Law of equilibrium and Equilibrium constant Questions in English

(B) For a general chemical 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 given reaction $3A + 2B \rightleftharpoons C$,the equilibrium constant expression is $K_c = \frac{[C]}{[A]^3 [B]^2}$.
Thus,the correct option is $B$ or $D$ (as both are identical).

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

(A) The chemical equation is: $H_2 + I_2 \rightleftharpoons 2HI$
Given equilibrium concentrations are: $[HI] = 0.80 \ mol/L$,$[H_2] = 0.10 \ mol/L$,and $[I_2] = 0.10 \ mol/L$.
The expression for the equilibrium constant $K_c$ is:
$K_c = \frac{[HI]^2}{[H_2][I_2]}$
Substituting the given values:
$K_c = \frac{(0.80)^2}{(0.10)(0.10)} = \frac{0.64}{0.01} = 64$
Therefore,the equilibrium constant is $64$.

(A) The equilibrium constant expression for the reaction $A_{(g)} + 2B_{(g)} \rightleftharpoons C_{(g)}$ is given by:
$K_{eq} = \frac{[C]}{[A][B]^2}$
Substituting the given equilibrium concentrations:
$K_{eq} = \frac{0.216}{(0.06) \times (0.12)^2}$
$K_{eq} = \frac{0.216}{0.06 \times 0.0144}$
$K_{eq} = \frac{0.216}{0.000864}$
$K_{eq} = 250$

(D) For a general reversible reaction $aA + bB \rightleftharpoons cC + dD$,the equilibrium constant $K_c$ 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 $H_2 + I_2 \rightleftharpoons 2HI$,the expression is:
$K_c = \frac{[HI]^2}{[H_2][I_2]}$
Therefore,the correct expression is given in option $D$.

(C) For a general reversible reaction $aA + bB \rightleftharpoons cC + dD$,the equilibrium constant $K_c$ is defined as $K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$.
Given the reaction $A + 2B \rightleftharpoons C$,the stoichiometric coefficients are $a=1$,$b=2$,and $c=1$.
Substituting these values into the expression,we get $K_c = \frac{[C]^1}{[A]^1 [B]^2} = \frac{[C]}{[A][B]^2}$.

(A) The dissociation reaction is: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$
Initial moles: $2$,$0$,$0$
Degree of dissociation $\alpha = 0.40$.
Moles at equilibrium:
$PCl_5 = 2(1 - 0.4) = 1.2 \ mol$
$PCl_3 = 2 \times 0.4 = 0.8 \ mol$
$Cl_2 = 2 \times 0.4 = 0.8 \ mol$
Concentrations at equilibrium (Volume $= 2 \ L$):
$[PCl_5] = \frac{1.2}{2} = 0.6 \ M$
$[PCl_3] = \frac{0.8}{2} = 0.4 \ M$
$[Cl_2] = \frac{0.8}{2} = 0.4 \ M$
Equilibrium constant $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{0.4 \times 0.4}{0.6} = \frac{0.16}{0.6} = 0.266$.

(D) For the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$,the change in the number of moles of gaseous products and reactants is $\Delta n = n_p - n_r = 2 - (1 + 1) = 0$.
The unit of the equilibrium constant $K_c$ is given by $(mol \ L^{-1})^{\Delta n}$.
Substituting $\Delta n = 0$,we get $(mol \ L^{-1})^0 = 1$.
Therefore,the equilibrium constant is dimensionless for this reaction.
Thus,the correct option is $D$.

(C) The reaction is $N_2O_4 \rightleftharpoons 2NO_2$.
First,calculate the molar concentrations of the species at equilibrium:
$[N_2O_4] = \frac{0.2 \ mol}{2 \ L} = 0.1 \ M$.
$[NO_2] = \frac{2 \times 10^{-3} \ mol}{2 \ L} = 10^{-3} \ M$.
The equilibrium constant expression is $K_c = \frac{[NO_2]^2}{[N_2O_4]}$.
Substituting the values: $K_c = \frac{(10^{-3})^2}{0.1} = \frac{10^{-6}}{10^{-1}} = 10^{-5}$.

(A) In the calculation of the equilibrium constant $(K_c)$,the concentration of a gas is expressed in terms of molarity,which is defined as the number of moles per litre $(mol \ L^{-1})$.

(B) The equilibrium constant $K$ is given by the expression $K = \frac{[C]}{[A][B]}$.
Substituting the units of concentration $(mol \ L^{-1})$,we get:
$K = \frac{mol \ L^{-1}}{(mol \ L^{-1})(mol \ L^{-1})} = \frac{1}{mol \ L^{-1}} = L \ mol^{-1}$.

(B) For the reaction $A + B \rightleftharpoons C + D$,the equilibrium constant $K_c$ is given by the expression:
$K_c = \frac{[C][D]}{[A][B]}$
Given concentrations are $[A] = 0.5 \ mol/L$,$[B] = 0.8 \ mol/L$,$[C] = 0.4 \ mol/L$,and $[D] = 1.0 \ mol/L$.
Substituting these values into the expression:
$K_c = \frac{0.4 \times 1.0}{0.5 \times 0.8} = \frac{0.4}{0.4} = 1$.
Thus,the equilibrium constant is $1$.

(C) The chemical equation is $A + B \rightleftharpoons C + D$.
Initial moles: $A = 1, B = 1, C = 0, D = 0$.
Moles at equilibrium: $A = 1 - 0.6 = 0.4, B = 1 - 0.6 = 0.4, C = 0.6, D = 0.6$.
The equilibrium constant $K_c$ is given by the expression: $K_c = \frac{[C][D]}{[A][B]}$.
Substituting the equilibrium concentrations: $K_c = \frac{0.6 \times 0.6}{0.4 \times 0.4} = \frac{0.36}{0.16} = 2.25$.

(A) For a general reversible reaction $aA + bB \rightleftharpoons cC + dD$,the equilibrium constant $K$ is defined as $K = \frac{[C]^c [D]^d}{[A]^a [B]^b}$.
Applying this to the given reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant expression is $K = \frac{[NH_3]^2}{[N_2][H_2]^3}$.
Therefore,the correct option is $A$.

(B) The reaction is $2NO_{(g)} + Cl_{2(g)} \rightleftharpoons 2NOCl_{(g)}$.
The equilibrium constant $(K_c)$ is defined as the ratio of the product of the equilibrium concentrations of the products to the product of the equilibrium concentrations of the reactants,with each concentration term raised to the power of its stoichiometric coefficient.
For the given reaction,the stoichiometric coefficient of $NO$ is $2$,$Cl_2$ is $1$,and $NOCl$ is $2$.
Therefore,the expression for the equilibrium constant is $K_c = \frac{[NOCl]^2}{[NO]^2[Cl_2]^1}$ or $K_c = \frac{[NOCl]^2}{[NO]^2[Cl_2]}$.

(A) The given reaction is $2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g)$.
The expression for the equilibrium constant $K_c$ is $K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$.
The change in the number of moles of gaseous products and reactants is $\Delta n_g = n_p - n_r = 2 - (2 + 1) = -1$.
The general unit for $K_c$ is $(mol \ L^{-1})^{\Delta n_g}$.
Substituting $\Delta n_g = -1$,the unit becomes $(mol \ L^{-1})^{-1} = L \ mol^{-1}$.

(C) The equilibrium reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$.
Given volume $V = 10 \ L$.
Equilibrium concentrations are:
$[PCl_5] = \frac{0.1 \ mol}{10 \ L} = 0.01 \ M$
$[PCl_3] = \frac{0.2 \ mol}{10 \ L} = 0.02 \ M$
$[Cl_2] = \frac{0.2 \ mol}{10 \ L} = 0.02 \ M$
The equilibrium constant $K_c$ is given by:
$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}$
$K_c = \frac{0.02 \times 0.02}{0.01} = \frac{0.0004}{0.01} = 0.04$.

(A) Let the initial concentration of $A$ and $B$ be $x \ mol/L$.
Reaction: $A + B \rightleftharpoons C + D$
Initial: $x, x, 0, 0$
At equilibrium: $(x-y), (x-y), y, y$
Given that at equilibrium,the concentration of $C$ is twice that of $A$: $y = 2(x-y)$
$y = 2x - 2y \implies 3y = 2x \implies y = \frac{2}{3}x$
Equilibrium concentrations:
$[A] = x - \frac{2}{3}x = \frac{1}{3}x$
$[B] = x - \frac{2}{3}x = \frac{1}{3}x$
$[C] = y = \frac{2}{3}x$
$[D] = y = \frac{2}{3}x$
$K_c = \frac{[C][D]}{[A][B]} = \frac{(\frac{2}{3}x)(\frac{2}{3}x)}{(\frac{1}{3}x)(\frac{1}{3}x)} = \frac{4/9}{1/9} = 4$

(A) The chemical equation is: ${H_2}_{(g)} + {I_2}_{(g)} \rightleftharpoons 2HI_{(g)}$
Initial moles: $4.5 \ mol$ of ${H_2}$ and $4.5 \ mol$ of ${I_2}$ are taken in a $10 \ L$ vessel.
Let $x$ be the extent of reaction such that at equilibrium,moles of $HI$ formed is $2x$.
Given $2x = 3$,so $x = 1.5 \ mol$.
Equilibrium moles:
${H_2} = 4.5 - 1.5 = 3.0 \ mol$
${I_2} = 4.5 - 1.5 = 3.0 \ mol$
$HI = 3.0 \ mol$
Equilibrium concentrations (moles / $10 \ L$):
$[H_2] = 3.0 / 10 = 0.3 \ M$
$[I_2] = 3.0 / 10 = 0.3 \ M$
$[HI] = 3.0 / 10 = 0.3 \ M$
Equilibrium constant ${K_c} = \frac{{[HI]}^2}{[H_2][I_2]} = \frac{(0.3)^2}{(0.3)(0.3)} = 1$.

(C) The given reaction is $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$.
Since the volume of the vessel is $1 L$,the molar concentrations are equal to the number of moles:
$[H_2S] = 0.5 mol L^{-1}$
$[H_2] = 0.10 mol L^{-1}$
$[S_2] = 0.4 mol L^{-1}$
The expression for the equilibrium constant $K_c$ is:
$K_c = \frac{[H_2]^2 [S_2]}{[H_2S]^2}$
Substituting the values:
$K_c = \frac{(0.10)^2 \times (0.4)}{(0.5)^2} = \frac{0.01 \times 0.4}{0.25} = \frac{0.004}{0.25} = 0.016 mol L^{-1}$.

(B) The equilibrium constant expression for the reaction $I_2 + H_2 \rightleftharpoons 2HI$ is given by:
$K_c = \frac{[HI]^2}{[H_2][I_2]}$
Substituting the given equilibrium concentrations:
$K_c = \frac{(0.7)^2}{(0.1)(0.1)}$
$K_c = \frac{0.49}{0.01} = 49$
Therefore,the correct option is $B$.

(C) The equilibrium constant expression for the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$ is given by:
$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$
Given $K_c = 2.37 \times 10^{-3}$,$[N_2] = 2 \ M$,and $[H_2] = 3 \ M$.
Substituting these values into the expression:
$2.37 \times 10^{-3} = \frac{[NH_3]^2}{(2)(3)^3}$
$2.37 \times 10^{-3} = \frac{[NH_3]^2}{2 \times 27}$
$2.37 \times 10^{-3} = \frac{[NH_3]^2}{54}$
$[NH_3]^2 = 2.37 \times 10^{-3} \times 54 = 0.12798$
$[NH_3] = \sqrt{0.12798} \approx 0.358 \ M$

(D) The chemical equation is $A + B \rightleftharpoons 2C$.
At equilibrium,the concentrations are given as $[A] = 0.20 \ mol \ L^{-1}$,$[B] = 0.20 \ mol \ L^{-1}$,and $[C] = 0.60 \ mol \ L^{-1}$.
The expression for the equilibrium constant $K_c$ is $K_c = \frac{[C]^2}{[A][B]}$.
Substituting the given values: $K_c = \frac{(0.60)^2}{(0.20)(0.20)} = \frac{0.36}{0.04} = 9$.
Thus,the equilibrium constant is $9$.

(A) The chemical equation for the reaction is: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$
Initial moles:
$H_2 = 15$,$I_2 = 5.2$,$HI = 0$
At equilibrium,the moles of $HI$ formed is $10$. According to the stoichiometry,$2$ moles of $HI$ are formed from $1$ mole of $H_2$ and $1$ mole of $I_2$. Therefore,$5$ moles of $H_2$ and $5$ moles of $I_2$ are consumed.
Equilibrium moles:
$H_2 = 15 - 5 = 10$
$I_2 = 5.2 - 5 = 0.2$
$HI = 10$
The equilibrium constant $K_C$ is given by:
$K_C = \frac{[HI]^2}{[H_2][I_2]} = \frac{10^2}{10 \times 0.2} = \frac{100}{2} = 50$

(B) For a reversible reaction $A \rightleftharpoons B$,the rate of forward reaction is $r_f = k_f[A]$ and the rate of reverse reaction is $r_b = k_b[B]$.
Given that at identical concentrations,the rate of forward reaction is two times the rate of reverse reaction,we have $r_f = 2r_b$.
Substituting the rate expressions: $k_f[A] = 2k_b[B]$.
Since the concentrations are identical,$[A] = [B]$,which implies $k_f = 2k_b$.
The equilibrium constant $K_{eq}$ is defined as the ratio of the rate constants: $K_{eq} = \frac{k_f}{k_b} = \frac{2}{1} = 2$.

(B) The equilibrium constant $K$ for a reversible reaction 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:
$K = \frac{[H_3O^{+}][CH_3COO^{-}]}{[CH_3COOH][H_2O]}$

(D) The balanced chemical equation is: $2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$
Initially,$2 \ mol$ of $NH_3$ are present in a $1 \ L$ vessel.
At equilibrium,$1 \ mol$ of $NH_3$ remains,meaning $1 \ mol$ has dissociated.
According to the stoichiometry,$2 \ mol$ of $NH_3$ produce $1 \ mol$ of $N_2$ and $3 \ mol$ of $H_2$.
Therefore,$1 \ mol$ of $NH_3$ dissociation produces $0.5 \ mol$ of $N_2$ and $1.5 \ mol$ of $H_2$.
The equilibrium concentrations are: $[NH_3] = 1 \ mol/L$,$[N_2] = 0.5 \ mol/L$,and $[H_2] = 1.5 \ mol/L$.
The equilibrium constant $K_c$ is given by: $K_c = \frac{[N_2][H_2]^3}{[NH_3]^2} = \frac{0.5 \times (1.5)^3}{1^2} = 0.5 \times 3.375 = 1.6875 = \frac{27}{16} \ mol^2 \ L^{-2}$.

(C) The equilibrium constant expression for the reaction is $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}$.
Given that $[PCl_5] = 0.4 \ mol/L$,$[PCl_3] = 0.2 \ mol/L$,and $K_c = 0.5$.
Let the concentration of $Cl_2$ be $x \ mol/L$.
Substituting the values into the expression: $0.5 = \frac{0.2 \times x}{0.4}$.
Solving for $x$: $x = \frac{0.5 \times 0.4}{0.2} = \frac{0.2}{0.2} = 1 \ mol/L$.

(C) The equilibrium constant $K_c$ for the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ is given by the expression:
$K_c = \frac{[NO_2]^2}{[N_2O_4]}$
Given:
$[NO_2] = 1.2 \times 10^{-2} \ mol \ L^{-1}$
$[N_2O_4] = 4.8 \times 10^{-2} \ mol \ L^{-1}$
Substituting the values:
$K_c = \frac{(1.2 \times 10^{-2})^2}{4.8 \times 10^{-2}} = \frac{1.44 \times 10^{-4}}{4.8 \times 10^{-2}}$
$K_c = 0.3 \times 10^{-2} = 3 \times 10^{-3} \ mol \ L^{-1}$

(D) Let the initial concentration of $A$ and $B$ be $1 \ M$ each.
At equilibrium,one-third of $A$ and $B$ are consumed,so $[A] = 1 - 1/3 = 2/3 \ M$ and $[B] = 1 - 1/3 = 2/3 \ M$.
Since the stoichiometry is $1:1$,the concentration of products formed is $[C] = 1/3 \ M$ and $[D] = 1/3 \ M$.
The equilibrium constant $K$ is given by $K = \frac{[C][D]}{[A][B]}$.
Substituting the values: $K = \frac{(1/3) \times (1/3)}{(2/3) \times (2/3)} = \frac{1/9}{4/9} = 1/4 = 0.25$.

(A) For the reaction $P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$,the equilibrium constant expression is given by the ratio of the product of concentrations of products to the reactants,each raised to the power of their stoichiometric coefficients.
$K_c = \frac{[P_4O_{10(s)}]}{[P_{4(s)}][O_{2(g)}]^5}$
Since $P_{4(s)}$ and $P_4O_{10(s)}$ are pure solids,their concentrations are taken as unity $(1)$.
Therefore,$K_c = \frac{1}{[O_2]^5}$.

(D) The equilibrium constant $K_1$ for the reaction $2HI \rightleftharpoons H_2 + I_2$ is given as $0.25$.
The second reaction $H_2 + I_2 \rightleftharpoons 2HI$ is the reverse of the first reaction.
For a reversed reaction,the new equilibrium constant $K_2$ is the reciprocal of the original equilibrium constant $K_1$.
Therefore,$K_2 = \frac{1}{K_1} = \frac{1}{0.25} = 4$.
Thus,the correct option is $(D)$.

(C) The given reaction is $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ with equilibrium constant $K_1 = 2.4 \times 10^{-3}$.
The second reaction is the reverse of the first reaction: $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$.
For a reversed reaction,the new equilibrium constant $K_2 = \frac{1}{K_1}$.
$K_2 = \frac{1}{2.4 \times 10^{-3}} = \frac{1000}{2.4} \approx 4.17 \times 10^{2}$.

(D) The synthesis reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$,for which $K_{c1} = 50$.
The dissociation reaction is $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$.
This reaction is the reverse of the synthesis reaction.
Therefore,the equilibrium constant $K_{c2}$ for the dissociation is given by $K_{c2} = \frac{1}{K_{c1}}$.
$K_{c2} = \frac{1}{50} = 0.02$.

(C) The equilibrium constant $K_p$ is a function of temperature only for a given reaction.
Since the temperature remains constant,the value of $K_p$ will not change regardless of changes in volume,pressure,or concentration.
Therefore,the value of $K_p$ remains $16$.

(B) The given reaction is $2NO \rightleftharpoons N_2 + O_2$ with $\Delta H^\circ = +43.5 \ kcal \ mol^{-1}$.
Since $\Delta H^\circ$ is positive,the reaction is endothermic.
According to the Van't Hoff equation,$\ln \frac{K_2}{K_1} = \frac{\Delta H^\circ}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$.
For an endothermic reaction,as temperature decreases $(T_2 < T_1)$,the value of the equilibrium constant $K$ decreases.
Therefore,option $B$ is correct.

(D) The synthesis reaction of $HI$ is given by: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ with equilibrium constant $K_c = 50$.
The dissociation reaction of $HI$ is the reverse of the synthesis reaction: $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$.
For a reverse reaction,the new equilibrium constant $K'$ is the reciprocal of the original equilibrium constant $K$.
$K' = \frac{1}{K_c} = \frac{1}{50} = 0.02$.

(A) The given reaction is $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$.
For heterogeneous equilibria,the concentration or partial pressure of pure solids and pure liquids is taken as unity $(1)$.
Therefore,the expression for the equilibrium constant $K_p$ is given by $K_p = P_{CO_2}$.
Since $CaCO_{3(s)}$ and $CaO_{(s)}$ are solids,their active masses are considered to be $1$.

(C) Given reactions:
$(1)$ ${H_2(g)} + \frac{1}{2}{S_2(s)} \rightleftharpoons {H_2S(g)}$; ${K_1} = \frac{[{H_2S}]}{[{H_2}][{S_2}]^{1/2}}$
$(2)$ ${H_2(g)} + {Br_2(g)} \rightleftharpoons 2{HBr(g)}$; ${K_2} = \frac{[{HBr}]^2}{[{H_2}][{Br_2}]}$
Target reaction: ${Br_2(g)} + {H_2S(g)} \rightleftharpoons 2{HBr(g)} + \frac{1}{2}{S_2(s)}$
To obtain the target reaction,we perform: (Reaction $2$) - (Reaction $1$).
Equilibrium constant for the target reaction $K_3 = \frac{K_2}{K_1}$.
$K_3 = \frac{[{HBr}]^2}{[{H_2}][{Br_2}]} \times \frac{[{H_2}][{S_2}]^{1/2}}{[{H_2S}]} = \frac{[{HBr}]^2[{S_2}]^{1/2}}{[{Br_2}][{H_2S}]}$.

(B) For the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$,the equilibrium constant is given by $k = \frac{[NH_3]^2}{[N_2][H_2]^3}$.
For the reaction $2N_2 + 6H_2 \rightleftharpoons 4NH_3$,the equilibrium constant $k'$ is given by $k' = \frac{[NH_3]^4}{[N_2]^2[H_2]^6}$.
Comparing the two expressions,we see that $k' = \left( \frac{[NH_3]^2}{[N_2][H_2]^3} \right)^2 = k^2$.

(A) The equilibrium constant $K_p$ is a function of temperature only for a given reaction.
It does not depend on the pressure,volume,or concentration of the reactants or products.
Therefore,a change in pressure will not affect the value of $K_p$.

(A) For the reaction $2AB \rightleftharpoons A_2 + B_2$,the equilibrium constant is given by $K_c = \frac{[A_2][B_2]}{[AB]^2} = 49$.
For the reaction $AB \rightleftharpoons \frac{1}{2}A_2 + \frac{1}{2}B_2$,the new equilibrium constant $K_c'$ is given by $K_c' = \frac{[A_2]^{1/2} [B_2]^{1/2}}{[AB]}$.
Comparing the two expressions,we see that $K_c' = \sqrt{K_c}$.
Therefore,$K_c' = \sqrt{49} = 7$.

(B) For the equilibrium $AB \rightleftharpoons A + B$,the equilibrium constant $K_c$ is given by $K_c = \frac{[A][B]}{[AB]}$.
Assuming the concentration of $AB$ remains constant or the system adjusts to maintain $K_c$,if the concentration of $A$ is doubled $([A]' = 2[A])$,then to keep $K_c$ constant,the concentration of $B$ must change to $[B]'$ such that $[A]'[B]' = [A][B]$.
Substituting the values: $(2[A]) \times [B]' = [A] \times [B]$.
Therefore,$[B]' = \frac{[A][B]}{2[A]} = \frac{1}{2}[B]$.
Thus,the concentration of $B$ becomes half of its original value.

(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 original reaction is reversed and then multiplied by a factor of $\frac{1}{2}$.
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 reaction is $AB + CD \rightleftharpoons AD + CB$.
At $t = 0$,the moles are $1 \ mol$ of $AB$ and $1 \ mol$ of $CD$,with $0 \ mol$ of products.
At equilibrium,$\frac{3}{4} \ mol$ of each reactant has reacted.
The moles at equilibrium are:
$n(AB) = 1 - \frac{3}{4} = \frac{1}{4} \ mol$
$n(CD) = 1 - \frac{3}{4} = \frac{1}{4} \ mol$
$n(AD) = \frac{3}{4} \ mol$
$n(CB) = \frac{3}{4} \ mol$
Since the volume $V$ is constant,the equilibrium constant $K_c$ is given by:
$K_c = \frac{[AD][CB]}{[AB][CD]} = \frac{(\frac{3/4}{V})(\frac{3/4}{V})}{(\frac{1/4}{V})(\frac{1/4}{V})} = \frac{3/4 \times 3/4}{1/4 \times 1/4} = \frac{9/16}{1/16} = 9$.

(D) The equilibrium constant $(K_p)$ for a given reaction depends only on temperature.
It is independent of the initial pressure $(P)$ and the extent of decomposition $(x)$.
Therefore,$K_p$ remains constant even if $P$ or $x$ changes at a constant temperature.

(B) The equilibrium constant $K$ is defined as $K = \frac{[Products]}{[Reactants]}$.
$1$. If $K$ is very small $(K < 10^{-3})$,the reaction mixture contains mostly reactants.
$2$. If $K$ is very large $(K > 10^3)$,the reaction mixture contains mostly products.
$3$. If $K$ is in the intermediate range $(10^{-3} < K < 10^3)$,the reaction mixture contains significant amounts of both reactants and products.
Comparing the given values:
$(1)$ $K = 0.1$ (Intermediate)
$(2)$ $K = 0.02$ (Intermediate)
$(3)$ $K = 0.0003$ ($3 \times 10^{-4}$,very small,mostly reactants)
$(4)$ $K = 0.002$ ($2 \times 10^{-3}$,intermediate/small)
To have high concentrations of reactants,we look for the smallest $K$ (Reaction $3$). To have high concentrations of products,we look for the largest $K$ (Reaction $1$). Thus,the pair is $(3, 1)$.

(C) The expression $[\text{Product}] / [\text{Reactant}]$ represents the equilibrium constant $K$.
Therefore,$[\text{Product}] / [\text{Reactant}]^{-1}$ is equivalent to $1/K$.
Since the value of the equilibrium constant $K$ depends only on temperature,the factor that affects the value of $K$ is temperature.

(D) The reaction is: $Fe^{3+} + SCN^{-} \rightleftharpoons FeSCN^{2+}$
Initial moles: $[Fe^{3+}] = 3.1 \ M$,$[SCN^{-}] = 3.2 \ M$,$[FeSCN^{2+}] = 0 \ M$
At equilibrium,$[FeSCN^{2+}] = 3.0 \ M$
Change in moles: $[Fe^{3+}] = 3.1 - 3.0 = 0.1 \ M$,$[SCN^{-}] = 3.2 - 3.0 = 0.2 \ M$
Equilibrium constant $K_c = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^{-}]}$
$K_c = \frac{3.0}{(0.1)(0.2)} = \frac{3.0}{0.02} = 150$