Law of equilibrium and Equilibrium constant Questions in English

(B) For the reaction:
$CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$
The equilibrium constant expression is:
$K_c = \frac{[COCl_2]}{[CO][Cl_2]} \dots (I)$
Given volume $V = 500 \ mL = 0.5 \ L$.
Concentrations at equilibrium are:
$[COCl_2] = \frac{0.3 \ mol}{0.5 \ L} = 0.6 \ mol \ L^{-1}$
$[CO] = \frac{0.1 \ mol}{0.5 \ L} = 0.2 \ mol \ L^{-1}$
$[Cl_2] = \frac{0.1 \ mol}{0.5 \ L} = 0.2 \ mol \ L^{-1}$
Substituting these values into expression $(I)$:
$K_c = \frac{0.6}{(0.2)(0.2)} = \frac{0.6}{0.04} = 15$

(A) Given the reaction: $SO_{3(g)} \rightleftharpoons SO_{2(g)} + 1/2 O_{2(g)}$ with $K_1 = 4.9 \times 10^{-2}$.
The target reaction is: $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.
First,reverse the given reaction: $SO_{2(g)} + 1/2 O_{2(g)} \rightleftharpoons SO_{3(g)}$. The equilibrium constant for this reversed reaction is $K_2 = 1 / K_1 = 1 / (4.9 \times 10^{-2})$.
Next,multiply the stoichiometric coefficients by $2$: $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$. The equilibrium constant for this reaction is $K_{eq} = (K_2)^2 = (1 / K_1)^2 = 1 / (K_1)^2$.
$K_{eq} = 1 / (4.9 \times 10^{-2})^2 = 1 / (24.01 \times 10^{-4}) = 1 / 0.002401 \approx 416.49$.
Thus,the value is approximately $416$.

(B) Given reactions:
$1) A + D \rightleftharpoons AD, K_1 = \frac{[AD]}{[A][D]}$
$2) AD + D \rightleftharpoons AD_2, K_2 = \frac{[AD_2]}{[AD][D]}$
$3) AD_2 + D \rightleftharpoons AD_3, K_3 = \frac{[AD_3]}{[AD_2][D]}$
Adding these three equations gives the net reaction:
$A + 3D \rightleftharpoons AD_3$
The equilibrium constant $K$ for the net reaction is the product of the individual equilibrium constants:
$K = K_1 \times K_2 \times K_3$
Taking the logarithm on both sides:
$\log K = \log(K_1 \times K_2 \times K_3)$
$\log K = \log K_1 + \log K_2 + \log K_3$

(A) The reaction is $A + 2B \rightleftharpoons 2C + D$.
Let the initial concentration of $A$ be $a_0$. Then the initial concentration of $B$ is $1.5a_0$.
At $t=0$: $[A] = a_0$,$[B] = 1.5a_0$,$[C] = 0$,$[D] = 0$.
At equilibrium $(t=t_{eqm})$: $[A] = a_0 - x$,$[B] = 1.5a_0 - 2x$,$[C] = 2x$,$[D] = x$.
Given that at equilibrium $[A] = [B]$:
$a_0 - x = 1.5a_0 - 2x \Rightarrow x = 0.5a_0$.
Substituting $x$ into the equilibrium concentrations:
$[A] = a_0 - 0.5a_0 = 0.5a_0$
$[B] = 1.5a_0 - 2(0.5a_0) = 0.5a_0$
$[C] = 2(0.5a_0) = a_0$
$[D] = x = 0.5a_0$
The equilibrium constant $K_C$ is given by:
$K_C = \frac{[C]^2 [D]}{[A][B]^2} = \frac{(a_0)^2 (0.5a_0)}{(0.5a_0)(0.5a_0)^2} = \frac{a_0^2 \times 0.5a_0}{0.5a_0 \times 0.25a_0^2} = \frac{1}{0.25} = 4$.

(C) The equilibrium constant expression for the reaction $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$ is given by $K_c = \frac{[H_2][I_2]}{[HI]^2}$.
Given $K_c = 5 \times 10^{3}$,$[H_2] = 2.2 \times 10^{-2} \ M$,and $[I_2] = 2.2 \times 10^{-4} \ M$.
Substituting the values: $5 \times 10^{3} = \frac{(2.2 \times 10^{-2})(2.2 \times 10^{-4})}{[HI]^2}$.
$[HI]^2 = \frac{4.84 \times 10^{-6}}{5 \times 10^{3}} = 0.968 \times 10^{-9} = 9.68 \times 10^{-10}$.
$[HI] = \sqrt{9.68 \times 10^{-10}} \approx 3.11 \times 10^{-5} \ M$.

(D) When multiple chemical reactions are added to obtain a target reaction,the equilibrium constant of the target reaction is the product of the equilibrium constants of the individual reactions.
Given:
$1) A \rightleftharpoons B, K_{C1} = 2$
$2) B \rightleftharpoons C, K_{C2} = 3$
$3) C \rightleftharpoons D + E, K_{C3} = 5$
Adding these reactions:
$A + B + C \rightleftharpoons B + C + D + E$
Canceling common species on both sides gives:
$A \rightleftharpoons D + E$
The equilibrium constant $K_C$ for the target reaction is:
$K_C = K_{C1} \times K_{C2} \times K_{C3}$
$K_C = 2 \times 3 \times 5 = 30$
Therefore,the correct option is $D$.

(B) The chemical equation is $A + B \rightleftharpoons C + D$.
Initially,the moles are: $n(A) = 1, n(B) = 1, n(C) = 0, n(D) = 0$.
At equilibrium,$0.5 \, mol$ of each product is formed.
So,the moles at equilibrium are: $n(C) = 0.5, n(D) = 0.5$.
Since $1 \, mol$ of $A$ and $B$ produce $1 \, mol$ of $C$ and $D$,the moles of reactants consumed are $0.5 \, mol$ each.
Thus,at equilibrium: $n(A) = 1 - 0.5 = 0.5$ and $n(B) = 1 - 0.5 = 0.5$.
The equilibrium constant $K_c$ is given by the expression: $K_c = \frac{[C][D]}{[A][B]}$.
Assuming the volume of the container is $V$,the concentrations are: $[C] = \frac{0.5}{V}, [D] = \frac{0.5}{V}, [A] = \frac{0.5}{V}, [B] = \frac{0.5}{V}$.
Substituting these values: $K_c = \frac{(0.5/V)(0.5/V)}{(0.5/V)(0.5/V)} = 1$.
Therefore,the equilibrium constant is $1$.

(A) Given:
$1) N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}; K_1$
$2) NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}; K_2$
$3) \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \rightleftharpoons NH_{3(g)}; K_3$
$4) 2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}; K_4$
From $(2)$,$K_2 = (K_1)^{1/2} \times (1/K_1)^{1/2}$ is not direct. Let's relate them to $K_1$:
$K_2 = (1/K_1)^{1/2} \Rightarrow K_2^{-2} = K_1$
$K_3 = (K_1)^{1/2} \Rightarrow K_3^2 = K_1$
$K_4 = (1/K_1) \Rightarrow K_4^{-1} = K_1$
Comparing $K_1 = K_2^x = K_3^y = K_4^z$:
$x = -2, y = 2, z = -1$.

(A) For the reversible reaction $A \underset{k_2}{\overset{k_1}{\longleftrightarrow}} B$,at equilibrium,the rate of the forward reaction equals the rate of the backward reaction.
$k_1 [A]_{eq} = k_2 [B]_{eq}$
Given $[A]_{eq} = (a - x)$ and $[B]_{eq} = (b + x)$,we have:
$k_1 (a - x) = k_2 (b + x)$
$k_1 a - k_1 x = k_2 b + k_2 x$
$k_1 a - k_2 b = k_1 x + k_2 x$
$k_1 a - k_2 b = x(k_1 + k_2)$
$x = \frac{k_1 a - k_2 b}{k_1 + k_2}$
Thus,the correct option is $A$.

(A) For the given reaction: $CaSO_4 \cdot 5H_2O_{(s)} \rightleftharpoons CaSO_4 \cdot 3H_2O_{(s)} + 2H_2O_{(g)}$
The equilibrium constant expression is given by the ratio of the concentration of products to reactants,raised to their stoichiometric coefficients.
$K_c = \frac{[CaSO_4 \cdot 3H_2O] \cdot [H_2O]^2}{[CaSO_4 \cdot 5H_2O]}$
Since $CaSO_4 \cdot 5H_2O$ and $CaSO_4 \cdot 3H_2O$ are pure solids,their active masses are taken as unity $(1)$.
Therefore,$K_c = 1 \cdot [H_2O]^2 / 1 = [H_2O]^2$.

(B) The equilibrium constant $K_p$ is a function of temperature only for a given reaction. According to the van't Hoff equation,the value of the equilibrium constant changes with a change in temperature $(T)$.

(C) The equilibrium constant expression for the reaction $2x + y \rightleftharpoons yx_2$ is given by:
$K_c = \frac{[yx_2]}{[x]^2[y]}$
Given equilibrium concentrations are $[x] = 4$,$[y] = 2$,and $[yx_2] = 2$.
Substituting these values into the expression:
$K_c = \frac{2}{4^2 \times 2} = \frac{2}{16 \times 2} = \frac{2}{32} = \frac{1}{16} = 0.0625$

(A) The reaction quotient $Q_c$ is given by the expression: $Q_c = \frac{[C]^2}{[A][B]^2}$.
Given concentrations in a $10 \, L$ vessel:
$[A] = \frac{2 \, \text{mol}}{10 \, L} = 0.2 \, M$
$[B] = \frac{3 \, \text{mol}}{10 \, L} = 0.3 \, M$
$[C] = \frac{1 \, \text{mol}}{10 \, L} = 0.1 \, M$
Substituting these values into the expression for $Q_c$:
$Q_c = \frac{(0.1)^2}{(0.2)(0.3)^2} = \frac{0.01}{0.2 \times 0.09} = \frac{0.01}{0.018} = \frac{10}{18} \approx 0.556$.
Since $Q_c (0.556) < K_c (3.6)$,the reaction will proceed in the forward direction.

(D) The equilibrium constant $K_c$ for the reaction $2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$ is given by the expression:
$K_c = \frac{[H_2][I_2]}{[HI]^2}$
Substituting the units of concentration (mol $L^{-1}$):
$K_c = \frac{(\text{mol L}^{-1}) \times (\text{mol L}^{-1})}{(\text{mol L}^{-1})^2} = \frac{(\text{mol L}^{-1})^2}{(\text{mol L}^{-1})^2} = 1$
Since the units cancel out, the equilibrium constant is dimensionless.
Therefore, the correct option is $D$.

(C) For a general chemical reaction $aA + bB \rightleftharpoons cC + dD$,the equilibrium constant $K_c$ is defined as the ratio of the product of the molar concentrations of the products to the product of the molar concentrations of the reactants,each raised to the power of their respective stoichiometric coefficients.
For the given reaction $3A + 2B \rightleftharpoons C$,the equilibrium constant expression is:
$K_c = \frac{[C]}{[A]^3[B]^2}$

(D) The balanced chemical equation is: $AB_{(g)} + CD_{(g)} \rightleftharpoons AD_{(g)} + CB_{(g)}$
Initial moles: $1 \quad 1 \quad 0 \quad 0$
Moles at equilibrium: $(1 - 3/4) \quad (1 - 3/4) \quad 3/4 \quad 3/4$
Equilibrium moles: $1/4 \quad 1/4 \quad 3/4 \quad 3/4$
Since the volume is constant,the concentration is proportional to the number of moles. The equilibrium constant $K_c$ is given by:
$K_c = \frac{[AD][CB]}{[AB][CD]} = \frac{(3/4) \times (3/4)}{(1/4) \times (1/4)} = \frac{9/16}{1/16} = 9$

(C) The dissociation reaction is: $2 HI(g) \rightleftharpoons H_2(g) + I_2(g)$
Initial moles: $2 \ mol$ of $HI$,$0 \ mol$ of $H_2$,$0 \ mol$ of $I_2$.
At equilibrium,$22\%$ of $HI$ dissociates,so the amount of $HI$ reacted is $2 \times 0.22 = 0.44 \ mol$.
Remaining $HI = 2 - 0.44 = 1.56 \ mol$.
Moles of $H_2$ formed = $0.44 / 2 = 0.22 \ mol$.
Moles of $I_2$ formed = $0.44 / 2 = 0.22 \ mol$.
Assuming volume $V$,the concentrations are $[HI] = 1.56/V$,$[H_2] = 0.22/V$,$[I_2] = 0.22/V$.
$K_c = \frac{[H_2][I_2]}{[HI]^2} = \frac{(0.22/V)(0.22/V)}{(1.56/V)^2} = \frac{0.22 \times 0.22}{1.56 \times 1.56} = \frac{0.0484}{2.4336} \approx 0.0199$.

(C) The reaction quotient $Q$ compares the current state of the reaction to the equilibrium constant $K_c$.
If $Q < K_c$,the ratio of products to reactants is less than the equilibrium ratio,so the reaction proceeds in the forward direction (left to right) to reach equilibrium.
If $Q > K_c$,the reaction proceeds in the reverse direction.
If $Q = K_c$,the system is at equilibrium.

(C) The reaction quotient $(Q_c)$ is defined as the ratio of the concentration of products to the concentration of reactants at any given time. Before the reaction starts,the concentration of products is $0$. Therefore,$Q_c = \frac{[Products]}{[Reactants]} = \frac{0}{[Reactants]} = 0$.

(D) The given reactions represent the decomposition of nitrogen oxides into $N_2$ and $O_2$.
Stability is inversely proportional to the equilibrium constant $(K_c)$ of the decomposition reaction.
$A$ higher $K_c$ value indicates a greater extent of decomposition,meaning the oxide is less stable.
$A$ lower $K_c$ value indicates a lesser extent of decomposition,meaning the oxide is more stable.
Comparing the given $K_c$ values: $1 \times 10^{15} < 1 \times 10^{30} < 1 \times 10^{32} < 1 \times 10^{34}$.
Since $2NO_2 \rightleftharpoons N_2 + 2O_2$ has the lowest $K_c$ value,$NO_2$ is the most stable among the given options.

(B) For the reaction: $cis-C_2H_2Cl_2 \rightleftharpoons trans-C_2H_2Cl_2$,the equilibrium constant is $K_c = 0.6$.
When a reaction is reversed,the new equilibrium constant $K'_c$ is the reciprocal of the original equilibrium constant.
Therefore,for the reaction $trans-C_2H_2Cl_2 \rightleftharpoons cis-C_2H_2Cl_2$,the equilibrium constant $K'_c = \frac{1}{K_c} = \frac{1}{0.6} = 1.67$.

(A) The equilibrium constant $K$ for the reaction ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$ is given by $K = \frac{[NH_3]^2}{[N_2][H_2]^3}$.
For the reaction $2{N_2} + 6{H_2} \rightleftharpoons 4N{H_3}$,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 $K' = \left( \frac{[NH_3]^2}{[N_2][H_2]^3} \right)^2 = K^2$.

(B) Given reaction $(i)$: $2AB \rightleftharpoons A_2 + B_2$ with $K_1 = 49$.
Target reaction $(ii)$: $AB \rightleftharpoons 1/2 A_2 + 1/2 B_2$ with $K_2 = ?$.
Reaction $(ii)$ is obtained by multiplying reaction $(i)$ by $1/2$.
Therefore,the equilibrium constant $K_2$ is related to $K_1$ as $K_2 = (K_1)^{1/2}$.
$K_2 = (49)^{1/2} = 7$.

(C) Given equilibria:
$(I)$ $N_2 + 3H_2 \rightleftharpoons 2NH_3 ; K_1$
$(II)$ $N_2 + O_2 \rightleftharpoons 2NO ; K_2$
$(III)$ $H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O ; K_3$
To obtain the target reaction $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$,we perform the following operations:
Reverse reaction $(I)$: $2NH_3 \rightleftharpoons N_2 + 3H_2$ with constant $K' = \frac{1}{K_1}$
Add reaction $(II)$: $N_2 + O_2 \rightleftharpoons 2NO$ with constant $K_2$
Add $3 \times$ reaction $(III)$: $3H_2 + \frac{3}{2}O_2 \rightleftharpoons 3H_2O$ with constant $K_3^3$
Summing these reactions:
$(2NH_3 + N_2 + O_2 + 3H_2 + \frac{3}{2}O_2) \rightleftharpoons (N_2 + 3H_2 + 2NO + 3H_2O)$
Canceling common terms gives: $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$
The equilibrium constant $K_c = \frac{1}{K_1} \times K_2 \times K_3^3 = \frac{K_2 K_3^3}{K_1}$.

(D) Let the given reactions be:
$(1) XeF_{6(g)} + H_2O_{(g)} \rightleftharpoons XeOF_{4(g)} + 2HF_{(g)} ; K_1$
$(2) XeO_{4(g)} + XeF_{6(g)} \rightleftharpoons XeOF_{4(g)} + XeO_3F_{2(g)} ; K_2$
To obtain the target reaction $XeO_{4(g)} + 2HF_{(g)} \rightleftharpoons XeO_3F_{2(g)} + H_2O_{(g)}$,we perform the operation: $(2) - (1)$.
This is equivalent to adding reaction $(2)$ and the reverse of reaction $(1)$.
The equilibrium constant for the reverse of reaction $(1)$ is $\frac{1}{K_1}$.
Therefore,the equilibrium constant $K$ for the target reaction is $K = K_2 \times \frac{1}{K_1} = \frac{K_2}{K_1}$.

(B) The balanced chemical equation for the reaction is: $3Fe_{(s)} + 4H_2O_{(g)} \leftrightarrow Fe_3O_{4(s)} + 4H_{2(g)}$.
In the expression for the equilibrium constant $(K_p)$,only the partial pressures of gaseous species are included.
Solid species like $Fe_{(s)}$ and $Fe_3O_{4(s)}$ are considered to have unit activity and are omitted from the expression.
Therefore,the equilibrium constant expression is $K_p = \frac{(P_{H_2})^4}{(P_{H_2O})^4}$.

(C) The reaction quotient $(Q_c)$ is defined for a general reaction $aA + bB \rightleftharpoons cC + dD$ as $Q_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$ at any stage of the reaction,which is the same expression as the equilibrium constant $(K_c)$. Thus,the Assertion is correct.
If $Q_c < K_c$,the ratio of products to reactants is less than the equilibrium ratio,so the reaction proceeds in the forward direction (towards products) to reach equilibrium. The Reason states the reaction moves towards reactants,which is incorrect. Therefore,the Assertion is correct but the Reason is incorrect.

(D) For a reversible reaction at equilibrium,the rate of the forward reaction $(r_f)$ is equal to the rate of the reverse reaction $(r_b)$.
The equilibrium constant expression is $K_{eq} = \frac{k_f}{k_b} = \frac{[N_2] [H_2O]^2}{[H_2]^2 [NO]^2}$.
Given the forward rate law: $r_f = k_f [NO]^2 [H_2]$.
At equilibrium,$r_f = r_b$,so $r_b = k_f [NO]^2 [H_2]$.
From the equilibrium expression,we can write $k_f [NO]^2 = \frac{k_b [N_2] [H_2O]^2}{[H_2]^2}$.
Substituting this into the forward rate expression: $r_b = \left( \frac{k_b [N_2] [H_2O]^2}{[H_2]^2} \right) [H_2] = \frac{k_b [N_2] [H_2O]^2}{[H_2]}$.

(A) From the figure,count the number of squares $(A)$ and circles $(B)$:
Number of squares $(A)$ = $4$
Number of circles $(B)$ = $8$
Assuming the reaction $A \rightleftharpoons B$,the equilibrium constant $K$ is given by:
$K = \frac{[B]}{[A]} = \frac{8}{4} = 2$
Therefore,the correct option is $A$.

(N/A) The balanced chemical equation for the formation of ammonia is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
The expression for the equilibrium constant $K_{c}$ is given by: $K_{c} = \frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}$
Substituting the given values: $K_{c} = \frac{(1.2 \times 10^{-2})^{2}}{(1.5 \times 10^{-2}) \times (3.0 \times 10^{-2})^{3}}$
$K_{c} = \frac{1.44 \times 10^{-4}}{(1.5 \times 10^{-2}) \times (27 \times 10^{-6})}$
$K_{c} = \frac{1.44 \times 10^{-4}}{40.5 \times 10^{-8}}$
$K_{c} = \frac{1.44}{40.5} \times 10^{4} \approx 0.03555 \times 10^{4} = 3.555 \times 10^{2}$

(0.622) The equilibrium constant $K_{c}$ for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ is given by the expression:
$K_{c} = \frac{[NO]^{2}}{[N_{2}][O_{2}]}$
Substituting the given equilibrium concentrations:
$K_{c} = \frac{(2.8 \times 10^{-3})^{2}}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}$
$K_{c} = \frac{7.84 \times 10^{-6}}{12.6 \times 10^{-6}}$
$K_{c} = 0.622$

(1.79) The equilibrium constant expression for the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is given by:
$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}$
Substituting the given equilibrium concentrations into the expression:
$K_c = \frac{(1.59 \ M)(1.59 \ M)}{1.41 \ M}$
$K_c = \frac{2.5281}{1.41} \approx 1.79$

(B) For the reaction $2 A \rightleftharpoons B + C$,the reaction quotient $Q_{c}$ is given by:
$Q_{c} = \frac{[ B ][ C ]}{[ A ]^{2}}$
Given $[ A ] = [ B ] = [ C ] = 3 \times 10^{-4} \ M$,we calculate $Q_{c}$:
$Q_{c} = \frac{(3 \times 10^{-4})(3 \times 10^{-4})}{(3 \times 10^{-4})^{2}} = 1$
Since $Q_{c} = 1$ and $K_{c} = 2 \times 10^{-3}$,we observe that $Q_{c} > K_{c}$.
Therefore,the reaction will proceed in the reverse direction to attain equilibrium.

(N/A) The equilibrium constant $(K_{c})$ for the given reaction is defined as:
$K_{c} = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}$
Substituting the given equilibrium concentrations:
$K_{c} = \frac{(1.90)^{2}}{(0.60)^{2} \times (0.82)}$
$K_{c} = \frac{3.61}{0.36 \times 0.82}$
$K_{c} = \frac{3.61}{0.2952} \approx 12.23 \, M^{-1}$
Thus,the equilibrium constant $K_{c}$ is approximately $12.23 \, M^{-1}$.

(N/A) The equilibrium constant $K_{c}$ is defined as the ratio of the product of molar concentrations of products to that of reactants,each raised to the power of their stoichiometric coefficients. Pure solids and liquids are taken as unity $(1)$.
$(i)$ $K_{c} = \frac{[NO]^{2} [Cl_{2}]}{[NOCl]^{2}}$
$(ii)$ Since $Cu(NO_{3})_{2(s)}$ and $CuO_{(s)}$ are solids,their concentrations are taken as $1$.
$K_{c} = [NO_{2}]^{4} [O_{2}]$
$(iii)$ Since $H_{2}O_{(l)}$ is a pure liquid,its concentration is taken as $1$.
$K_{c} = \frac{[CH_{3}COOH] [C_{2}H_{5}OH]}{[CH_{3}COOC_{2}H_{5}]}$
$(iv)$ Since $Fe(OH)_{3(s)}$ is a solid,its concentration is taken as $1$.
$K_{c} = \frac{1}{[Fe^{3+}] [OH^{-}]^{3}}$
$(v)$ Since $I_{2(s)}$ is a solid,its concentration is taken as $1$.
$K_{c} = \frac{[IF_{5}]^{2}}{[F_{2}]^{5}}$

(D) The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.
Given $K_{C} = 6.3 \times 10^{14}$ for the forward reaction.
The equilibrium constant for the reverse reaction is $K_{C}^{\prime} = \frac{1}{K_{C}}$.
$K_{C}^{\prime} = \frac{1}{6.3 \times 10^{14}} = 1.587 \times 10^{-15} \approx 1.59 \times 10^{-15}$.

(N/A) For a pure substance (both solids and liquids),the concentration is defined as:
$[Pure \text{ } substance] = \frac{\text{Number of moles}}{\text{Volume}}$
$= \frac{\text{Mass} / \text{Molecular mass}}{\text{Volume}}$
$= \frac{\text{Mass}}{\text{Volume} \times \text{Molecular mass}}$
$= \frac{\text{Density}}{\text{Molecular mass}}$
Since the density and molecular mass of a pure substance are constant at a given temperature,the concentration of a pure substance is a constant value.
This constant value is incorporated into the equilibrium constant $(K_c)$,and therefore,pure solids and liquids are omitted from the equilibrium constant expression.

(N/A) The equilibrium constant expression $K_{C}$ is defined as the ratio of the product of concentrations of products raised to their stoichiometric coefficients to the product of concentrations of reactants raised to their stoichiometric coefficients.
Given $K_{C} = \frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}$,the species in the numerator ($NH_{3}$ and $O_{2}$) are the products,and the species in the denominator ($NO$ and $H_{2}O$) are the reactants.
The stoichiometric coefficients correspond to the powers in the expression.
Thus,the balanced chemical equation is:
$4NO_{(g)} + 6H_{2}O_{(g)} \leftrightarrow 4NH_{3_{(g)}} + 5O_{2_{(g)}}$

(N/A) The given reaction is:
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}$
Initial concentration: $[ICl] = 0.78 \, M$,$[I_2] = 0 \, M$,$[Cl_2] = 0 \, M$
At equilibrium: $[ICl] = (0.78 - 2x) \, M$,$[I_2] = x \, M$,$[Cl_2] = x \, M$
The equilibrium constant expression is:
$K_c = \frac{[I_2][Cl_2]}{[ICl]^2} = 0.14$
Substituting the values:
$\frac{x \cdot x}{(0.78 - 2x)^2} = 0.14$
Taking the square root on both sides:
$\frac{x}{0.78 - 2x} = \sqrt{0.14} \approx 0.374$
$x = 0.374(0.78 - 2x)$
$x = 0.2917 - 0.748x$
$1.748x = 0.2917$
$x \approx 0.167 \, M$
Therefore,at equilibrium:
$[I_2] = [Cl_2] = 0.167 \, M$
$[ICl] = 0.78 - 2(0.167) = 0.446 \, M$

(A) Let the concentrations of both $PCl_{3}$ and $Cl_{2}$ at equilibrium be $x \, mol \, L^{-1}$.
The given reaction is:
$PCl_{5(g)} \longleftrightarrow PCl_{3(g)} + Cl_{2(g)}$
At equilibrium,the concentrations are:
$[PCl_{5}] = 0.5 \times 10^{-1} \, mol \, L^{-1}$
$[PCl_{3}] = x \, mol \, L^{-1}$
$[Cl_{2}] = x \, mol \, L^{-1}$
The equilibrium constant expression is:
$K_{c} = \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}$
Substituting the given values:
$8.3 \times 10^{-3} = \frac{x \times x}{0.5 \times 10^{-1}}$
$x^{2} = 8.3 \times 10^{-3} \times 0.5 \times 10^{-1}$
$x^{2} = 4.15 \times 10^{-4}$
$x = \sqrt{4.15 \times 10^{-4}} \approx 2.04 \times 10^{-2} \, mol \, L^{-1}$
Thus,$[PCl_{3}] = [Cl_{2}] \approx 0.02 \, mol \, L^{-1}$.

(A) The given reaction is: $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$
At a particular time,the concentrations are: $[N_{2}] = 3.0 \, mol \, L^{-1}$,$[H_{2}] = 2.0 \, mol \, L^{-1}$,$[NH_{3}] = 0.5 \, mol \, L^{-1}$
The reaction quotient $Q_{c}$ is calculated as:
$Q_{c} = \frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}$
Substituting the values:
$Q_{c} = \frac{(0.5)^{2}}{(3.0)(2.0)^{3}} = \frac{0.25}{3.0 \times 8} = \frac{0.25}{24} \approx 0.0104$
Given $K_{c} = 0.061$.
Since $Q_{c} \neq K_{c}$,the reaction is not at equilibrium.
Since $Q_{c} < K_{c}$,the reaction will proceed in the forward direction to reach equilibrium.

The given reaction is:
$3 O_{2(g)} \longleftrightarrow 2 O_{3(g)}$
The expression for the equilibrium constant is:
$K_{c} = \frac{[O_{3}]^{2}}{[O_{2}]^{3}}$
Given values are:
$K_{c} = 2.0 \times 10^{-50}$
$[O_{2}] = 1.6 \times 10^{-2} \, M$
Substituting the values into the expression:
$2.0 \times 10^{-50} = \frac{[O_{3}]^{2}}{(1.6 \times 10^{-2})^{3}}$
$[O_{3}]^{2} = 2.0 \times 10^{-50} \times (1.6 \times 10^{-2})^{3}$
$[O_{3}]^{2} = 2.0 \times 10^{-50} \times 4.096 \times 10^{-6}$
$[O_{3}]^{2} = 8.192 \times 10^{-56}$
Taking the square root:
$[O_{3}] = \sqrt{8.192 \times 10^{-56}}$
$[O_{3}] \approx 2.86 \times 10^{-28} \, M$
Thus,the concentration of $O_{3}$ is $2.86 \times 10^{-28} \, M$.

(D) Let the concentration of methane at equilibrium be $x$.
The equilibrium expression for the reaction $CO_{(g)} + 3H_{2(g)} \longleftrightarrow CH_{4(g)} + H_{2}O_{(g)}$ is:
$K_{c} = \frac{[CH_{4}][H_{2}O]}{[CO][H_{2}]^3}$
Given concentrations at equilibrium in a $1 \, L$ flask:
$[CO] = 0.30 \, M$,$[H_{2}] = 0.10 \, M$,$[H_{2}O] = 0.02 \, M$,$[CH_{4}] = x \, M$
Substituting the values into the $K_{c}$ expression:
$3.90 = \frac{x \times 0.02}{0.30 \times (0.10)^3}$
$3.90 = \frac{0.02x}{0.30 \times 0.001}$
$3.90 = \frac{0.02x}{0.0003}$
$x = \frac{3.90 \times 0.0003}{0.02}$
$x = \frac{0.00117}{0.02} = 0.0585 \, M$
Thus,the concentration of $CH_{4}$ is $5.85 \times 10^{-2} \, M$.

(N/A) For a general reversible reaction $aA + bB \rightleftharpoons cC + dD$,where $A$ and $B$ are reactants and $C$ and $D$ are products,the law of chemical equilibrium states that at a given temperature,the product of the concentrations of the products raised to their respective stoichiometric coefficients,divided by the product of the concentrations of the reactants raised to their respective stoichiometric coefficients,is a constant value known as the equilibrium constant $(K_c)$.
The mathematical expression is given by:
$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$
Here,$[A], [B], [C],$ and $[D]$ represent the molar concentrations of the species at equilibrium. This relationship is also referred to as the law of mass action.

(N/A) The equilibrium constant $(K_c)$ is defined as the ratio of the product of the molar concentrations of the products to the product of the molar concentrations of the reactants,each raised to the power of their stoichiometric coefficients.
$(i)$ For $4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_{2}O_{(g)}$,the expression is: $K_c = \frac{[NO]^4 [H_2O]^6}{[NH_3]^4 [O_2]^5}$
$(ii)$ For $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the expression is: $K_c = \frac{[HI]^2}{[H_2] [I_2]}$
$(iii)$ For $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the expression is: $K_c = \frac{[NH_3]^2}{[N_2] [H_2]^3}$

(N/A) The equilibrium of $HI$ synthesis at a definite temperature is as follows:
$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ $\quad (Eq.-I)$
The equilibrium constant $K_c$ is:
$K_c = \frac{[HI]^2}{[H_2][I_2]} = X$ $\quad (Eq.-II)$
On multiplying the $HI$ synthesis equation by a factor $n$,we get:
$nH_{2(g)} + nI_{2(g)} \rightleftharpoons 2nHI_{(g)}$ $\quad (Eq.-III)$
For this reaction,the new equilibrium constant $K'_c$ is:
$K'_c = \frac{[HI]^{2n}}{[H_2]^n[I_2]^n} = \left( \frac{[HI]^2}{[H_2][I_2]} \right)^n = X^n$ $\quad (Eq.-IV)$
Thus,$K'_c = (K_c)^n$.
It should be noted that because the equilibrium constants $K_c$ and $K'_c$ have different numerical values,it is important to specify the form of the balanced chemical equation when quoting the value of an equilibrium constant.

(N/A) The equilibrium constant expression for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is given by:
$K_c = \frac{[HI]^2}{[H_2][I_2]}$
Substituting the given equilibrium concentrations:
$K_c = \frac{(0.14)^2}{(0.6)(0.8)}$
$K_c = \frac{0.0196}{0.48}$
$K_c \approx 0.04083$ or $4.08 \times 10^{-2}$

(A) The equilibrium constant $K_c$ for the reaction $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$ is given by the expression:
$K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]}$
Given equilibrium concentrations:
$[PCl_3] = 1.59 \ M$
$[Cl_2] = 1.59 \ M$
$[PCl_5] = 1.41 \ M$
Substituting these values into the expression:
$K_c = \frac{1.41}{1.59 \times 1.59} = \frac{1.41}{2.5281} \approx 0.558 \ M^{-1}$

(N/A) The equilibrium constant expression is defined as 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.
From the given expression $K_{c} = \frac{[I_{2}][H_{5}IO_{6}]^{5}}{[IO_{3}^{-}]^{7}[H_{2}O]^{9}[H^{+}]^{7}}$,the products are $I_{2}$ and $H_{5}IO_{6}$ with coefficients $1$ and $5$ respectively.
The reactants are $IO_{3}^{-}$,$H_{2}O$,and $H^{+}$ with coefficients $7$,$9$,and $7$ respectively.
Thus,the balanced chemical equilibrium equation is:
$7IO_{3(aq)}^{-} + 9H_{2}O_{(l)} + 7H_{(aq)}^{+} \rightleftharpoons I_{2(aq)} + 5H_{5}IO_{6(aq)}$

(N/A) The equilibrium constant $K_c$ for the reaction $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2_{(g)}} + H_{2_{(g)}}$ is given by the expression:
$K_c = \frac{[CO_2][H_2]}{[CO][H_2O]}$
Given concentrations are:
$[CO] = 0.18 \ mol \ L^{-1}$
$[H_2O] = 0.0412 \ mol \ L^{-1}$
$[CO_2] = 0.15 \ mol \ L^{-1}$
$[H_2] = 0.2 \ mol \ L^{-1}$
Substituting these values into the expression:
$K_c = \frac{0.15 \times 0.2}{0.18 \times 0.0412}$
$K_c = \frac{0.03}{0.007416}$
$K_c \approx 4.045$