Law of equilibrium and Equilibrium constant Questions in English

(N/A) Given reaction: ${N_2}_{(g)} + 2{O_2}_{(g)} \rightleftharpoons 2{NO_2}_{(g)}$,$K = 100$.
$(1)$ The reaction $2{NO_2}_{(g)} \rightleftharpoons {N_2}_{(g)} + 2{O_2}_{(g)}$ is the reverse of the given reaction.
Therefore,$K_1 = \frac{1}{K} = \frac{1}{100} = 0.01$.
$(2)$ The reaction ${NO_2}_{(g)} \rightleftharpoons \frac{1}{2}{N_2}_{(g)} + {O_2}_{(g)}$ is half of the reverse reaction.
Therefore,$K_2 = (K_1)^{1/2} = (0.01)^{1/2} = 0.1$.

The equilibrium constant expression for the reaction is $K_c = \frac{[CH_3OH]}{[CO][H_2]^2}$.
Given $K_c = 0.5$,$[CO] = 0.18 \ mol \ L^{-1}$,and $[H_2] = 0.22 \ mol \ L^{-1}$.
Substituting the values into the expression: $0.5 = \frac{[CH_3OH]}{(0.18)(0.22)^2}$.
$[CH_3OH] = 0.5 \times 0.18 \times (0.22)^2$.
$[CH_3OH] = 0.5 \times 0.18 \times 0.0484$.
$[CH_3OH] = 0.004356 \ mol \ L^{-1}$ or $4.356 \times 10^{-3} \ mol \ L^{-1}$.

(A) For a reversible reaction,if the reaction is reversed,the new equilibrium constant $K'$ is the reciprocal of the original equilibrium constant $K$.
Given: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$K = 48$.
The reaction $2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$ is the reverse of the given reaction.
Therefore,$K' = \frac{1}{K} = \frac{1}{48} \approx 0.0208$.

(A) The equilibrium constant expression for the reaction $2NO_{2(g)} \rightleftharpoons N_{2}O_{4(g)}$ is given by:
$K_{c} = \frac{[N_{2}O_{4}]}{[NO_{2}]^2}$
Given:
$[N_{2}O_{4}] = 0.145 \ M$
$[NO_{2}] = 0.710 \ M$
Substituting the values:
$K_{c} = \frac{0.145}{(0.710)^2}$
$K_{c} = \frac{0.145}{0.5041}$
$K_{c} \approx 0.2876 \ M^{-1}$

(N/A) The molar concentration of a pure solid or liquid is constant,meaning it is independent of the amount present. For a substance '$X$',the values $[X_{(s)}]$ and $[X_{(l)}]$ remain constant regardless of the quantity taken.
$\therefore \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume}} = \frac{\text{Mass} / \text{Molar mass}}{\text{Volume}} = \frac{\text{Density}}{\text{Molar mass}}$
Since density and molar mass are constant for a pure substance at a given temperature,the concentration is constant.
Because these values are constant,they are incorporated into the equilibrium constant ($K_c$ or $K_p$). Thus,they are omitted from the expression.
Example: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$
The expression is $K_c = [CO_{2}]$.
Here,$[CaCO_{3(s)}]$ and $[CaO_{(s)}]$ are treated as $1$.

(N/A) For equilibrium constant expressions,the concentration of pure solids and pure liquids is taken as unity $(1)$ and is omitted from the expression.
$(a)$ For the reaction $Ni_{(s)} + 4CO_{(g)} \rightleftharpoons Ni(CO)_{4_{(g)}}$:
$K_{c} = \frac{[Ni(CO)_{4}]}{[CO]^{4}}$ and $K_{p} = \frac{p_{Ni(CO)_{4}}}{(p_{CO})^{4}}$
$(b)$ For the reaction $Ag_{2}O_{(s)} + 2HNO_{3_{(aq)}} \rightleftharpoons 2AgNO_{3_{(aq)}} + H_{2}O_{(l)}$:
Since $Ag_{2}O$ is a solid and $H_{2}O$ is a pure liquid,their concentrations are constant.
$K_{c} = \frac{[AgNO_{3}]^{2}}{[HNO_{3}]^{2}}$

(N/A) The equilibrium constant $(K)$ provides valuable information about a chemical reaction:
$1$. Predicting the extent of a reaction: If $K_c > 10^3$,the product predominates; if $K_c < 10^{-3}$,the reactant predominates; if $K_c$ is between $10^{-3}$ and $10^3$,both reactants and products are present in significant amounts.
$2$. Predicting the direction of the reaction: By comparing the reaction quotient $(Q_c)$ with the equilibrium constant $(K_c)$: if $Q_c < K_c$,the reaction proceeds in the forward direction; if $Q_c > K_c$,it proceeds in the reverse direction; if $Q_c = K_c$,the system is at equilibrium.
$3$. Calculating equilibrium concentrations: Given the initial concentrations and the value of $K_c$,the equilibrium concentrations of reactants and products can be calculated.

(N/A) The equilibrium constant $(K_c)$ is used for the following purposes:
$1$. Predicting the direction of the reaction: By calculating the reaction quotient $(Q_c)$,we can compare it with $K_c$. If $Q_c < K_c$,the reaction proceeds in the forward direction. If $Q_c > K_c$,the reaction proceeds in the reverse direction. If $Q_c = K_c$,the reaction is at equilibrium.
$2$. Calculating equilibrium concentrations: If the equilibrium constant and the initial concentrations of the reactants are known,the concentrations of all species at equilibrium can be calculated using the equilibrium expression.

(N/A) The numerical value of the equilibrium constant $(K_{c})$ for a reaction indicates the extent of the reaction,but it does not provide information about the rate of the reaction.
The magnitude of $K_{c}$ or $K_{p}$ is directly proportional to the concentrations of products and inversely proportional to the concentrations of the reactants. $A$ high value of $K$ indicates that the concentration of products is high,while a low value of $K$ indicates that the concentration of products is low.
Value of $K \propto \text{[Products]} \propto \frac{1}{\text{[Reactants]}}$
The following generalizations describe the composition of equilibrium mixtures:
$(a)$ If $K_{c} > 10^{3}$: Products predominate over reactants; i.e.,if $K_{c}$ is very large,the reaction proceeds nearly to completion. Examples:
$(i)$ The reaction $H_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons H_{2}O_{(g)}$ at $500 \ K$ has a very large equilibrium constant,$K_{c} = 2.4 \times 10^{47}$.
$(ii)$ $H_{2(g)} + Cl_{2(g)} \rightleftharpoons 2 HCl_{(g)}$ at $300 \ K$ has a very large $K_{c} = 4.0 \times 10^{31}$.
$(iii)$ $H_{2(g)} + Br_{2(g)} \rightleftharpoons 2 HBr_{(g)}$ at $300 \ K$ has a very large $K_{c} = 5.4 \times 10^{18}$.

(N/A) The equilibrium constants $K_{c}$ and $K_{p}$ help in predicting the direction in which a given reaction will proceed at any stage.
For this,we calculate the reaction quotient $Q$ ($Q_{c}$ for molar concentrations and $Q_{p}$ for partial pressures).
It is defined in the same way as $K_{c}$,except that the concentrations in $Q_{c}$ are not necessarily equilibrium values.
For a general reaction: $aA + bB \rightleftharpoons cC + dD$
$Q_{c} = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$
$(i)$ If $Q_{c} < K_{c}$,the reaction will proceed in the forward direction (towards products).
$(ii)$ If $Q_{c} > K_{c}$,the reaction will proceed in the reverse direction (towards reactants).
$(iii)$ If $Q_{c} = K_{c}$,the reaction mixture is at equilibrium.
Example: Consider the gaseous reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,where $K_{c} = 57.0$ at $700 \ K$. If the concentrations at an arbitrary time $t$ are $[H_{2}]_{t} = 0.1 \ M$,$[I_{2}]_{t} = 0.2 \ M$,and $[HI]_{t} = 0.40 \ M$,then:
$Q_{c} = \frac{[HI]^{2}}{[H_{2}][I_{2}]} = \frac{(0.40)^{2}}{(0.1)(0.2)} = \frac{0.16}{0.02} = 8.0$
Since $Q_{c} < K_{c}$,the reaction will proceed in the forward direction to form more $HI$ until $Q_{c} = K_{c}$.

(N/A) To calculate equilibrium concentrations when initial concentrations are known and the equilibrium constant $(K_c)$ is provided,follow these steps:
Step-$1$: Write the balanced chemical equation for the reaction.
Step-$2$: Create an $ICE$ (Initial,Change,Equilibrium) table under the balanced equation for each substance involved.
$(i)$ List the initial concentration.
$(ii)$ Define the change in concentration as $x$ (in $mol/L$) based on the stoichiometry of the reaction.
$(iii)$ Express the equilibrium concentration as the sum of the initial concentration and the change.
Step-$3$: Substitute these equilibrium concentrations into the equilibrium constant expression $(K_c)$ and solve the resulting algebraic equation for $x$. If a quadratic equation is obtained,select the root that is physically meaningful (i.e.,concentrations must be positive).
Step-$4$: Use the calculated value of $x$ to determine the actual equilibrium concentrations of all reactants and products.
Step-$5$: Verify the results by substituting the calculated equilibrium concentrations back into the $K_c$ expression to ensure they yield the given constant.

(B) The direction of a reaction is predicted by comparing the reaction quotient $Q_c$ with the equilibrium constant $K_c$.
$(i)$ If $Q_c < K_c$,the reaction proceeds in the forward direction.
$(ii)$ If $Q_c > K_c$,the reaction proceeds in the reverse direction.
$(iii)$ If $Q_c = K_c$,the reaction is at equilibrium.
Therefore,the value of $Q$ is used to predict the direction of the reaction.

(D) The equilibrium constant expression for the reaction is $K_c = \frac{[CH_3OH]}{[CO][H_2]^2}$.
Given $K_c = 0.5$,$[CO] = 0.18 \ M$,and $[H_2] = 0.22 \ M$.
Substituting the values: $0.5 = \frac{[CH_3OH]}{(0.18)(0.22)^2}$.
$[CH_3OH] = 0.5 \times 0.18 \times 0.0484$.
$[CH_3OH] = 0.004356 \ M$ or $4.356 \times 10^{-3} \ M$.

(N/A) The reaction quotient $Q_c$ is calculated as: $Q_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{0.042 \times 0.024}{0.005} = 0.2016$.
Since $Q_c = 0.2016$ and $K_c = 0.18$,we observe that $Q_c > K_c$.
Because $Q_c > K_c$,the reaction is not in equilibrium.
To reach equilibrium,the reaction will proceed in the reverse direction (towards the reactants) to decrease the value of $Q_c$ until it equals $K_c$.

(B) The reaction quotient $Q_c$ is calculated as: $Q_c = \frac{[Cu^{2+}]}{[Ag^+]^2} = \frac{1.8 \times 10^{-2}}{(3.0 \times 10^{-9})^2} = \frac{1.8 \times 10^{-2}}{9.0 \times 10^{-18}} = 2.0 \times 10^{15}$.
Since $Q_c = 2.0 \times 10^{15}$ and $K_c = 3.0 \times 10^{14}$,we observe that $Q_c > K_c$.
Because $Q_c > K_c$,the reaction is not in equilibrium.
Since $Q_c > K_c$,the reaction will proceed in the reverse direction to reach equilibrium.

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

(N/A) The direction of a reaction can be predicted by comparing the reaction quotient $Q_{c}$ with the equilibrium constant $K_{c}$:
$(i)$ If $Q_{c} < K_{c}$,the reaction proceeds in the forward direction (towards products).
$(ii)$ If $Q_{c} > K_{c}$,the reaction proceeds in the backward direction (towards reactants).
$(iii)$ If $Q_{c} = K_{c}$,the reaction is at equilibrium,and no net reaction occurs.

(N/A) For the given reaction: $3Fe_{(s)} + 4H_2O_{(g)} \rightleftharpoons Fe_3O_{4(s)} + 4H_{2(g)}$.
$1$. The expression for $K_{C}$ is given by the ratio of the molar concentrations of products to reactants,raised to the power of their stoichiometric coefficients. Pure solids are excluded from the expression as their activity is taken as $1$.
$K_{C} = \frac{[H_2]^4}{[H_2O]^4}$
$2$. The expression for $K_{P}$ is given by the ratio of the partial pressures of gaseous products to gaseous reactants,raised to the power of their stoichiometric coefficients.
$K_{P} = \frac{(P_{H_2})^4}{(P_{H_2O})^4}$

(A) Given reaction: ${N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ with ${K_p = 35}$.
$(i)$ For the reaction ${2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)}$,this is the reverse of the given reaction. Therefore,${K_{p1} = \frac{1}{K_p} = \frac{1}{35} \approx 0.0286}$.
$(ii)$ For the reaction ${\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightleftharpoons NH_3(g)}$,this is the original reaction multiplied by a factor of $\frac{1}{2}$. Therefore,${K_{p2} = (K_p)^{1/2} = \sqrt{35} \approx 5.916}$.

(B) For the first reaction: $A \rightleftharpoons B + C$,$K_{eq}^{(1)} = \frac{[B][C]}{[A]}$ $(1)$
For the second reaction: $B + C \rightleftharpoons P$,$K_{eq}^{(2)} = \frac{[P]}{[B][C]}$ $(2)$
For the overall reaction: $A \rightleftharpoons P$,the equilibrium constant $K_{eq} = \frac{[P]}{[A]}$
Multiplying equation $(1)$ and $(2)$:
$K_{eq}^{(1)} \times K_{eq}^{(2)} = \frac{[B][C]}{[A]} \times \frac{[P]}{[B][C]} = \frac{[P]}{[A]} = K_{eq}$
Therefore,$K_{eq} = K_{eq}^{(1)} \times K_{eq}^{(2)}$.

(B) For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant is $K_{C} = 64$.
When the reaction is reversed,the new equilibrium constant becomes $K'_{C} = \frac{1}{K_{C}} = \frac{1}{64}$.
When the reaction is multiplied by a factor of $\frac{1}{2}$,the new equilibrium constant becomes $K''_{C} = (K'_{C})^{1/2} = \left(\frac{1}{64}\right)^{1/2} = \frac{1}{8}$.

(B) The reaction is $A + B \rightleftharpoons C + D$ with $K_{eq} = 100$.
Initial concentrations are $[A] = 1 \ M, [B] = 1 \ M, [C] = 1 \ M, [D] = 1 \ M$.
Calculate the reaction quotient $Q_{c} = \frac{[C][D]}{[A][B]} = \frac{1 \times 1}{1 \times 1} = 1$.
Since $Q_{c} < K_{eq}$,the reaction proceeds in the forward direction.
Let $x$ be the change in concentration at equilibrium:
$[A] = 1-x, [B] = 1-x, [C] = 1+x, [D] = 1+x$.
$K_{eq} = \frac{(1+x)(1+x)}{(1-x)(1-x)} = 100$.
Taking the square root: $\frac{1+x}{1-x} = 10$.
$1+x = 10 - 10x$ $\Rightarrow 11x = 9$ $\Rightarrow x = \frac{9}{11} \approx 0.818$.
Equilibrium concentration of $D = 1 + x = 1 + \frac{9}{11} = \frac{20}{11} \approx 1.8181 \ M$.
Expressing in terms of $10^{-2} \ M$: $1.8181 \times 10^{0} = 181.81 \times 10^{-2} \ M$.
Rounding to the nearest integer,we get $182 \times 10^{-2} \ M$.

(D) The equilibrium constant expression for the reaction $3 O_{2(g)} \rightleftharpoons 2 O_{3(g)}$ is given by $K_c = \frac{[O_3]^2}{[O_2]^3}$.
Given $K_c = 3.0 \times 10^{-59}$ and $[O_2] = 0.040 \ M = 4 \times 10^{-2} \ M$.
Substituting the values into the expression:
$3.0 \times 10^{-59} = \frac{[O_3]^2}{(4 \times 10^{-2})^3}$
$[O_3]^2 = 3.0 \times 10^{-59} \times (64 \times 10^{-6})$
$[O_3]^2 = 192 \times 10^{-65} = 1.92 \times 10^{-63}$
$[O_3] = \sqrt{1.92 \times 10^{-63}} = \sqrt{19.2 \times 10^{-64}} \approx 4.38 \times 10^{-32} \ M$.

(B) The reaction is: $2 NOCl_{(g)} \rightleftharpoons 2 NO_{(g)} + Cl_{2(g)}$
Initial concentration: $[NOCl] = 2.0 \ M$,$[NO] = 0 \ M$,$[Cl_2] = 0 \ M$
At equilibrium: $[NOCl] = (2 - x) \ M$,$[NO] = x \ M$,$[Cl_2] = \frac{x}{2} \ M$
Given that at equilibrium,$[NO] = 0.4 \ M$,so $x = 0.4 \ M$.
Therefore,$[NOCl]_{eq} = 2 - 0.4 = 1.6 \ M$ and $[Cl_2]_{eq} = \frac{0.4}{2} = 0.2 \ M$.
The equilibrium constant $K_c$ is given by:
$K_c = \frac{[NO]^2 [Cl_2]}{[NOCl]^2} = \frac{(0.4)^2 \times (0.2)}{(1.6)^2}$
$K_c = \frac{0.16 \times 0.2}{2.56} = \frac{0.032}{2.56} = 0.0125$
$K_c = 125 \times 10^{-4}$

(D) The ionization constant for reaction $(a)$ is $K_{a_1} = \frac{[H^{+}][HC_2O_4^-]}{[H_2C_2O_4]}$.
The ionization constant for reaction $(b)$ is $K_{a_2} = \frac{[H^{+}][C_2O_4^{2-}]}{[HC_2O_4^-]}$.
The reaction $(c)$ is the sum of reactions $(a)$ and $(b)$.
For reaction $(c)$,the equilibrium constant $K_{a_3} = \frac{[H^{+}]^2[C_2O_4^{2-}]}{[H_2C_2O_4]}$.
Multiplying $K_{a_1}$ and $K_{a_2}$ gives: $K_{a_1} \times K_{a_2} = \frac{[H^{+}][HC_2O_4^-]}{[H_2C_2O_4]} \times \frac{[H^{+}][C_2O_4^{2-}]}{[HC_2O_4^-]} = \frac{[H^{+}]^2[C_2O_4^{2-}]}{[H_2C_2O_4]} = K_{a_3}$.
Thus,$K_{a_3} = K_{a_1} \times K_{a_2}$.

(A) For the reaction,$2 A \rightleftharpoons B + C$,the reaction quotient is given by $Q_C = \frac{[B][C]}{[A]^2}$.
Given,$K_C = 0.5$.
The reaction proceeds in the backward direction if $Q_C > K_C$.
Let us calculate $Q_C$ for each option:
$(A)$ $Q_C = \frac{(10^{-2})(10^{-2})}{(10^{-3})^2} = \frac{10^{-4}}{10^{-6}} = 100$. Since $100 > 0.5$,the reaction proceeds in the backward direction.
$(B)$ $Q_C = \frac{(10^2)(10^2)}{(10^{-1})^2} = \frac{10^4}{10^{-2}} = 10^6$. Since $10^6 > 0.5$,the reaction proceeds in the backward direction. (Note: Option $A$ is the standard intended answer for this problem type).
$(C)$ $Q_C = \frac{(10^{-2})(10^{-3})}{(10^{-2})^2} = \frac{10^{-5}}{10^{-4}} = 0.1$. Since $0.1 < 0.5$,the reaction proceeds in the forward direction.
$(D)$ $Q_C = \frac{(10^{-3})(10^{-3})}{(10^{-2})^2} = \frac{10^{-6}}{10^{-4}} = 0.01$. Since $0.01 < 0.5$,the reaction proceeds in the forward direction.
Thus,option $A$ is the correct choice.

(D) The given reactions are:
$(i) \, CO_2 \rightleftharpoons CO + \frac{1}{2} O_2, \, K_{C_1} = 9.1 \times 10^{-12}$
$(ii) \, H_2O \rightleftharpoons H_2 + \frac{1}{2} O_2, \, K_{C_2} = 7.1 \times 10^{-12}$
We want to find the equilibrium constant $K$ for the reaction:
$CO_2 + H_2 \rightleftharpoons CO + H_2O \quad (iii)$
To obtain reaction $(iii)$,we reverse reaction $(ii)$ and add it to reaction $(i)$:
Reverse of $(ii): \, H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O, \, K_{C_3} = \frac{1}{K_{C_2}} = \frac{1}{7.1 \times 10^{-12}}$
Adding $(i)$ and reversed $(ii)$:
$CO_2 + H_2 + \frac{1}{2} O_2 \rightleftharpoons CO + \frac{1}{2} O_2 + H_2O$
$CO_2 + H_2 \rightleftharpoons CO + H_2O$
The equilibrium constant $K$ is given by $K = K_{C_1} \times \frac{1}{K_{C_2}} = \frac{9.1 \times 10^{-12}}{7.1 \times 10^{-12}} = 1.28$.

(C) Given the reaction: $N_2 + 3H_2 \rightleftharpoons 2NH_3$ with $K_C = 41 \dots (i)$
For the reaction: $\frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3 \dots (ii)$
Equation $(ii)$ is obtained by dividing equation $(i)$ by $2$.
Therefore,the new equilibrium constant $K_C^{\prime}$ is given by $K_C^{\prime} = (K_C)^{1/2} = \sqrt{K_C}$.
$K_C^{\prime} = \sqrt{41} \approx 6.4$.

(A) To obtain the target reaction $CuCl_4^{2-} + 3Br^{-} \rightleftharpoons CuClBr_3^{2-} + 3Cl^{-}$,we add the first three given reactions:
$(i)$ $CuCl_4^{2-} + Br^{-} \rightleftharpoons CuCl_3Br^{2-} + Cl^{-}$; $K_1$
$(ii)$ $CuCl_3Br^{2-} + Br^{-} \rightleftharpoons CuCl_2Br_2^{2-} + Cl^{-}$; $K_2$
$(iii)$ $CuCl_2Br_2^{2-} + Br^{-} \rightleftharpoons CuClBr_3^{2-} + Cl^{-}$; $K_3$
When reactions are added,their equilibrium constants are multiplied.
Therefore,$K = K_1 \times K_2 \times K_3$.

(D) For the reaction,$3 C_2H_{2(g)} \rightleftharpoons C_6H_{6(g)}$
The expression for the equilibrium constant is $K_c = \frac{[C_6H_6]}{[C_2H_2]^3}$.
Given $K_c = 4 \, L^2 \, mol^{-2}$ and $[C_6H_6] = 0.5 \, mol \, L^{-1}$.
Substituting the values: $4 = \frac{0.5}{[C_2H_2]^3}$.
Rearranging for $[C_2H_2]^3$: $[C_2H_2]^3 = \frac{0.5}{4} = \frac{1}{8} = 0.125$.
Taking the cube root: $[C_2H_2] = \sqrt[3]{0.125} = 0.5 \, mol \, L^{-1}$.

(B) The correct option is $B$.
For a general reversible reaction,$A + B \rightleftharpoons C + D$,the reaction quotient is defined as $Q_C = \frac{[C][D]}{[A][B]}$.
$1$. If $Q_C = K_C$,the reaction is at equilibrium.
$2$. If $Q_C > K_C$,the reaction proceeds in the backward direction (towards reactants).
$3$. If $Q_C < K_C$,the reaction proceeds in the forward direction (towards products).
Therefore,the reaction proceeds in the direction of the products when $Q_C < K_C$.

(A) The chemical equation is: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$
Initial moles: $H_2 = 4.5$,$I_2 = 4.5$,$HI = 0$
At equilibrium,$3 \text{ moles}$ of $HI$ are formed. Since $2 \text{ moles}$ of $HI$ are produced from $1 \text{ mole}$ of $H_2$ and $1 \text{ mole}$ of $I_2$,the amount of $H_2$ and $I_2$ consumed is $3/2 = 1.5 \text{ moles}$.
Equilibrium moles: $H_2 = 4.5 - 1.5 = 3$,$I_2 = 4.5 - 1.5 = 3$,$HI = 3$
Volume of vessel = $10 \text{ L}$.
Equilibrium concentrations: $[H_2] = 3/10$,$[I_2] = 3/10$,$[HI] = 3/10$
$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(3/10)^2}{(3/10)(3/10)} = \frac{9/100}{9/100} = 1$

(C) The balanced chemical equation for the formation of $NH_3$ is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.
The expression for the equilibrium constant $K_c$ is: $K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$.
Substituting the given equilibrium concentrations:
$K_c = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2}) \times (3 \times 10^{-2})^3}$.
$K_c = \frac{2.25 \times 10^{-4}}{(2 \times 10^{-2}) \times (27 \times 10^{-6})}$.
$K_c = \frac{2.25 \times 10^{-4}}{54 \times 10^{-8}}$.
$K_c = \frac{2.25}{54} \times 10^4 = 0.04166 \times 10^4 = 416.66 \approx 417$.

(A) The equilibrium constant $K_{C}$ for a reversible reaction 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.
For the reaction $Fe_{(aq)}^{3+} + SCN_{(aq)}^{-} \rightleftharpoons (FeSCN)_{(aq)}^{2+}$,the expression is:
$K_{C} = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^{-}]}$

(C) The given reactions are:
$1) X \rightleftharpoons Y ; K_1 = 1.0$
$2) Y \rightleftharpoons Z ; K_2 = 2.0$
$3) Z \rightleftharpoons W ; K_3 = 4.0$
To find the equilibrium constant for the overall reaction $X \rightleftharpoons W$,we add the three reactions:
$(X \rightleftharpoons Y) + (Y \rightleftharpoons Z) + (Z \rightleftharpoons W) \implies X \rightleftharpoons W$
When reactions are added,their equilibrium constants are multiplied:
$K_{eq} = K_1 \times K_2 \times K_3$
$K_{eq} = 1.0 \times 2.0 \times 4.0 = 8.0$

(B) The reaction is $2\,A \rightleftharpoons B + C$ with $K_c = 4 \times 10^{-3}$.
At a given time $t$, the reaction quotient $Q_c$ is calculated as:
$Q_c = \frac{[B][C]}{[A]^2} = \frac{(2 \times 10^{-3})(2 \times 10^{-3})}{(2 \times 10^{-3})^2} = 1$.
Comparing $Q_c$ with $K_c$:
$Q_c = 1$ and $K_c = 4 \times 10^{-3}$.
Since $Q_c > K_c$, the reaction will proceed in the backward direction to reach equilibrium.

(C) For the reaction $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$,the equilibrium constant is $K = \frac{[SO_3]^2}{[SO_2]^2 [O_2]}$.
For the reaction $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$,the equilibrium constant $K'$ is given by $K' = \frac{[SO_2] [O_2]^{1/2}}{[SO_3]}$.
Comparing $K$ and $K'$,we can see that $K' = \sqrt{\frac{1}{K}} = \frac{1}{\sqrt{K}}$.

(A) The equilibrium constant expression for the reaction $A_{(g)} + B_{(g)} \rightleftharpoons 2 C_{(g)}$ is given by:
$K_{c} = \frac{[C]^2}{[A][B]}$
Given that $K_{c} = 4$,$[A] = 0.1 \ M$,and $[B] = 0.4 \ M$:
$4 = \frac{[C]^2}{0.1 \times 0.4}$
$4 = \frac{[C]^2}{0.04}$
$[C]^2 = 4 \times 0.04 = 0.16$
$[C] = \sqrt{0.16} = 0.4 \ M$

(A) Given reactions:
$(i) N_2 + 3 H_2 \rightleftharpoons 2 NH_3 ; k_1$
$(ii) N_2 + O_2 \rightleftharpoons 2 NO ; k_2$
$(iii) H_2 + 1/2 O_2 \rightleftharpoons H_2 O ; k_3$
Target reaction: $NH_3 + 5/4 O_2 \rightleftharpoons NO + 3/2 H_2 O$
To obtain the target reaction,we perform the following operations:
$1/2 \times (ii) + 3/2 \times (iii) - 1/2 \times (i)$
$1/2 (N_2 + O_2 \rightleftharpoons 2 NO) \Rightarrow 1/2 N_2 + 1/2 O_2 \rightleftharpoons NO$ $(k_2^{1/2})$
$3/2 (H_2 + 1/2 O_2 \rightleftharpoons H_2 O) \Rightarrow 3/2 H_2 + 3/4 O_2 \rightleftharpoons 3/2 H_2 O$ $(k_3^{3/2})$
$-1/2 (N_2 + 3 H_2 \rightleftharpoons 2 NH_3) \Rightarrow NH_3 \rightleftharpoons 1/2 N_2 + 3/2 H_2$ $(k_1^{-1/2})$
Adding these gives: $NH_3 + 5/4 O_2 \rightleftharpoons NO + 3/2 H_2 O$
The equilibrium constant $K = \frac{k_2^{1/2} \times k_3^{3/2}}{k_1^{1/2}}$.

(C) The equilibrium constant expression for the reaction $2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g)$ is given by $K_C = \frac{[NO_2]^2}{[NO]^2 [O_2]}$.
Given $K_C = 3 \times 10^6$,which is a very large value $(K_C \gg 1)$.
$A$ large value of $K_C$ indicates that the equilibrium position lies far to the right,meaning the concentration of the product $(NO_2)$ is much higher than the concentrations of the reactants ($NO$ and $O_2$) at equilibrium.
Therefore,the concentrations of either $NO$ or $O_2$ (or possibly both) will be much smaller than the concentration of $NO_2$.

(C) Given:
$A \rightleftharpoons B, K_1 = 2.0$
$B \rightleftharpoons C, K_2 = 0.01$
We need the equilibrium constant for $2C \rightleftharpoons 2A$.
First,reverse the reaction $B \rightleftharpoons C$ to get $C \rightleftharpoons B$,the constant becomes $K_3 = \frac{1}{K_2} = \frac{1}{0.01} = 100$.
Next,reverse the reaction $A \rightleftharpoons B$ to get $B \rightleftharpoons A$,the constant becomes $K_4 = \frac{1}{K_1} = \frac{1}{2.0} = 0.5$.
Adding the reactions $C \rightleftharpoons B$ and $B \rightleftharpoons A$ gives $C \rightleftharpoons A$ with $K_{net} = K_3 \times K_4 = 100 \times 0.5 = 50$.
For the reaction $2C \rightleftharpoons 2A$,the equilibrium constant is $K = (K_{net})^2 = (50)^2 = 2500 = 2.5 \times 10^3$.

(B) For the first reaction: $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$; $K_{1} = \frac{[SO_{3}]}{[SO_{2}][O_{2}]^{1/2}}$
For the second reaction: $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$; $K_{2} = \frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}$
Comparing the two expressions,we can see that $K_{2} = \left( \frac{[SO_{2}][O_{2}]^{1/2}}{[SO_{3}]} \right)^{2} = \left( \frac{1}{K_{1}} \right)^{2} = \frac{1}{K_{1}^{2}}$
Therefore,$K_{1}^{2} = \frac{1}{K_{2}}$.

(B) For the given reversible reaction,$\Delta G^{\circ} = -350 \ kJ$.
Since $\Delta G^{\circ} < 0$,the reaction is thermodynamically feasible.
We know the relationship $\Delta G^{\circ} = -RT \ln K$.
Substituting the value,$-350 \times 10^3 = -RT \ln K$,which implies $\ln K > 0$,so $K > 1$.
Therefore,the equilibrium constant is greater than one.
Regarding entropy,the reaction involves the conversion of $1 \ mol$ of solid and $1 \ mol$ of gas into $2 \ mol$ of gas,leading to an increase in disorder,so $\Delta S$ is positive.

(A) The given reactions are:
$I$) $\frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3$ $(K_1)$
$II$) $2 NO \rightleftharpoons N_2 + O_2$ $(K_2)$
$III$) $H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2 O$ $(K_3)$
We want to find the equilibrium constant $K$ for the reaction:
$2 NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2 NO + 3 H_2 O$
To obtain this,we manipulate the given reactions:
$1$. Reverse reaction $(I)$ and multiply by $2$: $2 NH_3 \rightleftharpoons N_2 + 3 H_2$ $(K_{new1} = \frac{1}{K_1^2})$
$2$. Reverse reaction $(II)$: $N_2 + O_2 \rightleftharpoons 2 NO$ $(K_{new2} = \frac{1}{K_2})$
$3$. Multiply reaction $(III)$ by $3$: $3 H_2 + \frac{3}{2} O_2 \rightleftharpoons 3 H_2 O$ $(K_{new3} = K_3^3)$
Adding these three reactions:
$(2 NH_3) + (N_2 + O_2) + (3 H_2 + \frac{3}{2} O_2) \rightleftharpoons (N_2 + 3 H_2) + (2 NO) + (3 H_2 O)$
Canceling common species ($N_2$ and $3 H_2$): $2 NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2 NO + 3 H_2 O$
The equilibrium constant $K$ is the product of the constants of the manipulated reactions:
$K = K_{new1} \times K_{new2} \times K_{new3} = \frac{1}{K_1^2} \times \frac{1}{K_2} \times K_3^3 = \frac{K_3^3}{K_1^2 \times K_2}$

(D) Let the initial concentration of $A$ be $a$ and $B$ be $1.5a$.
Reaction: $A + 2B \rightleftharpoons 2C + D$
Initial: $a, 1.5a, 0, 0$
At equilibrium: $(a-x), (1.5a-2x), 2x, x$
Given that at equilibrium,$[A] = [B]$,so $a-x = 1.5a-2x$.
Solving for $x$: $x = 0.5a$.
Equilibrium concentrations: $[A] = a - 0.5a = 0.5a$,$[B] = 1.5a - 2(0.5a) = 0.5a$,$[C] = 2(0.5a) = a$,$[D] = 0.5a$.
Equilibrium constant $K_c = \frac{[C]^2 [D]}{[A] [B]^2} = \frac{(a)^2 (0.5a)}{(0.5a) (0.5a)^2} = \frac{a^2 \times 0.5a}{0.5a \times 0.25a^2} = \frac{0.5a^3}{0.125a^3} = 4$.

(D) Given $K_c = 99.0$.
Initial moles of $A_2 = 2 \ mol$,Volume = $1 \ L$,so $[A_2]_{initial} = 2 \ mol \ L^{-1}$.
Let $x$ be the concentration of $B_2$ formed at equilibrium.
Reaction: $A_{2(g)} \rightleftharpoons B_{2(g)}$
Initial: $2 \ 0$
Equilibrium: $(2-x) \ x$
$K_c = \frac{[B_2]}{[A_2]} = \frac{x}{2-x} = 99.0$.
$x = 99(2-x) = 198 - 99x$.
$100x = 198 \implies x = 1.98 \ mol \ L^{-1}$.
Concentration of $B_2$ at equilibrium = $1.98 \ mol \ L^{-1}$.
Concentration of $A_2$ at equilibrium = $2 - 1.98 = 0.02 \ mol \ L^{-1}$.
Thus,the concentrations of $A_2$ and $B_2$ are $0.02 \ mol \ L^{-1}$ and $1.98 \ mol \ L^{-1}$ respectively.

(A) For the reaction $a A_{(g)} \rightleftharpoons b B_{(g)}$,the equilibrium constant is $K_c = \frac{[B]^b}{[A]^a}$.
For the reaction $2a A_{(g)} \rightleftharpoons 2b B_{(g)}$,the equilibrium constant is $K_c^{\prime} = \frac{[B]^{2b}}{[A]^{2a}}$.
Comparing the two expressions,we can see that $K_c^{\prime} = \left( \frac{[B]^b}{[A]^a} \right)^2$.
Therefore,$K_c^{\prime} = (K_c)^2$.

(A) The balanced chemical equation for the formation of ammonia is:
$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$
According to the law of chemical equilibrium,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.
Therefore,$K_{C} = \frac{[NH_3]^2}{[N_2][H_2]^3}$.

(B) The chemical equation for the formation of ammonia is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
For a reversible reaction,the direction of the reaction is determined by the relationship between the reaction quotient $(Q)$ and the equilibrium constant $(K_C)$.
If $Q > K_C$,the system is not at equilibrium and the reaction proceeds in the backward direction to form reactants from products.
Therefore,for the backward reaction to occur,the condition is $Q > K_C$.