dots (ii)$
For reaction $(III): NO_2 \rightleftharpoons \frac{1}{2} N_2 + O_2$,the equilibrium constant is $K_3 = \frac{[N_2]^{1/2}[O_2]}{[NO_2]}$
Squaring $K_3$,we get $(K_3)^2 = \frac{[N_2][O_2]^2}{[NO_2]^2} = K_2 = \frac{1}{K_1} \dots (iii)$
Thus,$K_1 = \frac{1}{K_2} = \frac{1}{(K_3)^2}$.