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

(A) According to the law of mass action,the rate of a chemical reaction is directly proportional to the concentration of the reactants. Therefore,as the concentration of the substances involved in a reaction increases,the speed of the reaction increases. Thus,statement $(A)$ is false.

(A) The balanced chemical equation is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$
Initial moles: $n(N_2) = \frac{28 \, g}{28 \, g/mol} = 1 \, mol$,$n(H_2) = \frac{6 \, g}{2 \, g/mol} = 3 \, mol$,$n(NH_3) = 0 \, mol$
Equilibrium moles of $NH_3 = \frac{27.54 \, g}{17 \, g/mol} = 1.62 \, mol$
According to stoichiometry,$2 \, mol$ of $NH_3$ are formed from $1 \, mol$ of $N_2$ and $3 \, mol$ of $H_2$.
So,$1.62 \, mol$ of $NH_3$ are formed from $0.81 \, mol$ of $N_2$ and $2.43 \, mol$ of $H_2$.
Equilibrium moles: $n(N_2) = 1 - 0.81 = 0.19 \, mol$,$n(H_2) = 3 - 2.43 = 0.57 \, mol$,$n(NH_3) = 1.62 \, mol$
Since the volume is $1 \, L$,concentration $[X] = n(X) \, mol/L$.
$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(1.62)^2}{(0.19)(0.57)^3} \approx \frac{2.6244}{0.19 \times 0.185} \approx \frac{2.6244}{0.03515} \approx 74.66 \approx 75$.

(C) The reaction is $A + 2B \rightleftharpoons 2C$.
Initial moles: $A = 2$,$B = 3$,$C = 2$.
Initial concentrations $(V = 2.0 \ L)$: $[A]_0 = 1 \ M$,$[B]_0 = 1.5 \ M$,$[C]_0 = 1 \ M$.
At equilibrium,$[C] = 0.5 \ M$.
Change in concentration of $C = 0.5 - 1 = -0.5 \ M$.
Since $2$ moles of $C$ are formed from $1$ mole of $A$ and $2$ moles of $B$,the change in $[A] = +0.25 \ M$ and $[B] = +0.5 \ M$.
Equilibrium concentrations: $[A] = 1 + 0.25 = 1.25 \ M$,$[B] = 1.5 + 0.5 = 2.0 \ M$,$[C] = 0.5 \ M$.
$K_c = \frac{[C]^2}{[A][B]^2} = \frac{(0.5)^2}{(1.25)(2.0)^2} = \frac{0.25}{1.25 \times 4} = \frac{0.25}{5} = 0.05$.

(A) The chemical equation is: $2HI \rightleftharpoons H_2 + I_2$
Let the initial concentration of $HI$ be $1 \ M$ and $H_2$ and $I_2$ be $0 \ M$.
At equilibrium,$50\%$ of $HI$ is dissociated,so the amount of $HI$ remaining is $1 - 0.5 = 0.5 \ M$.
According to the stoichiometry,$2 \ moles$ of $HI$ produce $1 \ mole$ of $H_2$ and $1 \ mole$ of $I_2$.
Therefore,the concentration of $H_2$ formed is $0.5 / 2 = 0.25 \ M$ and $I_2$ formed is $0.25 \ M$.
The equilibrium constant $K_c$ is given by:
$K_c = \frac{[H_2][I_2]}{[HI]^2} = \frac{0.25 \times 0.25}{(0.5)^2} = \frac{0.0625}{0.25} = 0.25$.

(D) The given reaction is $C_6H_{12}O_6 \rightleftharpoons 6HCHO$.
The formation reaction is $6HCHO \rightleftharpoons C_6H_{12}O_6$ with $K_1 = 6 \times 10^{22}$.
The dissociation reaction is the reverse,so $K_2 = \frac{1}{K_1} = \frac{1}{6 \times 10^{22}}$.
For the reaction $C_6H_{12}O_6 \rightleftharpoons 6HCHO$,the equilibrium constant $K = (K_2)^{1/6} = (\frac{1}{6 \times 10^{22}})^{1/6} \approx 1.6 \times 10^{-4}$.
Since the initial concentration of glucose is $1 \, M$,at equilibrium $[C_6H_{12}O_6] \approx 1 \, M$.
Thus,$[HCHO]^6 = K \times [C_6H_{12}O_6] = 1.6 \times 10^{-4} \times 1$.
$[HCHO] = (1.6 \times 10^{-4})^{1/6} \approx 0.26 \, M$.
Note: Given the options,the question implies calculating the equilibrium constant for the dissociation step directly as $K = (6 \times 10^{22})^{-1/6} \approx 1.6 \times 10^{-4}$.

(C) The balanced chemical equation is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$
Initial moles: $n(N_2) = \frac{56 \, g}{28 \, g/mol} = 2 \, mol$,$n(H_2) = \frac{8 \, g}{2 \, g/mol} = 4 \, mol$,$n(NH_3) = 0 \, mol$.
At equilibrium,$34 \, g$ of $NH_3$ is present,which is $n(NH_3) = \frac{34 \, g}{17 \, g/mol} = 2 \, mol$.
According to the stoichiometry of the reaction,$2 \, mol$ of $NH_3$ are produced from $1 \, mol$ of $N_2$ and $3 \, mol$ of $H_2$.
Equilibrium moles of $N_2 = 2 \, mol - 1 \, mol = 1 \, mol$.
Equilibrium moles of $H_2 = 4 \, mol - 3 \, mol = 1 \, mol$.
Equilibrium moles of $NH_3 = 2 \, mol$.
Thus,the moles are $1, 1, 2$.

(D) The rate of reaction is given by the rate law expression: $R = k[SO_2]^2[O_2]$.
Assuming $1 \ mol$ of each reactant is present in both vessels:
For the $1 \ dm^3$ vessel,the concentration is $1 \ mol/dm^3$. Thus,$R_1 = k(1)^2(1) = k$.
For the $2 \ dm^3$ vessel,the concentration is $0.5 \ mol/dm^3$. Thus,$R_2 = k(0.5)^2(0.5) = k(0.25)(0.5) = 0.125k = \frac{k}{8}$.
The ratio of reaction velocities $R_1 : R_2 = k : \frac{k}{8} = 8 : 1$.

(B) The reaction for the decomposition of calcium carbonate is:
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$ ; $K_{p2} = pCO_2 = 8 \times 10^{-2}$
The reaction for the formation of carbon monoxide is:
$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ ; $K_{p1} = \frac{(pCO)^2}{pCO_2} = 2$
Substituting the value of $pCO_2$ from the first equation into the second:
$(pCO)^2 = K_{p1} \times pCO_2$
$(pCO)^2 = 2 \times (8 \times 10^{-2}) = 16 \times 10^{-2}$
$pCO = \sqrt{16 \times 10^{-2}} = 4 \times 10^{-1} = 0.4$

(D) The reaction is $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$.
Initially: $0.6 \ mol$ of $NH_3$,$0 \ mol$ of $N_2$,$0 \ mol$ of $H_2$.
At equilibrium: $NH_3 = 0.6 - 2x$,$N_2 = x$,$H_2 = 3x$.
Given $3x = 0.15 \ mol$,so $x = 0.05 \ mol$.
Amount of $NH_3$ at equilibrium $= 0.6 - 2(0.05) = 0.5 \ mol$.
Concentration of $NH_3$ at equilibrium $= \frac{0.5 \ mol}{2 \ dm^3} = 0.25 \ mol \ dm^{-3}$.
Thus,the correct statement is that the concentration of $NH_3$ at equilibrium is $0.25 \ mol \ dm^{-3}$.

(B) The balanced chemical equation is:
$SO_2(g) + \frac{1}{2} O_2(g) \leftrightarrow SO_3(g)$
Initial moles:
$SO_2 = 5 \ \text{mol}$,$O_2 = 5 \ \text{mol}$,$SO_3 = 0 \ \text{mol}$
At equilibrium,$60\%$ of $SO_2$ is consumed:
Amount of $SO_2$ reacted $= 5 \times 0.6 = 3 \ \text{mol}$
Remaining $SO_2 = 5 - 3 = 2 \ \text{mol}$
According to the stoichiometry,$1 \ \text{mol}$ of $SO_2$ reacts with $0.5 \ \text{mol}$ of $O_2$ to produce $1 \ \text{mol}$ of $SO_3$.
Amount of $O_2$ reacted $= 3 \times 0.5 = 1.5 \ \text{mol}$
Remaining $O_2 = 5 - 1.5 = 3.5 \ \text{mol}$
Amount of $SO_3$ formed $= 3 \ \text{mol}$
Total moles at equilibrium $= 2 + 3.5 + 3 = 8.5 \ \text{mol}$.

(D) The reaction is an esterification reaction: $CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 + H_2O$.
For this reaction,the equilibrium constant $K_c$ is $4$.
Let $x$ be the number of moles of ethyl acetate formed at equilibrium.
Initial moles: $1$ mole of acid,$1$ mole of alcohol,$0$ mole of ester,$0$ mole of water.
At equilibrium: $(1-x)$ moles of acid,$(1-x)$ moles of alcohol,$x$ moles of ester,$x$ moles of water.
$K_c = \frac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]} = \frac{x \times x}{(1-x)(1-x)} = 4$.
Taking the square root on both sides: $\frac{x}{1-x} = 2$.
$x = 2 - 2x \implies 3x = 2 \implies x = 2/3$.
Thus,$2/3$ mole of ethyl acetate is formed.

(D) catalyst increases the rate of both the forward and backward reactions to the same extent.
It helps the system reach the state of equilibrium faster but does not shift the position of the equilibrium or favor one direction over the other.
Therefore,the statement that the forward reaction is favoured by the addition of a catalyst is incorrect.

(B) The reaction is $2NH_3 \rightleftharpoons N_2 + 3H_2$.
Initial moles: $a$,$0$,$0$.
Moles at equilibrium: $(a - 2x)$,$x$,$3x$.
Total moles at equilibrium: $(a - 2x) + x + 3x = a + 2x$.
Initial pressure of $a$ moles of $NH_3$ at $27 \ ^{\circ}C$ $(300 \ K)$ is $15 \ atm$.
Let $p$ be the pressure of $a$ moles of $NH_3$ at $347 \ ^{\circ}C$ $(620 \ K)$.
Using $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ (at constant volume),$\frac{15}{300} = \frac{p}{620}$,which gives $p = 31 \ atm$.
Since $P \propto n$ at constant $V$ and $T$,the initial pressure $P_i = 31 \ atm$ corresponds to $a$ moles.
The final pressure $P_f = 50 \ atm$ corresponds to $(a + 2x)$ moles.
Therefore,$\frac{a + 2x}{a} = \frac{50}{31}$.
$1 + \frac{2x}{a} = \frac{50}{31} \implies \frac{2x}{a} = \frac{50}{31} - 1 = \frac{19}{31}$.
Percentage of $NH_3$ decomposed $= \frac{2x}{a} \times 100 = \frac{19}{31} \times 100 \approx 61.29 \% \approx 61.3 \%$.

(D) The equilibrium constant expression for the reaction is $K_c = \frac{[SO_2]^2 [O_2]}{[SO_3]^2}$.
Given that $[SO_3] = [SO_2]$,the expression simplifies to $K_c = \frac{[SO_2]^2 [O_2]}{[SO_2]^2} = [O_2]$.
Since $K_c = 100$,it follows that $[O_2] = 100 \ M$.
However,the initial concentrations of the reactants are not provided in the problem statement.
Without the initial concentration or the total volume of the system,the absolute concentration of $O_2$ cannot be determined.
Therefore,the data provided is insufficient to calculate a specific numerical value for the concentration.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.
For the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$,the change in the number of gaseous moles is $\Delta n_g = n_p - n_r = 2 - (1 + 3) = 2 - 4 = -2$.
Substituting this value into the equation,we get: $\Delta H = \Delta E + (-2)RT = \Delta E - 2RT$.

(B) The Van't Hoff equation relates the equilibrium constant $(K)$ to temperature $(T)$: $\ln(\frac{K_2}{K_1}) = \frac{-\Delta H}{R} (\frac{1}{T_2} - \frac{1}{T_1})$.
Given $T_2 > T_1$ $(1000 \ K > 298 \ K)$ and $K_2 < K_1$ $(2 \times 10^{-5} < 5 \times 10^{-3})$,the ratio $\frac{K_2}{K_1}$ is less than $1$,making $\ln(\frac{K_2}{K_1})$ negative.
Since $(\frac{1}{T_2} - \frac{1}{T_1})$ is negative,the equation implies that $\Delta H$ must be negative for the equality to hold.
Therefore,the reaction is exothermic,and $\Delta H < 0$.

(C) The Nernst distribution law (or partition law) states that a solute distributes itself between two immiscible solvents in a constant ratio of concentrations at a constant temperature.
$(i)$ The two solvents must be mutually immiscible (or sparingly miscible).
$(ii)$ The temperature must remain constant.
$(iii)$ The solute must exist in the same molecular state in both solvents.
$(iv)$ The solution must be dilute.
Among the given options,water and amyl alcohol are mutually immiscible,making them a suitable pair for the application of the distribution law. Therefore,the correct answer is $C$.

(A) The stability of a compound is inversely proportional to the equilibrium constant $(K)$ of its decomposition reaction.
$A$ lower value of $K$ indicates that the equilibrium lies far to the left,meaning the compound has a lower tendency to decompose into its elements.
Comparing the given equilibrium constants:
$K(NO_2) = 6.7 \times 10^{16}$
$K(NO) = 2.2 \times 10^{30}$
$K(N_2O_5) = 1.2 \times 10^{34}$
$K(N_2O) = 3.5 \times 10^{33}$
Since $NO_2$ has the smallest equilibrium constant value,it is the most stable oxide among the given options.
Therefore,option $A$ is the correct answer.

(C) In a reversible reaction,a catalyst provides an alternate pathway with lower activation energy for both the forward and backward reactions.
As a result,it increases the rate of both the forward and backward reactions equally.
It does not change the equilibrium constant of the reaction.

(B) For option $A$: An octadecapeptide consists of $18$ amino acid residues linked by $17$ peptide bonds. This statement is correct.
For option $B$: Addition of an inert gas at constant volume does not change the partial pressures of the reactants or products,so the equilibrium does not shift. The original statement claiming it shifts to the left is incorrect.
For option $C$: Gold dissolves in aqua regia $(HCl + HNO_3)$ to form chloroauric acid $(HAuCl_4)$. This statement is correct.
For option $D$: The relationship between $pH$ and $[H^{+}]$ is $[H^{+}] = 10^{-pH}$. If $pH$ changes from $5$ to $3$,the concentration changes by a factor of $10^{(5-3)} = 10^2 = 100$. This statement is correct.

(D) The relationship between the equilibrium constant $K$ and standard Gibbs free energy change is $\Delta G^\circ = -RT \ln K$.
Rearranging this,we get $\ln K = -\frac{\Delta G^\circ}{RT}$,which implies $K = e^{-\Delta G^\circ / RT}$. Thus,option $A$ is correct.
For any process,$\Delta G_{\text{system}} = -T\Delta S_{\text{total}}$. Therefore,$\frac{\Delta G_{\text{system}}}{\Delta S_{\text{total}}} = -T$. Thus,option $B$ is correct.
For an isothermal process,the definition $\Delta G = \Delta H - T\Delta S$ is correct.
Comparing this with the expression in option $D$,we know $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Substituting this into $\Delta G^\circ = -RT \ln K$,we get $-RT \ln K = \Delta H^\circ - T\Delta S^\circ$,which simplifies to $\ln K = -\frac{\Delta H^\circ - T\Delta S^\circ}{RT}$.
The expression given in option $D$ is $\ln K = \frac{\Delta H^\circ - T\Delta S^\circ}{RT}$,which lacks the negative sign. Therefore,option $D$ is the incorrect equation.

(D) The decomposition of $PCl_5$ is an endothermic process,which requires the absorption of heat.
Therefore,$\Delta H > 0$.
Additionally,the change in the number of gaseous moles is $\Delta n_g = n_p - n_r = (1 + 1) - 1 = 1$.
Since $\Delta n_g > 0$,the entropy of the system increases,meaning $\Delta S > 0$.

(B) Initial state: $P_1 = 1 \ atm$,$T_1 = 300 \ K$.
After heating to $T_2 = 600 \ K$ (assuming no dissociation initially),the pressure $P_2$ is calculated using Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ $\Rightarrow \frac{1}{300} = \frac{P_2}{600}$ $\Rightarrow P_2 = 2 \ atm$.
Now,consider the dissociation: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$.
At $t=0$,$P(N_2O_4) = 2 \ atm$,$P(NO_2) = 0$.
At equilibrium,$20 \%$ dissociation means $P' = 0.20 \times 2 = 0.4 \ atm$.
$P(N_2O_4) = 2 - 0.4 = 1.6 \ atm$ and $P(NO_2) = 2 \times 0.4 = 0.8 \ atm$.
Total pressure $P_{total} = 1.6 + 0.8 = 2.4 \ atm$.

(D) The chemical reaction is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Given that $2x = 0.8 \ mol$,therefore $x = 0.4 \ mol$.

	$N_2$	$3H_2$	$2NH_3$
Initial moles	$1$	$2$	$0$
Equilibrium moles	$1 - x = 0.6$	$2 - 3x = 2 - 1.2 = 0.8$	$2x = 0.8$
Equilibrium concentration ($1 \ L$ vessel)	$0.6 \ M$	$0.8 \ M$	$0.8 \ M$

The concentration of $H_2$ at equilibrium is $0.8 \ M$.

(C) The reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$

Initial moles	$a$	$0$	$0$
Equilibrium moles	$(a - x)$	$x$	$x$

Given that $x = 0.1$ mole and $V = 1 \ L$,$K_c = 0.04$.
$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(x/V)(x/V)}{(a-x)/V} = \frac{x^2}{(a-x)V}$
Substituting the values:
$0.04 = \frac{(0.1)^2}{(a - 0.1) \times 1}$
$0.04 = \frac{0.01}{a - 0.1}$
$a - 0.1 = \frac{0.01}{0.04} = 0.25$
$a = 0.25 + 0.1 = 0.35 \ mol/L$

(B) The reaction is $CO + H_2O \rightleftharpoons CO_2 + H_2$.
Initially,we have $1 \ mol$ of $CO$,$1 \ mol$ of $H_2O$,$1 \ mol$ of $CO_2$,and $0 \ mol$ of $H_2$.
At equilibrium,the moles are: $CO = 1 - x$,$H_2O = 1 - x$,$CO_2 = 1 + x$,and $H_2 = x$.
Since the reaction proceeds to establish equilibrium,$x$ must be a positive value less than $1$ (as the initial concentration of reactants is $1 \ mol$).
Therefore,$CO = 1 - x < 1$,$H_2O = 1 - x < 1$,and $H_2 = x < 1$.
Thus,$CO$,$H_2O$,and $H_2$ are all less than $1 \ mol$.

(D) The reaction is $I_{2(g)} \rightleftharpoons 2I_{(g)}$.
Initial moles: $1 \ mol$ of $I_2$,$0 \ mol$ of $I$.
At equilibrium,let $x$ be the degree of dissociation. The moles are $(1-x)$ for $I_2$ and $2x$ for $I$.
Total moles at equilibrium = $(1-x) + 2x = 1 + x$.
Since $K_c = 10^{-6}$,which is very small,the equilibrium lies far to the left,meaning the concentration of reactant is much greater than the product.
Therefore,$[I_{2(g)}] \gg [I_{(g)}]$ and the total moles expression is correct.
Thus,both conditions are satisfied.

(D) $1$. The equilibrium constant $K_p$ depends only on temperature. Since the temperature is constant,$K_p$ remains unchanged.
$2$. For the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$,the equilibrium constant $K_p$ in terms of degree of dissociation $\alpha$ and total pressure $P$ is given by $K_p = \frac{4\alpha^2 P}{1 - \alpha^2}$.
$3$. When the volume of the vessel is halved,the total pressure $P$ increases. To keep $K_p$ constant at a fixed temperature,the degree of dissociation $\alpha$ must decrease to compensate for the increase in pressure.
$4$. Therefore,$K_p$ remains constant,but $\alpha$ changes.

(C) The balanced dissociation equation is: $A_2B_3 \rightarrow 2A^{3+} + 3B^{2-}$.
From the stoichiometry of the reaction,$2$ moles of $A^{3+}$ ions are produced for every $3$ moles of $B^{2-}$ ions.
Therefore,the number of $A^{3+}$ ions = $\frac{2}{3}$ of the number of $B^{2-}$ ions.

(D) The reaction is $N_2 + O_2 \rightleftharpoons 2NO$.
At equilibrium,the concentration of $NO$ is $2x = 1.0 \ mol/L$,so $x = 0.50 \ mol/L$.

Initial concentration	$a$	$b$	$0$
Equilibrium concentration	$(a - x)$	$(b - x)$	$2x$

Given that at equilibrium:
$[N_2] = a - x = 0.25 \ mol/L$
$[O_2] = b - x = 0.05 \ mol/L$
Substituting $x = 0.50 \ mol/L$:
$a = 0.25 + 0.50 = 0.75 \ mol/L$
$b = 0.05 + 0.50 = 0.55 \ mol/L$
Therefore,the initial concentrations of $N_2$ and $O_2$ are $0.75 \ mol/L$ and $0.55 \ mol/L$ respectively.

(A) The reaction is $2P_{(g)} + Q_{(g)} \rightleftharpoons 3R_{(g)} + S_{(g)}$.
Initial moles: $P = 2, Q = 2, R = 0, S = 0$.
At equilibrium,let the extent of reaction be $x$.
Moles at equilibrium: $P = 2 - 2x, Q = 2 - x, R = 3x, S = x$.
Since $x > 0$,we compare the concentrations of $P$ and $Q$:
$[P] = 2 - 2x$ and $[Q] = 2 - x$.
Subtracting the two: $[Q] - [P] = (2 - x) - (2 - 2x) = x$.
Since $x > 0$,$[Q] - [P] > 0$,which implies $[Q] > [P]$ or $[P] < [Q]$.

(B) The rate law for the reaction $2NO + O_2 \rightleftharpoons 2NO_2$ is given by $r = k[NO]^2[O_2]$.
When the volume of the container is reduced to half,the concentration of each reactant doubles because $[C] = \frac{n}{V}$.
If the initial concentrations are $[NO] = x$ and $[O_2] = y$,the initial rate is $r_1 = k(x)^2(y) = kx^2y$.
After reducing the volume to half,the new concentrations become $[NO] = 2x$ and $[O_2] = 2y$.
The new rate is $r_2 = k(2x)^2(2y) = k(4x^2)(2y) = 8kx^2y$.
Therefore,the rate of the reaction becomes $8$ times the initial rate.

(A) The reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$.
Initial moles: $H_2 = 1$,$I_2 = 2$,$HI = 0$.
At equilibrium,moles of $H_2 = 0.2$.
Change in moles of $H_2 = 1 - 0.2 = 0.8 \ mol$.
According to the stoichiometry of the reaction:
$I_2$ consumed $= 0.8 \ mol$.
$HI$ produced $= 2 \times 0.8 = 1.6 \ mol$.
At equilibrium:
Moles of $I_2 = 2 - 0.8 = 1.2 \ mol$.
Moles of $HI = 1.6 \ mol$.
Thus,the moles of $I_2$ and $HI$ are $1.2$ and $1.6$ respectively.

(A) The reaction $A_{(g)} \rightleftharpoons B_{(g)} + \text{Heat}$ is exothermic. According to Le Chatelier's principle,exothermic reactions are favored at low temperatures. So,$A-ii$.
$(B)$ The ratio of the rate of backward reaction to the rate of forward reaction $(r_b/r_f)$ is equal to the inverse of the equilibrium constant $(K_c^{-1})$. So,$B-iii$.
$(C)$ The ratio of the rate of forward reaction to the rate of backward reaction $(r_f/r_b)$ is equal to the equilibrium constant $(K_c)$. So,$C-i$.
$(D)$ For the reaction $2A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$,the change in moles of gas is $\Delta n = 1 - (2+1) = -2$. Since $\Delta n < 0$,this matches $(V)$. So,$D-v$.
$(E)$ The effect of pressure is significant when $\Delta n \neq 0$. The reaction $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$ has $\Delta n = (1+1) - (1+1) = 0$,meaning pressure has no effect. However,in the context of the options provided,$E$ corresponds to the reaction where pressure change is relevant or the general principle. Given the options,$E-iv$ is the intended match as it represents a system where pressure change is discussed.
Therefore,the correct sequence is $A-ii, B-iii, C-i, D-v, E-iv$.

(C) The balanced chemical equation is: $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
Initial moles: $SO_2 = 5$,$O_2 = 5$,$SO_3 = 0$
Since $60\%$ of $SO_2$ is consumed,the amount reacted is $5 \times 0.60 = 3 \ mol$.
At equilibrium:
$SO_2 = 5 - 3 = 2 \ mol$
$O_2 = 5 - (3/2) = 3.5 \ mol$
$SO_3 = 3 \ mol$
Total moles at equilibrium = $2 + 3.5 + 3 = 8.5 \ mol$
Partial pressure of $O_2$ $(P_{O_2})$ = $(\text{mole fraction of } O_2) \times \text{Total Pressure}$
$P_{O_2} = (3.5 / 8.5) \times 1 \ atm \approx 0.41 \ atm$.

(D) For a heterogeneous equilibrium,the concentration of pure solids is taken as unity $(1)$.
The equilibrium constant expression for $K_c$ is given by: $K_c = \frac{[CuSO_4 \cdot 3H_2O] [H_2O]^2}{[CuSO_4 \cdot 5H_2O]}$.
Since $CuSO_4 \cdot 3H_2O$ and $CuSO_4 \cdot 5H_2O$ are solids,their concentrations are $1$. Thus,$K_c = [H_2O]^2$.
For $K_p$,only the gaseous species are considered: $K_p = (P_{H_2O})^2$.
The relationship between $K_p$ and $K_c$ is $K_p = K_c(RT)^{\Delta n_g}$.
Here,$\Delta n_g = 2 - 0 = 2$.
Therefore,$K_p = K_c(RT)^2$.
Since all three expressions are correct,the answer is $D$.

(B) The reaction is $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$.
Initial moles: $NH_3 = 2, N_2 = 0, H_2 = 0$.
At equilibrium,$NH_3$ remaining is $1 \, \text{mol}$.
Change in moles: $2 - 2x = 1 \implies 2x = 1 \implies x = 0.5$.
Equilibrium moles: $NH_3 = 1, N_2 = x = 0.5, H_2 = 3x = 1.5$.
Volume $V = 500 \, \text{mL} = 0.5 \, \text{L}$.
Equilibrium concentrations: $[NH_3] = 1/0.5 = 2 \, \text{M}, [N_2] = 0.5/0.5 = 1 \, \text{M}, [H_2] = 1.5/0.5 = 3 \, \text{M}$.
$K_c = \frac{[N_2][H_2]^3}{[NH_3]^2} = \frac{1 \times (3)^3}{(2)^2} = \frac{27}{4} = 6.75$.

(A) Active mass is defined as the molar concentration of a substance,so $(A-ii)$.
$(B)$ The equilibrium constant is a characteristic of chemical equilibrium,so $(B-v)$.
$(C)$ For an endothermic reaction $(A + \text{Heat} \rightleftharpoons B)$,increasing the temperature shifts the equilibrium to the right (favoured),so $(C-iv)$.
$(D)$ For the reaction $2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)}$,the change in moles of gas $\Delta n = (3) - (2+1) = 0$,so $(D-i)$.
Therefore,the correct matching is $(A-ii, B-v, C-iv, D-i)$.

(A) The reaction is $C_{(s)} + S_{2_{(g)}} \rightleftharpoons CS_{2_{(g)}}$.
Calculate the initial moles of reactants and products:
Moles of $S_2 = \frac{64 \ g}{64 \ g/mol} = 1 \ mol$.
Moles of $CS_2 = \frac{76 \ g}{76 \ g/mol} = 1 \ mol$.
Since $C_{(s)}$ is a solid,its amount does not affect the gas pressure.
The total moles of gaseous species present in the vessel are $n_{total} = n_{S_2} + n_{CS_2} = 1 + 1 = 2 \ mol$.
Given temperature $T = 1027 \ ^oC = 1027 + 273 = 1300 \ K$ and volume $V = 13 \ L$.
Using the ideal gas equation $PV = nRT$:
$P = \frac{nRT}{V} = \frac{2 \times R \times 1300}{13} = 200 \ R$.

(A) The total volume of the system after connecting the flasks is $V = 1 \ L + 2 \ L = 3 \ L$.
Initial moles: $n(X_2) = 1 \ mol$,$n(Y_2) = 2 \ mol$.
Initial concentrations: $[X_2]_0 = \frac{1}{3} \ M$,$[Y_2]_0 = \frac{2}{3} \ M$.
Let the change in concentration be $x$.
Reaction: $X_{2(g)} + Y_{2(g)} \rightleftharpoons 2XY_{(g)}$
At equilibrium: $[X_2] = \frac{1}{3} - x$,$[Y_2] = \frac{2}{3} - x$,$[XY] = 2x$.
Given $[XY] = 0.6 \ M$,so $2x = 0.6 \implies x = 0.3 \ M$.
Therefore,equilibrium concentrations are $[X_2] = (\frac{1}{3} - 0.3) \ M$ and $[Y_2] = (\frac{2}{3} - 0.3) \ M$.

(A) For reaction $1$: $X \rightleftharpoons 2Y$. Let initial moles be $1$,degree of dissociation be $\alpha$. At equilibrium: $X = 1-\alpha$,$Y = 2\alpha$. Total moles $= 1+\alpha$. $K_{p1} = \frac{(2\alpha/(1+\alpha))^2 P_1}{(1-\alpha)/(1+\alpha)} = \frac{4\alpha^2 P_1}{1-\alpha^2}$.
For reaction $2$: $Z \rightleftharpoons P + Q$. Let initial moles be $1$,degree of dissociation be $\alpha$. At equilibrium: $Z = 1-\alpha$,$P = \alpha$,$Q = \alpha$. Total moles $= 1+\alpha$. $K_{p2} = \frac{(\alpha/(1+\alpha))^2 P_2}{(1-\alpha)/(1+\alpha)} = \frac{\alpha^2 P_2}{1-\alpha^2}$.
Given $\frac{K_{p1}}{K_{p2}} = \frac{1}{9}$.
$\frac{4\alpha^2 P_1 / (1-\alpha^2)}{\alpha^2 P_2 / (1-\alpha^2)} = \frac{1}{9} \implies \frac{4P_1}{P_2} = \frac{1}{9}$.
Therefore,$\frac{P_1}{P_2} = \frac{1}{36}$.

(A) The reaction is $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$.

Initial moles	$1$	$0$	$0$
Moles at equilibrium	$0.3$	$0.7$	$0.7$

Total moles at equilibrium = $0.3 + 0.7 + 0.7 = 1.7 \ mol$.
Concentration of $PCl_3 = \frac{0.7 \ mol}{5 \ L} = 0.14 \ M$.
$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(0.7/5) \times (0.7/5)}{(0.3/5)} = \frac{0.49}{1.5} = \frac{49}{150}$.

(C) The equilibrium constant expression for the reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2_{(g)}} + S_{2_{(g)}}$ is given by:
$K = \frac{[H_2]^2 [S_2]}{[H_2S]^2}$
Given concentrations in a $1 \ L$ vessel:
$[H_2S] = 0.5 \ mol \ L^{-1}$,$[H_2] = 0.10 \ mol \ L^{-1}$,$[S_2] = 0.4 \ mol \ L^{-1}$
Substituting these values into the expression:
$K = \frac{(0.10)^2 \times (0.4)}{(0.5)^2} = \frac{0.01 \times 0.4}{0.25} = \frac{0.004}{0.25} = 0.016$

(A) The dissociation reaction is: $PCl_5 \rightleftharpoons PCl_3 + Cl_2$

Initial moles	$1, 1, 0$
Moles at equilibrium	$(1 - x), (1 + x), x$
Concentration (moles / $10 \ L$)	$0.1(1 - x), 0.1(1 + x), 0.1x$

Thus,the concentrations are $0.1(1 - x)$,$0.1(1 + x)$,and $0.1x$ respectively.

(A) The balanced chemical equation is: $2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g)$.
Initial moles: $CO_2 = 2$,$CO = 0$,$O_2 = 0$.
Degree of dissociation is $40\%$,so the amount of $CO_2$ dissociated is $2 \times 0.4 = 0.8 \ mol$.
At equilibrium:
$CO_2 = 2 - 0.8 = 1.2 \ mol$
$CO = 0.8 \ mol$
$O_2 = \frac{0.8}{2} = 0.4 \ mol$
Total moles at equilibrium = $1.2 + 0.8 + 0.4 = 2.4 \ mol$.

(D) $K_c = \frac{[NO]^2}{[N_2][O_2]} = \frac{(2x)^2}{(a-x)(a-x)} = \frac{4x^2}{(a-x)^2} = 9 \times 10^{-4}$
Taking the square root on both sides:
$\frac{2x}{a-x} = \sqrt{9 \times 10^{-4}} = 3 \times 10^{-2} = 0.03$
$2x = 0.03(a - x)$
$2x = 0.03a - 0.03x$
$2.03x = 0.03a$
$x = \frac{0.03a}{2.03} \approx 0.014778a \approx 0.0148a$
The concentration of $NO$ at equilibrium is $2x$.
$[NO] = 2 \times 0.0148a = 0.0296a$

Initial moles	$a$ (for $N_2$),$a$ (for $O_2$),$0$ (for $NO$)
Equilibrium moles	$(a-x)$ (for $N_2$),$(a-x)$ (for $O_2$),$2x$ (for $NO$)

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

(D) The reaction is $A + B \rightleftharpoons C + D$.
Let the initial concentration of $A$ and $B$ be $1 \ M$ each.
At equilibrium,let the concentration of $C$ be $x$. Then the concentration of $D$ is also $x$.
The remaining concentrations of $A$ and $B$ will be $(1 - x)$ each.
According to the problem,at equilibrium,$[C] = 2[A]$.
So,$x = 2(1 - x)$.
$x = 2 - 2x \implies 3x = 2 \implies x = 2/3$.
At equilibrium,$[A] = 1 - 2/3 = 1/3$,$[B] = 1/3$,$[C] = 2/3$,$[D] = 2/3$.
$K_c = \frac{[C][D]}{[A][B]} = \frac{(2/3)(2/3)}{(1/3)(1/3)} = \frac{4/9}{1/9} = 4$.