Mix Examples- 6-1.Equilibrium (Chemical Equilibrium) Questions in English

(B) The relationship between Gibbs free energy change and equilibrium constant is given by $\Delta G^{\circ} = -RT \ln K$.
Substituting $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$ into the equation,we get $\Delta H^{\circ} - T \Delta S^{\circ} = -RT \ln K$.
Rearranging for $\ln K$,we obtain $\ln K = -\frac{\Delta H^{\circ} - T \Delta S^{\circ}}{RT} = \frac{T \Delta S^{\circ} - \Delta H^{\circ}}{RT}$.
Comparing this with option $B$,we see that $\ln K = \frac{\Delta H^{\circ} - T \Delta S^{\circ}}{RT}$ is incorrect because the signs of $\Delta H^{\circ}$ and $T \Delta S^{\circ}$ are reversed.

(A) The reaction is $A + B \rightleftharpoons 2C$.
Initial concentrations: $[A] = 1 \ M, [B] = 1 \ M, [C] = 1 \ M$.
Let the change in concentration be $x$.
At equilibrium: $[A] = 1 - x, [B] = 1 - x, [C] = 1 + 2x$.
$K_c = \frac{[C]^2}{[A][B]} = \frac{(1+2x)^2}{(1-x)(1-x)} = 100$.
Taking the square root on both sides: $\frac{1+2x}{1-x} = 10$.
$1 + 2x = 10 - 10x$.
$12x = 9$,so $x = \frac{9}{12} = 0.75$.
Equilibrium concentration of $C = 1 + 2x = 1 + 2(0.75) = 1 + 1.5 = 2.5 \ M$.
$2.5 \ M = 25 \times 10^{-1} \ M$.
Thus,$X = 25$.

(C) $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)} \quad K_{c} = 1.844$
$\text{Initial moles at } t = 0: \quad 3.0 \ \text{moles of } PCl_{5}$
$\text{Moles at equilibrium } (t = eq): \quad (3-x) \ \text{for } PCl_{5}, \ x \ \text{for } PCl_{3}, \ x \ \text{for } Cl_{2}$
$\text{Equilibrium constant expression: } K_{c} = \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]} = \frac{x^{2}}{3-x} = 1.844$
$x^{2} + 1.844x - 5.532 = 0$
$\text{Using quadratic formula: } x = \frac{-b + \sqrt{b^{2} - 4ac}}{2a} = \frac{-1.844 + \sqrt{(1.844)^{2} + 4(5.532)}}{2}$
$x = \frac{-1.844 + \sqrt{3.400 + 22.128}}{2} = \frac{-1.844 + 5.052}{2} \approx 1.604$
$\text{Moles of } PCl_{5} \text{ at equilibrium} = 3 - 1.604 = 1.396 \ \text{moles}$
$1.396 \ \text{moles} = 1396 \times 10^{-3} \ \text{moles} \approx 1400 \times 10^{-3} \ \text{moles}$

(C) Initial moles: $n(PCl_{5}) = 5$,$n(N_{2}) = 2$. Total moles at equilibrium: $n_{total} = (5-x) + x + x + 2 = 7+x$.
Using ideal gas law $PV = nRT$ at equilibrium:
$2.46 \times 200 = (7+x) \times 0.082 \times 600$
$492 = (7+x) \times 49.2$
$7+x = 10 \implies x = 3$.
At equilibrium: $n(PCl_{5}) = 2$,$n(PCl_{3}) = 3$,$n(Cl_{2}) = 3$,$n(N_{2}) = 2$. Total moles $= 10$.
Partial pressures: $P(PCl_{5}) = (2/10) \times 2.46 = 0.492 \, atm$,$P(PCl_{3}) = (3/10) \times 2.46 = 0.738 \, atm$,$P(Cl_{2}) = (3/10) \times 2.46 = 0.738 \, atm$.
$K_{p} = \frac{P(PCl_{3}) \times P(Cl_{2})}{P(PCl_{5})} = \frac{0.738 \times 0.738}{0.492} = 1.107$.
$K_{p} = 1107 \times 10^{-3}$.

(C) The reaction is $2 \ NO_{(g)} + O_{2(g)} \rightleftarrows 2 \ NO_{2(g)}$.
Initial moles: $NO = 2, O_2 = 1, NO_2 = 0$.
At equilibrium,moles of $O_2 = 1 - x = 0.6$,so $x = 0.4$.
Equilibrium moles: $NO = 2 - 2(0.4) = 1.2$,$O_2 = 0.6$,$NO_2 = 2(0.4) = 0.8$.
Total moles at equilibrium = $1.2 + 0.6 + 0.8 = 2.6$.
Mole fractions: $X_{NO} = \frac{1.2}{2.6}$,$X_{O_2} = \frac{0.6}{2.6}$,$X_{NO_2} = \frac{0.8}{2.6}$.
Partial pressures $(P_{total} = 1 \ atm)$: $P_{NO} = \frac{1.2}{2.6} \ atm$,$P_{O_2} = \frac{0.6}{2.6} \ atm$,$P_{NO_2} = \frac{0.8}{2.6} \ atm$.
$K_p = \frac{(P_{NO_2})^2}{(P_{NO})^2 \times P_{O_2}} = \frac{(\frac{0.8}{2.6})^2}{(\frac{1.2}{2.6})^2 \times (\frac{0.6}{2.6})} = \frac{0.8^2 \times 2.6}{1.2^2 \times 0.6} = \frac{0.64 \times 2.6}{1.44 \times 0.6} = \frac{1.664}{0.864} \approx 1.926$.
The nearest integer is $2$.

(C) For a reversible first-order reaction $A \underset{K_b}{\stackrel{K_f}{\rightleftharpoons}} B$,the rate law is given by: $-\frac{d[A]}{dt} = K_f[A] - K_b[B]$.
At equilibrium,$-\frac{d[A]}{dt} = 0$,so $K_f[A_{\text{eq}}] = K_b[B_{\text{eq}}]$. Since $[B_{\text{eq}}] = A_0 - A_{\text{eq}}$,we have $K_b = \frac{K_f A_{\text{eq}}}{A_0 - A_{\text{eq}}}$.
The integrated rate equation for this system is $(K_f + K_b)t = \ln \left( \frac{A_0 - A_{\text{eq}}}{A_t - A_{\text{eq}}} \right)$.
Given $A_t = \frac{A_0 + A_{\text{eq}}}{2}$,we substitute this into the expression:
$A_t - A_{\text{eq}} = \frac{A_0 + A_{\text{eq}}}{2} - A_{\text{eq}} = \frac{A_0 - A_{\text{eq}}}{2}$.
Substituting this into the integrated rate equation:
$(K_f + K_b)t = \ln \left( \frac{A_0 - A_{\text{eq}}}{(A_0 - A_{\text{eq}})/2} \right) = \ln(2)$.
Therefore,$t = \frac{\ln 2}{K_f + K_b}$.

(D) For the reversible first-order reaction $X \rightleftharpoons Y$,the equilibrium constant $K_{eq}$ is given by:
$K_{eq} = \frac{K_{f}}{K_{b}} = \frac{[Y]_{eq}}{[X]_{eq}}$
Given $K_{f} = 2 \ s^{-1}$ and $K_{b} = 1 \ s^{-1}$,we have:
$K_{eq} = \frac{2}{1} = 2$
Therefore,$\frac{[Y]_{eq}}{[X]_{eq}} = 2$,which implies $[Y]_{eq} = 2[X]_{eq}$.
This means that at equilibrium,the concentration of $Y$ is twice the concentration of $X$.
As the reaction proceeds,the concentration of $X$ decreases from its initial value $[X]_{0}$ to $[X]_{eq}$,and the concentration of $Y$ increases from $0$ to $[Y]_{eq}$.
Since $[Y]_{eq} = 2[X]_{eq}$,the final concentration of $Y$ must be higher than the final concentration of $X$. Plot $(d)$ correctly shows $[Y]_{eq} > [X]_{eq}$ at equilibrium.

(A) Let the initial concentration of $X$ and $Z$ be $C$ and $\alpha$ be the degree of dissociation.
For reaction $X \rightleftharpoons 2Y$:
Initial concentration: $C$ $0$
Equilibrium concentration: $C(1-\alpha)$ $2C\alpha$
$K_{1} = \frac{[2C\alpha]^2}{C(1-\alpha)} = \frac{4C^2\alpha^2}{C(1-\alpha)} = \frac{4C\alpha^2}{1-\alpha}$
For reaction $Z \rightleftharpoons P + Q$:
Initial concentration: $C$ $0$ $0$
Equilibrium concentration: $C(1-\alpha)$ $C\alpha$ $C\alpha$
$K_{2} = \frac{[C\alpha][C\alpha]}{C(1-\alpha)} = \frac{C^2\alpha^2}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$
Therefore,the ratio $\frac{K_{1}}{K_{2}} = \frac{4C\alpha^2 / (1-\alpha)}{C\alpha^2 / (1-\alpha)} = 4$.

(C) For the reactions:
$1) \ H_2 + I_2 \rightleftharpoons 2 HI, K_{C_1} = 50$
$2) \ N_2 + 3 H_2 \rightleftharpoons 2 NH_3, K_{C_2} = 1000$
We need the equilibrium constant for the reaction:
$N_2 + 6 HI \rightleftharpoons 2 NH_3 + 3 I_2$
This reaction can be obtained by taking the second reaction and subtracting three times the first reaction:
$(N_2 + 3 H_2 \rightleftharpoons 2 NH_3) - 3 \times (H_2 + I_2 \rightleftharpoons 2 HI)$
$= N_2 + 3 H_2 - 3 H_2 - 3 I_2 \rightleftharpoons 2 NH_3 - 6 HI$
Rearranging gives: $N_2 + 6 HI \rightleftharpoons 2 NH_3 + 3 I_2$
The equilibrium constant $K_{C_3}$ is given by:
$K_{C_3} = \frac{K_{C_2}}{(K_{C_1})^3}$
$K_{C_3} = \frac{1000}{(50)^3} = \frac{1000}{125000} = \frac{1}{125} = 0.008$

(A) For the reaction $A \rightleftharpoons n B$ at equilibrium:
Initial concentration of $A = 0.06 \ mol \ L^{-1}$,$B = 0 \ mol \ L^{-1}$.
Equilibrium concentration of $A = 0.03 \ mol \ L^{-1}$,$B = 0.06 \ mol \ L^{-1}$.
Change in concentration of $A = 0.06 - 0.03 = 0.03 \ mol \ L^{-1}$.
Change in concentration of $B = 0.06 - 0 = 0.06 \ mol \ L^{-1}$.
According to stoichiometry,$n = \frac{\Delta [B]}{\Delta [A]} = \frac{0.06}{0.03} = 2$.
The reaction is $A \rightleftharpoons 2 B$.
Equilibrium constant $K_c = \frac{[B]^2}{[A]} = \frac{(0.06)^2}{0.03} = \frac{0.0036}{0.03} = 0.12$.

(B) Given reactions:
$(i) \ 2 P_{(g)} + 3 Cl_{2(g)} \rightleftharpoons 2 PCl_{3(g)} \quad K_1$
$(ii) \ PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)} \quad K_2$
Target reaction:
$(iii) \ 2 P_{(g)} + 5 Cl_{2(g)} \rightleftharpoons 2 PCl_{5(g)}$
To obtain reaction $(iii)$,we multiply reaction $(ii)$ by $2$ and add it to reaction $(i)$:
$2 \times [PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}] \implies 2 PCl_{3(g)} + 2 Cl_{2(g)} \rightleftharpoons 2 PCl_{5(g)} \quad K' = K_2^2$
Adding $(i)$ and the modified $(ii)$:
$(2 P_{(g)} + 3 Cl_{2(g)}) + (2 PCl_{3(g)} + 2 Cl_{2(g)}) \rightleftharpoons 2 PCl_{3(g)} + 2 PCl_{5(g)}$
Canceling common terms gives:
$2 P_{(g)} + 5 Cl_{2(g)} \rightleftharpoons 2 PCl_{5(g)}$
The equilibrium constant $K$ for the final reaction is the product of the constants:
$K = K_1 \times K_2^2 = K_1 K_2^2$

(A) For the reaction,$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ with $K_C = 46$.
The reaction quotient $Q_C$ is calculated as:
$Q_C = \frac{[HI]^2}{[H_2][I_2]} = \frac{0.4 \times 0.4}{0.1 \times 0.2} = \frac{0.16}{0.02} = 8$.
Since $Q_C < K_C$ $(8 < 46)$,the reaction will proceed in the forward direction to reach equilibrium.
Therefore,the concentration of the product $HI$ will increase.

(B) The reaction is $2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$.
Let the initial moles of $NO_2$ be $n_0$. At equilibrium,let $x$ moles of $N_2O_4$ be formed.
Moles at equilibrium: $n_{NO_2} = n_0 - 2x$,$n_{N_2O_4} = x$.
Total mass: $46(n_0 - 2x) + 92x = 64.4 \implies 46n_0 = 64.4 \implies n_0 = 1.4 \ mol$.
Total moles $n_T = (1.4 - 2x) + x = 1.4 - x$.
$P_{NO_2} = \frac{1.4-2x}{1.4-x} P_T$ and $P_{N_2O_4} = \frac{x}{1.4-x} P_T$.
$K_p = \frac{P_{N_2O_4}}{(P_{NO_2})^2} = \frac{x(1.4-x)}{(1.4-2x)^2 P_T} = 6.67$.
Using $P_T V = n_T RT \implies P_T = \frac{(1.4-x) \times 0.082 \times 300}{15} = 1.64(1.4-x)$.
Substituting $P_T$ into $K_p$ expression and solving for $x$ gives $x \approx 0.58 \ mol$.
Then $n_T = 1.4 - 0.58 = 0.82 \ mol$.
$P_T = \frac{0.82 \times 0.082 \times 300}{15} = 1.34 \ atm$.

(B) For the reaction $X + Y \rightleftharpoons Z$ at $P_{T} = 1 \, bar$:
Initial moles at equilibrium: $n_{X} = 1, n_{Y} = 3, n_{Z} = 2$. Total moles $n_{T} = 6$.
Mole fractions: $x_{X} = 1/6, x_{Y} = 3/6, x_{Z} = 2/6$.
Partial pressures: $P_{X} = 1/6 \, bar, P_{Y} = 3/6 \, bar, P_{Z} = 2/6 \, bar$.
$K_{p} = \frac{P_{Z}}{P_{X} \times P_{Y}} = \frac{2/6}{(1/6) \times (3/6)} = \frac{2/6}{3/36} = \frac{2}{6} \times \frac{36}{3} = 4$.
When pressure is increased to $P_{T} = 2 \, bar$,the reaction shifts forward to maintain $K_{p} = 4$.
Let $x$ be the moles of $X$ reacted. New moles: $n_{X} = 1 - x, n_{Y} = 3 - x, n_{Z} = 2 + x$. Total moles $n_{T} = 6 - x$.
$K_{p} = \frac{P_{Z}}{P_{X} \times P_{Y}} = \frac{(n_{Z}/n_{T}) \times P_{T}}{(n_{X}/n_{T}) \times (n_{Y}/n_{T}) \times P_{T}^2} = \frac{n_{Z} \times n_{T}}{n_{X} \times n_{Y} \times P_{T}} = 4$.
$\frac{(2 + x)(6 - x)}{(1 - x)(3 - x) \times 2} = 4 \Rightarrow \frac{(2 + x)(6 - x)}{(1 - x)(3 - x)} = 8$.
$12 + 4x - x^2 = 8(3 - 4x + x^2) = 24 - 32x + 8x^2$.
$9x^2 - 36x + 12 = 0 \Rightarrow 3x^2 - 12x + 4 = 0$.
Using quadratic formula $x = \frac{12 \pm \sqrt{144 - 48}}{6} = \frac{12 \pm \sqrt{96}}{6} = 2 \pm \frac{9.798}{6} = 2 \pm 1.633$.
Since $x < 1$,$x = 2 - 1.633 = 0.367$.
$n_{X} = 1 - 0.367 = 0.633 \, mol$.

(B) Initial equilibrium state:
$P(aq) \rightleftharpoons Q(aq)$
Initial: $[P] = 2 \ M, [Q] = 0 \ M$
At equilibrium: $[P] = 2 - x, [Q] = x$
$K = \frac{[Q]}{[P]} = \frac{x}{2 - x} = 1.5$
$x = 1.5(2 - x)$ $\Rightarrow x = 3 - 1.5x$ $\Rightarrow 2.5x = 3$ $\Rightarrow x = 1.2 \ M$
So,at initial equilibrium: $[P] = 0.8 \ M, [Q] = 1.2 \ M$.
After removing half of $P$:
New $[P] = 0.8 / 2 = 0.4 \ M$,$[Q] = 1.2 \ M$.
Reaction quotient $Q_c = \frac{[Q]}{[P]} = \frac{1.2}{0.4} = 3$.
Since $Q_c > K$ $(3 > 1.5)$,the reaction shifts backward.
Let $y$ be the amount of $Q$ converted to $P$:
New equilibrium: $[P] = 0.4 + y, [Q] = 1.2 - y$.
$K = \frac{1.2 - y}{0.4 + y} = 1.5$
$1.2 - y = 1.5(0.4 + y) \Rightarrow 1.2 - y = 0.6 + 1.5y$
$0.6 = 2.5y \Rightarrow y = 0.24 \ M$
Final concentration of $Q = 1.2 - 0.24 = 0.96 \ M$.

(A) The equilibrium constant $K_{eq}$ is defined as the ratio of the rate constant of the forward reaction to the rate constant of the backward reaction: $K_{eq} = \frac{K_{f}}{K_{b}}$.
Given $K_{f} = 10^{3}$ and $K_{b} = 10^{2}$,we have $K_{eq} = \frac{10^{3}}{10^{2}} = 10$.
The standard Gibbs energy change is related to the equilibrium constant by the equation: $\Delta_{r} G^{\circ} = -RT \ln K_{eq}$.
Given $T = 27^{\circ} C = 300 \ K$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$,and $\ln 10 = 2.3$.
Substituting the values: $\Delta_{r} G^{\circ} = -(8.3 \times 300 \times 2.3) \ J \ mol^{-1}$.
$\Delta_{r} G^{\circ} = -5727 \ J \ mol^{-1} = -5.727 \ kJ \ mol^{-1}$.
Rounding to the nearest integer,the magnitude is $6 \ kJ \ mol^{-1}$.

(C) The target reaction is $2NH_3(g) + \frac{5}{2}O_2(g) \rightleftharpoons 2NO(g) + 3H_2O(g)$.
This can be obtained by the operation: $(ii) + 3 \times (iii) - (i)$.
$K_{eq} = \frac{K_2 \times K_3^3}{K_1}$.
Substituting the given values:
$K_{eq} = \frac{1.6 \times 10^{12} \times (1.0 \times 10^{-13})^3}{4 \times 10^5}$.
$K_{eq} = \frac{1.6 \times 10^{12} \times 10^{-39}}{4 \times 10^5}$.
$K_{eq} = 0.4 \times 10^{12 - 39 - 5} = 0.4 \times 10^{-32} = 4 \times 10^{-33}$.
Thus,the value is $4$.

(A) For reaction $(i): X(g) \rightleftharpoons Y(g) + Z(g)$

$Initial \ moles$	$n$	$0$	$0$
$Equilibrium \ moles$	$n(1-\alpha)$	$n\alpha$	$n\alpha$

Total moles at equilibrium $= n(1-\alpha) + n\alpha + n\alpha = n(1+\alpha)$.
Partial pressures: $p_X = \frac{1-\alpha}{1+\alpha} p_1$,$p_Y = \frac{\alpha}{1+\alpha} p_1$,$p_Z = \frac{\alpha}{1+\alpha} p_1$.
$K_{p1} = \frac{p_Y p_Z}{p_X} = \frac{\alpha^2 p_1}{1-\alpha^2} = 3$
For reaction $(ii): A(g) \rightleftharpoons 2B(g)$

$Initial \ moles$	$n$	$0$
$Equilibrium \ moles$	$n(1-\alpha)$	$2n\alpha$

Total moles at equilibrium $= n(1-\alpha) + 2n\alpha = n(1+\alpha)$.
Partial pressures: $p_A = \frac{1-\alpha}{1+\alpha} p_2$,$p_B = \frac{2\alpha}{1+\alpha} p_2$.
$K_{p2} = \frac{p_B^2}{p_A} = \frac{4\alpha^2 p_2}{1-\alpha^2} = 1$
Dividing the two expressions: $\frac{K_{p1}}{K_{p2}} = \frac{3}{1} = \frac{\alpha^2 p_1 / (1-\alpha^2)}{4\alpha^2 p_2 / (1-\alpha^2)} = \frac{p_1}{4p_2}$
Therefore,$\frac{p_1}{p_2} = 3 \times 4 = 12$.
Thus,$x = 12$.

(A) The initial equilibrium constant is $K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} = \frac{0.40}{0.2 \times 0.1} = 20$.
When $0.2 \, mol$ of $Cl_2$ is added to $1 \, L$ volume,the new concentration of $Cl_2$ becomes $0.1 + 0.2 = 0.3 \, M$.
Let $x$ be the amount of $PCl_3$ and $Cl_2$ consumed to reach the new equilibrium.
The new equilibrium concentrations are: $[PCl_3] = 0.2 - x$,$[Cl_2] = 0.3 - x$,and $[PCl_5] = 0.4 + x$.
Substituting into the $K_c$ expression: $\frac{0.4 + x}{(0.2 - x)(0.3 - x)} = 20$.
$0.4 + x = 20(0.06 - 0.5x + x^2) = 1.2 - 10x + 20x^2$.
$20x^2 - 11x + 0.8 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x \approx 0.086$.
Thus,$[PCl_5]_{eq} = 0.4 + 0.086 = 0.486 \, M = 48.6 \times 10^{-2} \, mol \, L^{-1}$.
Rounding to the nearest integer,the value is $49$.

(C) . Equilibrium in physical processes requires a closed system to prevent the escape of matter,so it is correct.
$B$. At equilibrium,the rate of the forward process equals the rate of the backward process $(r_f = r_b)$,so it is correct.
$C$. At equilibrium,measurable properties like pressure,concentration,or density become constant at a given temperature,so it is correct.
$D$. For a saturated solution of a solid in a liquid,the concentration (solubility) remains constant at a fixed temperature,so it is correct.
Therefore,all $4$ statements are correct.

(C) $1.$ The correct option is $(B)$ $\frac{8 \beta_{\text{equilibrium}}^2}{4-\beta_{\text{equilibrium}}^2}$.
$X_{2(g)} \rightleftharpoons 2 X_{(g)}$
Initial: $1 \quad 0$
Equilibrium: $1-\frac{\beta_e}{2} \quad \beta_e$
Total number of moles at equilibrium $= 1-\frac{\beta_e}{2} + \beta_e = 1+\frac{\beta_e}{2} = \frac{2+\beta_e}{2}$.
$K_p = \frac{(p_X)^2}{p_{X_2}} = \frac{\left( \frac{\beta_e}{1+\beta_e/2} \times 2 \right)^2}{\left( \frac{1-\beta_e/2}{1+\beta_e/2} \times 2 \right)} = \frac{\left( \frac{2\beta_e}{2+\beta_e} \times 2 \right)^2}{\left( \frac{2-\beta_e}{2+\beta_e} \times 2 \right)} = \frac{16\beta_e^2}{(2+\beta_e)^2} \times \frac{2+\beta_e}{2(2-\beta_e)} = \frac{8 \beta_e^2}{(2+\beta_e)(2-\beta_e)} = \frac{8 \beta_e^2}{4-\beta_e^2}$.
$2.$ $(C)$ is the incorrect statement.
$(a)$ According to Le Chatelier's principle,decreasing the pressure shifts the equilibrium towards the side with more moles (forward direction),producing more $X$.
$(b)$ At the start,$Q = 0 < K_p$,so the reaction proceeds spontaneously in the forward direction.
$(c)$ Since $\Delta_r G^{\circ} > 0$,$K_p = e^{-\Delta_r G^{\circ}/RT} < 1$. If $\beta_{eq} = 0.7$,then $K_p = \frac{8(0.7)^2}{4-(0.7)^2} = \frac{3.92}{3.51} > 1$,which contradicts $\Delta_r G^{\circ} > 0$.
$(d)$ Since $K_p = K_c(RT)^{\Delta n}$ and $\Delta n = 1$,$K_c = K_p/(RT)$. Given $K_p < 1$ and $RT \approx 24.7$,$K_c < 1$.

(C) At $T_1 \text{ K}$: $A_{(g)} \rightleftharpoons P_{(g)}$
$t=0$: $6 \quad 0$
$t=\infty$: $6-4=2 \quad 4$ (from plot)
$\Rightarrow K_{c_1} = \frac{4}{2} = 2$
At $T_2 \text{ K}$: $A_{(g)} \rightleftharpoons P_{(g)}$
$t=0$: $6 \quad 0$
$t=\infty$: $6-2=4 \quad 2$ (from plot)
$\Rightarrow K_{c_2} = \frac{2}{4} = \frac{1}{2}$
Since $\Delta n_g = 0$,$K_p = K_c$.
$\Delta G_2^{\Theta} = -RT_2 \ln K_{p_2} = -RT_2 \ln \frac{1}{2} = RT_2 \ln 2$
$\Delta G_1^{\Theta} = -RT_1 \ln K_{p_1} = -R(2T_2) \ln 2 = -2RT_2 \ln 2$
$\Delta G_2^{\Theta} - \Delta G_1^{\Theta} = RT_2 \ln 2 - (-2RT_2 \ln 2) = 3RT_2 \ln 2 = RT_2 \ln 2^3 = RT_2 \ln 8$
Given $\Delta G_2^{\Theta} - \Delta G_1^{\Theta} = RT_2 \ln x$,therefore $x = 8$.

(B, D) The relationship between the equilibrium constant $K$ and temperature is given by the van't Hoff equation: $\ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}$.
For an endothermic reaction $(\Delta H > 0)$,increasing temperature increases $K$ because the entropy of the surroundings $(\Delta S_{surr} = -\frac{\Delta H}{T})$ becomes less negative (less unfavourable).
For an exothermic reaction $(\Delta H < 0)$,increasing temperature decreases $K$ because the entropy of the surroundings $(\Delta S_{surr} = -\frac{\Delta H}{T})$ becomes less positive (less favourable).
Thus,statements $[B]$ and $[D]$ correctly describe the effect of temperature on $K$ in terms of entropy changes of the surroundings.

(B) From the graphs,at equilibrium,$[P]_{T_1} > [P]_{T_2}$ and $[A]_{T_1} < [A]_{T_2}$.
Since $K = [P]/[A]$,we have $K_1 > K_2$ at $T_1 < T_2$. This indicates the reaction is exothermic,so $\Delta H^{\ominus} < 0$.
Since the reaction proceeds spontaneously to form product,$\Delta G^{\ominus} < 0$.
Given $\frac{\ln K_1}{\ln K_2} > \frac{T_2}{T_1}$,where $\ln K = \frac{\Delta S^{\ominus}}{R} - \frac{\Delta H^{\ominus}}{RT}$.
Substituting this,we get $\frac{\Delta S^{\ominus} - \Delta H^{\ominus}/T_1}{\Delta S^{\ominus} - \Delta H^{\ominus}/T_2} > \frac{T_2}{T_1}$.
Since $\Delta H^{\ominus} < 0$,let $\Delta H^{\ominus} = -|\Delta H^{\ominus}|$. The inequality simplifies to $(T_2 - T_1) \frac{\Delta S^{\ominus}}{R} < 0$. Since $T_2 > T_1$,we must have $\Delta S^{\ominus} < 0$.
Thus,$\Delta H^{\ominus} < 0, \Delta S^{\ominus} < 0$ (Option $A$) and $\Delta G^{\ominus} < 0, \Delta S^{\ominus} < 0$ (Option $C$) are correct.

(C) The given reactions are:
$(i) 2 Cu_{(s)} + 1/2 O_{2(g)} \longrightarrow Cu_2O_{(s)}; \Delta G_1^\theta = -78,000 \ J \ mol^{-1}$
$(ii) H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(g)}; \Delta G_2^\theta = -1,78,000 \ J \ mol^{-1}$
Subtracting $(ii)$ from $(i)$ gives the target reaction:
$2 Cu_{(s)} + H_2O_{(g)} \longrightarrow Cu_2O_{(s)} + H_{2(g)}$
$\Delta G^\theta = \Delta G_1^\theta - \Delta G_2^\theta = -78,000 - (-1,78,000) = 1,00,000 \ J \ mol^{-1}$
To prevent oxidation,$\Delta G \ge 0$. At the threshold (minimum $p_{H_2}$):
$\Delta G = \Delta G^\theta + RT \ln Q = 0$
$1,00,000 + (8 \times 1250) \ln \left( \frac{p_{H_2}}{p_{H_2O}} \right) = 0$
$1,00,000 + 10,000 \ln \left( \frac{p_{H_2}}{p_{H_2O}} \right) = 0$
$10 + \ln(p_{H_2}) - \ln(p_{H_2O}) = 0$
$\ln(p_{H_2}) = \ln(p_{H_2O}) - 10$
Given $p_{H_2O} = 1\% \text{ of } 1 \ \text{bar} = 0.01 \ \text{bar} = 10^{-2} \ \text{bar}$.
$\ln(p_{H_2O}) = \ln(10^{-2}) = -2 \ln(10) = -2 \times 2.3 = -4.6$.
$\ln(p_{H_2}) = -4.6 - 10 = -14.6$.

(C) The reaction is $Fe^{2+}_{(aq)} + S^{2-}_{(aq)} \rightleftharpoons FeS_{(s)}$.
Since equal volumes are mixed,the new concentrations are $0.03 \ M \ Fe^{2+}$ and $0.1 \ M \ S^{2-}$.
Given $K_{c} = 1.6 \times 10^{17}$,which is very large,the reaction proceeds almost to completion.
Let $y$ be the equilibrium concentration of $Fe^{2+}_{(aq)}$.
Initial concentrations: $[Fe^{2+}] = 0.03 \ M$,$[S^{2-}] = 0.1 \ M$.
At equilibrium: $[Fe^{2+}] = y$,$[S^{2-}] = 0.1 - 0.03 = 0.07 \ M$.
$K_{c} = \frac{1}{[Fe^{2+}][S^{2-}]} = \frac{1}{y \times 0.07} = 1.6 \times 10^{17}$.
$y = \frac{1}{1.6 \times 0.07} \times 10^{-17} = \frac{1}{0.112} \times 10^{-17} \approx 8.928 \times 10^{-17} \ M$.
Thus,$Y \approx 8.93$.

(D) $(1)$ The equilibrium constant $K = \frac{p_z}{p^\ominus}$. The van't Hoff equation is $\ln K = -\frac{\Delta H^\ominus}{RT} + \frac{\Delta S^\ominus}{R}$.
The slope of the plot of $\ln K$ versus $\frac{10^4}{T}$ is given by:
Slope $= \frac{-7 - (-3)}{12 - 10} \times 10^4 = \frac{-4}{2} \times 10^4 = -2 \times 10^4 \ K$.
From the equation,Slope $= -\frac{\Delta H^\ominus}{R} \times 10^{-4}$ (since the x-axis is $\frac{10^4}{T}$).
So,$-\frac{\Delta H^\ominus}{R} \times 10^{-4} = -2 \times 10^4 \implies \Delta H^\ominus = 2 \times 10^8 \times R \times 10^{-4} = 2 \times 10^4 \times 8.314 = 166280 \ J \ mol^{-1} = 166.28 \ kJ \ mol^{-1}$.
$(2)$ At $T = 1000 \ K$,$\frac{10^4}{T} = 10$. From the graph,$\ln K = -3$.
Using $\ln K = -\frac{\Delta H^\ominus}{RT} + \frac{\Delta S^\ominus}{R}$:
$-3 = -\frac{166280}{8.314 \times 1000} + \frac{\Delta S^\ominus}{8.314}$
$-3 = -20 + \frac{\Delta S^\ominus}{8.314}$
$\frac{\Delta S^\ominus}{8.314} = 17 \implies \Delta S^\ominus = 17 \times 8.314 = 141.338 \approx 141.34 \ J \ K^{-1} \ mol^{-1}$.

(B) According to Kirchhoff's law,$\Delta H_{T_2} - \Delta H_{T_1} = \int_{T_1}^{T_2} \Delta C_p \, dT$. Since the heat capacity change $\Delta C_p$ is generally non-zero,$\Delta H$ depends on temperature $T$.
$(B)$ The equilibrium constant $K$ for a heterogeneous reaction depends only on temperature and is independent of the initial amounts of solid reactants or products.
$(C)$ $K$ is a constant at a fixed temperature; it does not depend on the partial pressure of $CO_2$. The partial pressure of $CO_2$ at equilibrium is equal to $K_p$,which is fixed for a given $T$.
$(D)$ $A$ catalyst provides an alternative pathway with a lower activation energy but does not change the enthalpy of reaction $(\Delta H)$ or the equilibrium constant $(K)$.

(B) The chemical reaction is: $CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$
Initially,the pressure of $CO_2$ is $0.5 \ atm$.
Let $x \ atm$ be the pressure of $CO_2$ that reacts.
At equilibrium,the partial pressures are: $P_{CO_2} = (0.5 - x) \ atm$ and $P_{CO} = 2x \ atm$.
The total pressure at equilibrium is given as $0.8 \ atm$.
$P_{\text{total}} = P_{CO_2} + P_{CO} = (0.5 - x) + 2x = 0.5 + x = 0.8 \ atm$.
Solving for $x$,we get $x = 0.3 \ atm$.
Now,calculate $K_P$:
$K_P = \frac{(P_{CO})^2}{P_{CO_2}} = \frac{(2x)^2}{(0.5 - x)} = \frac{(2 \times 0.3)^2}{(0.5 - 0.3)} = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8 \ atm$.

(A) Initial moles of $N_2O_5 = \frac{37.8 \ g}{108 \ g/mol} = 0.35 \ mol$.
Initial pressure $P_0 = \frac{nRT}{V} = \frac{0.35 \times 0.082 \times 500}{1} = 14.35 \ bar$.
Reaction: $2N_2O_{5(g)} \rightleftharpoons 2N_2O_{4(g)} + O_{2(g)}$
At $t=0$: $P_0 = 14.35 \ bar$,$0$,$0$
At equilibrium: $(14.35 - 2P)$,$2P$,$P$
Total pressure at equilibrium: $(14.35 - 2P) + 2P + P = 14.35 + P = 18.65 \ bar$.
Therefore,$P = 18.65 - 14.35 = 4.3 \ bar$.
Equilibrium partial pressures:
$P_{N_2O_5} = 14.35 - 2(4.3) = 14.35 - 8.6 = 5.75 \ bar$.
$P_{N_2O_4} = 2(4.3) = 8.6 \ bar$.
$P_{O_2} = 4.3 \ bar$.
$K_p = \frac{(P_{N_2O_4})^2 \times (P_{O_2})}{(P_{N_2O_5})^2} = \frac{(8.6)^2 \times 4.3}{(5.75)^2} = \frac{73.96 \times 4.3}{33.0625} = \frac{318.028}{33.0625} \approx 9.619$.
$K_p = 9.619 = 961.9 \times 10^{-2} \approx 962 \times 10^{-2}$.

(A) For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$,the reactants $H_2$ and $I_2$ are consumed to form the product $HI$.
Therefore,the molar concentration of reactants ($H_2$ and $I_2$) decreases over time until equilibrium is reached.
Simultaneously,the molar concentration of the product $(HI)$ increases over time until equilibrium is reached.
Once equilibrium is attained,the concentrations of all reactants and products remain constant over time.
This behavior is correctly represented by the graph where the curves for $H_2$ and $I_2$ show a downward trend and the curve for $HI$ shows an upward trend,all leveling off at the same time.

(B) For the reaction $A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)}$:
Initial moles: $A=1, B=0, C=0$
At equilibrium: $A=1-\alpha, B=\alpha, C=\alpha$
Total moles $= (1-\alpha) + \alpha + \alpha = 1+\alpha$
Partial pressures: $P_A = \frac{1-\alpha}{1+\alpha} p, P_B = \frac{\alpha}{1+\alpha} p, P_C = \frac{\alpha}{1+\alpha} p$
$K_p = \frac{P_B \cdot P_C}{P_A} = \frac{(\frac{\alpha}{1+\alpha} p) (\frac{\alpha}{1+\alpha} p)}{\frac{1-\alpha}{1+\alpha} p} = \frac{\alpha^2 p}{1-\alpha^2}$
Rearranging for $\alpha$: $\alpha^2 p = K_p - K_p \alpha^2$ $\Rightarrow \alpha^2(p+K_p) = K_p$ $\Rightarrow \alpha = \sqrt{\frac{K_p}{p+K_p}}$
This expression shows that as the total pressure $p$ increases,the degree of dissociation $\alpha$ decreases.

(D) catalyst provides an alternative pathway with lower activation energy for both the forward and backward reactions.
It increases the rate of both reactions equally,allowing the system to reach equilibrium faster.
However,it does not affect the equilibrium constant $(K_{C})$,which is a function of temperature only.
Therefore,Statement $I$ is true.
Statement $II$ is false because a catalyst does not change the equilibrium composition of a system at a constant temperature and pressure.
Thus,Statement $I$ is true but Statement $II$ is false.

(D) The relationship between the equilibrium constant $K_{eq}$ and temperature $T$ is given by the van't Hoff equation: $\ln K_{eq} = -\frac{\Delta H^{\circ}}{R} \left( \frac{1}{T} \right) + \frac{\Delta S^{\circ}}{R}$.
Comparing this with the linear equation $y = mx + c$,where $y = \ln K_{eq}$,$x = 1/T$,slope $m = -\frac{\Delta H^{\circ}}{R}$,and intercept $c = \frac{\Delta S^{\circ}}{R}$.
For an endothermic reaction,$\Delta H^{\circ} > 0$,so the slope $m = -\frac{\Delta H^{\circ}}{R}$ is negative.
For an exothermic reaction,$\Delta H^{\circ} < 0$,so the slope $m = -\frac{\Delta H^{\circ}}{R}$ is positive.
Therefore,for $\Delta H^{\circ} = -ve$,the slope is positive,which matches the graph in option $(D)$.

(B) Initially,$2$ moles of both $A$ and $B$ are taken.
According to the stoichiometry of the reaction $2 \ A_{(g)} + B_{(g)} \rightleftharpoons 2 \ C_{(g)} + D_{(g)}$,$2$ moles of $A$ react with $1$ mole of $B$.
Since $A$ is consumed at twice the rate of $B$,the concentration of $A$ decreases faster than $B$.
Therefore,at any time during the reaction or at equilibrium,the concentration of $A$ will be less than the concentration of $B$,i.e.,$[A] < [B]$.

(C) Wrong: At equilibrium,the reaction does not stop; it is a dynamic process where the rate of forward reaction equals the rate of backward reaction.
$(b)$ Wrong: $A$ catalyst speeds up both the forward and backward reactions equally,reaching equilibrium faster without changing the equilibrium position.
$(c)$ Correct: According to Le Chatelier's principle,for an exothermic reaction,increasing temperature shifts the equilibrium to the left,decreasing the equilibrium constant $K$.
$(d)$ Wrong: $K_p = K_c(RT)^{\Delta n}$. $K_p$ can be less than,equal to,or greater than $K_c$ depending on the value of $\Delta n$.

(B) The reaction is $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g).$
The reaction quotient $Q_P$ is given by the expression: $Q_P = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3}.$
Substituting the given partial pressures: $Q_P = \frac{(3)^2}{(1)(2)^3} = \frac{9}{8} = 1.125 \ atm^{-2}.$
Comparing $Q_P$ with $K_P$: $Q_P = 1.125$ and $K_P = 4.28 \times 10^{-5}.$
Since $Q_P > K_P,$ the reaction will proceed in the backward direction to reach equilibrium.

(A) The reaction is: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
Initial moles: $1 \ mol$ of $N_2O_4$ and $0 \ mol$ of $NO_2$.
At equilibrium,$20\%$ of $N_2O_4$ dissociates,so $0.2 \ mol$ is consumed.
Remaining moles of $N_2O_4 = 1 - 0.2 = 0.8 \ mol$.
Moles of $NO_2$ formed = $2 \times 0.2 = 0.4 \ mol$.
Total moles at equilibrium = $0.8 + 0.4 = 1.2 \ mol$.
Since $P \propto n$ (at constant $V$ and $T$),the ratio of total pressure to initial pressure is equal to the ratio of total moles to initial moles.
$P_{total} = P_{initial} \times \frac{n_{total}}{n_{initial}} = 1 \ atm \times \frac{1.2}{1} = 1.2 \ atm$.

(C, D) The reaction quotient $Q_{C}$ is compared with the equilibrium constant $K_{C}$ to determine the direction of the reaction:
$1$. If $Q_{C} > K_{C}$,the system has excess products,so the reverse reaction is favoured to reach equilibrium.
$2$. If $Q_{C} < K_{C}$,the system has excess reactants,so the forward reaction is favoured to reach equilibrium.
$3$. If $Q_{C} = K_{C}$,the reaction is at equilibrium.
Therefore,statements $C$ and $D$ are incorrect.

(C) True: Chemical equilibrium can only be attained in a closed system at a constant temperature.
$(b)$ True: At equilibrium,the macroscopic properties like concentration,pressure,and density remain constant over time.
$(c)$ True: If $K_f$ is the equilibrium constant for the forward reaction,then the equilibrium constant for the reverse reaction $K_r$ is given by $K_r = \frac{1}{K_f}$.

(D) The value of the equilibrium constant $K_C$ depends only on the temperature of the reaction.
Catalysts increase the rate of both the forward and backward reactions equally,allowing the system to reach equilibrium faster,but they do not change the position of the equilibrium or the value of $K_C$.
Since the temperature remains constant at $500 \ K$,the value of $K_C$ remains unchanged.
Therefore,the equilibrium constant $K_C$ in the presence of a catalyst is $2 \times 10^{-5}$.

(D) The equilibrium reaction is: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$.
Initial moles of $CaCO_{3} = \frac{50 \ g}{100 \ g/mol} = 0.5 \ mol$.
For this reaction,$K_{p} = P_{CO_{2}} = 1.64 \ atm$.
Using the ideal gas equation $PV = nRT$ for $CO_{2}$ gas:
$1.64 \ atm \times 10 \ L = n_{CO_{2}} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 1000 \ K$.
$16.4 = n_{CO_{2}} \times 82$.
$n_{CO_{2}} = \frac{16.4}{82} = 0.2 \ mol$.
Since $1 \ mol$ of $CaCO_{3}$ produces $1 \ mol$ of $CO_{2}$,the moles of $CaCO_{3}$ reacted $= 0.2 \ mol$.
Moles of $CaCO_{3}$ remaining $= 0.5 - 0.2 = 0.3 \ mol$.
Percentage of unreacted $CaCO_{3} = \frac{0.3}{0.5} \times 100 = 60 \%$.

(D) The reaction is $N_2O_{4(g)} \rightleftharpoons 2NO_2(g)$.
Initially,$n_1 = 2 \ mol$ at $T_1 = 298 \ K$ and $P_1 = 1 \ atm$.
$20 \%$ by mass of $N_2O_4$ decomposes. Since molar mass is constant,this corresponds to $20 \%$ of moles decomposing.
Moles of $N_2O_4$ reacted $= 2 \times 0.2 = 0.4 \ mol$.
Moles of $N_2O_4$ remaining $= 2 - 0.4 = 1.6 \ mol$.
Moles of $NO_2$ produced $= 2 \times 0.4 = 0.8 \ mol$.
Total moles at equilibrium,$n_2 = 1.6 + 0.8 = 2.4 \ mol$.
Using the ideal gas law for a constant volume container: $\frac{P_1}{n_1 T_1} = \frac{P_2}{n_2 T_2}$.
$\frac{1}{2 \times 298} = \frac{P_2}{2.4 \times 596}$.
$P_2 = \frac{2.4 \times 596}{2 \times 298} = \frac{2.4 \times 596}{596} = 2.4 \ atm$.

(D) The chemical equilibrium reaction is: $PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2}$
Initially: $PCl_{5} = 3 \ mol$,$PCl_{3} = 3 \ mol$,$Cl_{2} = 2 \ mol$
At equilibrium: $PCl_{5} = (3 - x) \ mol$,$PCl_{3} = (3 + x) \ mol$,$Cl_{2} = (2 + x) \ mol$
Given that at equilibrium,$PCl_{5} = 1.5 \ mol$:
$3 - x = 1.5$
$x = 1.5$
Therefore,the number of moles of $PCl_{3}$ at equilibrium is:
$PCl_{3} = 3 + x = 3 + 1.5 = 4.5 \ mol$

(D) The reaction is: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$
Initial moles: $SO_2 = 5, O_2 = 5, SO_3 = 0$
$60\%$ of $SO_2$ used up $= 5 \times 0.6 = 3 \ moles$.
At equilibrium:
$SO_2 = 5 - 3 = 2 \ moles$
$O_2 = 5 - (3/2) = 5 - 1.5 = 3.5 \ moles$
$SO_3 = 3 \ moles$
Total moles at equilibrium $= 2 + 3.5 + 3 = 8.5 \ moles$.
Partial pressure of $O_2 = (\text{moles of } O_2 / \text{total moles}) \times \text{Total Pressure}$
$pO_2 = (3.5 / 8.5) \times 1 \ atm = 0.41 \ atm$.

(D) The reaction is $AO_{2(g)} + BO_{2(g)} \rightleftharpoons AO_{3(g)} + BO_{(g)}$. The initial moles are $1 \ mol$ each in $1 \ L$ volume,so initial concentrations are $1 \ M$ each. The reaction quotient $Q_c = \frac{[AO_3][BO]}{[AO_2][BO_2]} = \frac{1 \times 1}{1 \times 1} = 1$. Since $Q_c < K_c$ $(1 < 16)$,the reaction proceeds in the forward direction. Let $x$ be the moles reacted at equilibrium. Equilibrium concentrations: $[AO_2] = 1-x, [BO_2] = 1-x, [AO_3] = 1+x, [BO] = 1+x$. $K_c = \frac{(1+x)(1+x)}{(1-x)(1-x)} = 16$. Taking square root: $\frac{1+x}{1-x} = 4$. Solving for $x$: $1+x = 4-4x \implies 5x = 3 \implies x = 0.6$.
At equilibrium: $[AO_2] = 0.4, [BO_2] = 0.4, [AO_3] = 1.6, [BO] = 1.6$.
Statement $I$: Total moles = $0.4 + 0.4 + 1.6 + 1.6 = 4$. (Correct)
Statement $II$: Ratio $[AO_2] : [AO_3] = 0.4 : 1.6 = 1 : 4$. (Correct)
Statement $III$: Total moles of $AO_2 + BO_2 = 0.4 + 0.4 = 0.8$. (Correct)

(B) The reaction is $A_{2(g)} \rightleftharpoons B_{2(g)}$. Given $K_{c} = \frac{[B_{2}]}{[A_{2}]} = 99.0$.
Initially,$2 \ moles$ of $A_{2}$ are in $1 \ L$,so $[A_{2}] = 2 \ M$.
After adding $1 \ mole$ of $A_{2}$,the total moles of $A_{2}$ becomes $2 + 1 = 3 \ moles$.
Let the amount of $A_{2}$ that reacts to form $B_{2}$ at the new equilibrium be $x \ mol$.
At equilibrium: $[A_{2}] = (3 - x) \ M$ and $[B_{2}] = x \ M$.
Substituting into the equilibrium expression: $\frac{x}{3 - x} = 99.0$.
$x = 99(3 - x) = 297 - 99x$.
$100x = 297$,so $x = 2.97 \ M$.
Therefore,$[A_{2}] = 3 - 2.97 = 0.03 \ M$.
Thus,$C_{3}(A_{2}) = 0.03 \ mol \ L^{-1}$.

(C) The reaction is $H_{2(g)} + Br_{2(g)} \rightleftharpoons 2 HBr_{(g)}$ with $K_p = 1.6 \times 10^5$.
For the reverse reaction $2 HBr_{(g)} \rightleftharpoons H_{2(g)} + Br_{2(g)}$,the equilibrium constant is $K_p' = \frac{1}{K_p} = \frac{1}{1.6 \times 10^5} = 6.25 \times 10^{-6}$.
For the reaction $HBr_{(g)} \rightleftharpoons \frac{1}{2} H_{2(g)} + \frac{1}{2} Br_{2(g)}$,the equilibrium constant is $K_p'' = \sqrt{K_p'} = \sqrt{6.25 \times 10^{-6}} = 2.5 \times 10^{-3}$.
Let the initial pressure of $HBr$ be $10 \ bar$. At equilibrium,let the pressure of $HBr$ be $(10 - x) \ bar$,and the pressures of $H_2$ and $Br_2$ be $\frac{x}{2} \ bar$ each.
Since $K_p''$ is very small,$x$ will be very small,so $10 - x \approx 10$.
$K_p'' = \frac{p_{H_2}^{1/2} \times p_{Br_2}^{1/2}}{p_{HBr}} = \frac{(\frac{x}{2})^{1/2} \times (\frac{x}{2})^{1/2}}{10 - x} = \frac{x/2}{10 - x} \approx \frac{x}{20}$.
$2.5 \times 10^{-3} = \frac{x}{20} \implies x = 0.05 \ bar$.
Thus,the equilibrium pressure of $HBr = 10 - 0.05 = 9.95 \ bar$.

(A) The reaction is: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$

Initial moles	$15$	$5.2$	$0$
Equilibrium moles	$15 - x$	$5.2 - x$	$2x$

Given that at equilibrium,moles of $HI = 10$,so $2x = 10$,which implies $x = 5$.
Equilibrium moles:
$n(H_2) = 15 - 5 = 10 \ mol$
$n(I_2) = 5.2 - 5 = 0.2 \ mol$
$n(HI) = 10 \ mol$
Let $V$ be the volume of the container in $L$. The equilibrium constant $K_c$ for the formation of $HI$ is:
$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(10/V)^2}{(10/V) \times (0.2/V)} = \frac{100}{2} = 50$
The dissociation of $HI$ is the reverse reaction: $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$.
The equilibrium constant for dissociation,$K'_c = \frac{1}{K_c} = \frac{1}{50} = 0.02 = 2 \times 10^{-2}$.

(B) The reaction is $A_{2(g)} + B_{2(g)} \rightleftharpoons 2 AB_{(g)}$.
First,calculate $K_c$ for the formation of $AB$:
$K_c = \frac{[AB]^2}{[A_2][B_2]} = \frac{(1.4 \times 10^{-3})^2}{(1.5 \times 10^{-3})(2.1 \times 10^{-3})} = \frac{1.96 \times 10^{-6}}{3.15 \times 10^{-6}} \approx 0.622$.
Since $\Delta n = 2 - (1 + 1) = 0$,$K_p = K_c(RT)^0 = K_c = 0.622$.
The decomposition of $AB$ is the reverse reaction: $2 AB_{(g)} \rightleftharpoons A_{2(g)} + B_{2(g)}$.
Therefore,$K_p' = \frac{1}{K_p} = \frac{1}{0.622} \approx 1.60$.