Kp and Kc Relationship Questions in English

(D) For the reaction $A + B \rightleftharpoons C + D$,the equilibrium constant expression is $K_c = \frac{[C][D]}{[A][B]}$.
Given that at equilibrium,$[C] = 0.8 \ mol/L$ and $[D] = 0.8 \ mol/L$.
Assuming initial concentrations $[A]_0 = 1 \ mol/L$ and $[B]_0 = 1 \ mol/L$,the concentration of $A$ and $B$ at equilibrium will be $[A] = [A]_0 - [C] = 1 - 0.8 = 0.2 \ mol/L$ and $[B] = [B]_0 - [D] = 1 - 0.8 = 0.2 \ mol/L$.
Substituting these values into the expression: $K_c = \frac{0.8 \times 0.8}{0.2 \times 0.2} = \frac{0.64}{0.04} = 16$.

(B) For the reaction: $A + 2B \rightleftharpoons C + 3D$
The expression for the equilibrium constant $(K_p)$ is given by:
$K_p = \frac{P_C \times P_D^3}{P_A \times P_B^2}$
Substituting the given partial pressures:
$K_p = \frac{0.30 \times (0.50)^3}{0.20 \times (0.10)^2}$
$K_p = \frac{0.30 \times 0.125}{0.20 \times 0.01}$
$K_p = \frac{0.0375}{0.002} = 18.75$

(D) The equilibrium constant $K_c$ depends on the units of concentration if the change in the number of moles of gaseous species,$\Delta n_g$,is non-zero.
For option $A$: $\Delta n_g = (0.5 + 0.5) - 1 = 0$.
For option $B$: $\Delta n_g = 0$ (as only aqueous species are involved in the expression).
For option $C$: $\Delta n_g = 0$ (as liquid species are not included in the $K_c$ expression).
For option $D$: $\Delta n_g = (1 + 1) - 1 = 1$.
Since $\Delta n_g \neq 0$ for reaction $D$,the equilibrium constant $K_c$ depends on the units of concentration.

(D) For the reaction: $A + B \rightleftharpoons C + D$
Let the equilibrium concentration of $A$ be $[A] = x$.
Since the concentrations of $A$ and $B$ are equal at equilibrium,$[B] = x$.
Given that the equilibrium concentration of $D$ is twice that of $A$,$[D] = 2x$.
According to the stoichiometry of the reaction,if $1 \ mol$ of $A$ and $1 \ mol$ of $B$ react to form $1 \ mol$ of $C$ and $1 \ mol$ of $D$,then $[C] = [D] = 2x$.
The equilibrium constant $K_c$ is given by:
$K_c = \frac{[C][D]}{[A][B]} = \frac{(2x)(2x)}{(x)(x)} = \frac{4x^2}{x^2} = 4$.

(B) The equilibrium constant $K_c$ for the reaction $C_2H_4(g) + H_2(g) \rightleftharpoons C_2H_6(g)$ is given by the expression:
$K_c = \frac{[C_2H_6]}{[C_2H_4][H_2]}$
Substituting the units of concentration (molarity,$mole \, litre^{-1}$):
$K_c = \frac{mole \, litre^{-1}}{(mole \, litre^{-1})(mole \, litre^{-1})} = \frac{1}{mole \, litre^{-1}} = litre \, mole^{-1}$
Therefore,the correct unit is $litre \, mole^{-1}$.

(B) The reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$.
Initial moles: $H_2 = 0.3$,$I_2 = 0.3$,$HI = 0$.
Let the amount of $H_2$ reacted be $x \ mol$. At equilibrium,moles are: $H_2 = (0.3 - x)$,$I_2 = (0.3 - x)$,$HI = 2x$.
Since the volume is $10 \ L$,the concentrations are $[H_2] = \frac{0.3-x}{10}$,$[I_2] = \frac{0.3-x}{10}$,$[HI] = \frac{2x}{10}$.
$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(\frac{2x}{10})^2}{(\frac{0.3-x}{10})(\frac{0.3-x}{10})} = \frac{4x^2}{(0.3-x)^2} = 64$.
Taking the square root on both sides: $\frac{2x}{0.3-x} = 8$.
$2x = 8(0.3 - x) \implies 2x = 2.4 - 8x \implies 10x = 2.4 \implies x = 0.24 \ mol$.
Amount of unreacted $I_2$ at equilibrium $= 0.3 - x = 0.3 - 0.24 = 0.06 \ mol$.

(C) The relationship between the equilibrium constant $(K_c)$,the forward rate constant $(K_f)$,and the backward rate constant $(K_b)$ is given by:
$K_c = \frac{K_f}{K_b}$
Given:
$K_b = 7.5 \times 10^{-4}$
$K_c = 1.5$
Therefore,the forward rate constant is:
$K_f = K_c \times K_b$
$K_f = 1.5 \times 7.5 \times 10^{-4}$
$K_f = 1.125 \times 10^{-3}$

(B) For the reaction $2X + Y \rightleftharpoons YX_2$,the equilibrium constant expression is given by:
$K_c = \frac{[YX_2]}{[X]^2 [Y]}$
Substituting the given equilibrium concentrations:
$[X] = 4 \ mol/L$,$[Y] = 2 \ mol/L$,$[YX_2] = 2 \ mol/L$
$K_c = \frac{2}{(4)^2 \times 2} = \frac{2}{16 \times 2} = \frac{2}{32} = \frac{1}{16} = 0.0625$.

(D) The decomposition reaction is: $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$
Initial pressure: $P_{NH_3} = 0.50 \ atm$,$P_{H_2S} = 0 \ atm$
At equilibrium,let the increase in pressure due to decomposition be $x \ atm$ for each gas.
$P_{NH_3} = 0.50 + x$
$P_{H_2S} = x$
Total pressure $P_T = P_{NH_3} + P_{H_2S} = (0.50 + x) + x = 0.84 \ atm$
$0.50 + 2x = 0.84$ $\Rightarrow 2x = 0.34$ $\Rightarrow x = 0.17 \ atm$
Equilibrium pressures: $P_{NH_3} = 0.50 + 0.17 = 0.67 \ atm$ and $P_{H_2S} = 0.17 \ atm$
$K_p = P_{NH_3} \times P_{H_2S} = 0.67 \times 0.17 = 0.1139 \approx 0.11$

(B) The chemical equation is: $CO + Cl_2 \rightleftharpoons COCl_2$
First,calculate the molar concentrations at equilibrium:
$[CO] = \frac{0.1 \ mol}{0.5 \ L} = 0.2 \ M$
$[Cl_2] = \frac{0.1 \ mol}{0.5 \ L} = 0.2 \ M$
$[COCl_2] = \frac{0.2 \ mol}{0.5 \ L} = 0.4 \ M$
The expression for the equilibrium constant $K_c$ is:
$K_c = \frac{[COCl_2]}{[CO][Cl_2]}$
Substituting the values:
$K_c = \frac{0.4}{0.2 \times 0.2} = \frac{0.4}{0.04} = 10$

(A) The equilibrium constant $K_p$ for the reaction $2CO_2 \rightleftharpoons 2CO + O_2$ is given by the expression:
$K_p = \frac{(P_{CO})^2 \cdot (P_{O_2})}{(P_{CO_2})^2}$
Substituting the given equilibrium pressures:
$K_p = \frac{(0.4)^2 \cdot (0.2)}{(0.6)^2}$
$K_p = \frac{0.16 \cdot 0.2}{0.36} = \frac{0.032}{0.36} \approx 0.0888$
Rounding to three decimal places,we get $0.089$.

(D) The equilibrium constant $(K_c)$ is defined as the ratio of the rate constant of the forward reaction $(K_f)$ to the rate constant of the backward reaction $(K_b)$.
Given: $K_f = 1.1 \times 10^{-2} \text{ min}^{-1}$ and $K_b = 1.5 \times 10^{-3} \text{ min}^{-1}$.
$K_c = \frac{K_f}{K_b} = \frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}} = \frac{11 \times 10^{-3}}{1.5 \times 10^{-3}} = \frac{11}{1.5} = 7.33$.
Therefore,the correct option is $(D)$.

(D) The reaction is $A_{(g)} + B_{(g)} \rightleftharpoons 2C_{(g)}$.
Initial moles: $A = 3$,$B = 1$,$C = 0$.
At equilibrium,$1.5 \ mol$ of $C$ are formed. Since the stoichiometry of $C$ is $2$,the moles of $A$ and $B$ reacted are $1.5 / 2 = 0.75 \ mol$ each.
Equilibrium moles: $A = 3 - 0.75 = 2.25 \ mol$,$B = 1 - 0.75 = 0.25 \ mol$,$C = 1.5 \ mol$.
Since the volume is $1 \ L$,the concentrations are equal to the number of moles.
$K_c = \frac{[C]^2}{[A][B]} = \frac{(1.5)^2}{(2.25)(0.25)} = \frac{2.25}{0.5625} = 4$.

(B) The chemical equation for the reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$.
The expression for the equilibrium constant $K_c$ is given by $K_c = \frac{[HI]^2}{[H_2][I_2]}$.
Given values are $[H_2] = 8.0 \ mol \ L^{-1}$,$[I_2] = 3.0 \ mol \ L^{-1}$,and $[HI] = 28.0 \ mol \ L^{-1}$.
Substituting these values into the expression:
$K_c = \frac{(28.0)^2}{8.0 \times 3.0} = \frac{784}{24} = 32.66$.

(B) The relationship between the equilibrium constant $(K_c)$,forward rate constant $(K_f)$,and backward rate constant $(K_b)$ is given by the expression: $K_c = \frac{K_f}{K_b}$.
Given that $K_c = 100$ and $K_f = 10^5$,we can rearrange the formula to solve for $K_b$:
$K_b = \frac{K_f}{K_c} = \frac{10^5}{100} = \frac{10^5}{10^2} = 10^3$.
Therefore,the rate constant for the backward reaction is $10^3$.

(C) $1$. Initial moles of $N_2O_4 = \frac{9.2 \ g}{92 \ g \ mol^{-1}} = 0.1 \ mol$.
$2$. Since the volume of the vessel is $1 \ L$,the initial concentration $[N_2O_4]_0 = 0.1 \ mol \ L^{-1}$.
$3$. At equilibrium,$50\%$ of $N_2O_4$ is dissociated,so the amount dissociated is $0.1 \times 0.5 = 0.05 \ mol$.
$4$. The equilibrium concentrations are: $[N_2O_4]_{eq} = 0.1 - 0.05 = 0.05 \ mol \ L^{-1}$ and $[NO_2]_{eq} = 2 \times 0.05 = 0.1 \ mol \ L^{-1}$.
$5$. The equilibrium constant $K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.1)^2}{0.05} = \frac{0.01}{0.05} = 0.2 \ mol \ L^{-1}$.

(B) For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the equilibrium constant is $K_c = \frac{[NO]^2}{[N_2][O_2]} = 4 \times 10^{-4}$.
For the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}$,the equilibrium constant $K'_c$ is given by $K'_c = \frac{[N_2]^{1/2}[O_2]^{1/2}}{[NO]}$.
Comparing the two expressions,we see that $K'_c = \frac{1}{\sqrt{K_c}}$.
Substituting the value of $K_c$,we get $K'_c = \frac{1}{\sqrt{4 \times 10^{-4}}} = \frac{1}{2 \times 10^{-2}} = \frac{100}{2} = 50$.

(D) The reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$.
Initial moles: $H_2 = 0.4 \ mol$,$I_2 = 0.4 \ mol$,$HI = 0 \ mol$.
At equilibrium,$0.5 \ mol$ of $HI$ is formed. Since $2 \ mol$ of $HI$ are produced from $1 \ mol$ of $H_2$ and $1 \ mol$ of $I_2$,the moles of $H_2$ and $I_2$ consumed are $0.5/2 = 0.25 \ mol$ each.
Equilibrium moles: $H_2 = 0.4 - 0.25 = 0.15 \ mol$,$I_2 = 0.4 - 0.25 = 0.15 \ mol$,$HI = 0.5 \ mol$.
Equilibrium concentrations in $2 \ L$ volume: $[H_2] = 0.15/2 \ M$,$[I_2] = 0.15/2 \ M$,$[HI] = 0.5/2 \ M$.
$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.5/2)^2}{(0.15/2) \times (0.15/2)} = \frac{0.5^2}{0.15 \times 0.15} = \frac{0.25}{0.0225} \approx 11.11$.

(A) The balanced chemical equation is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Initial moles: $N_2 = 2 \ mol$,$H_2 = 6 \ mol$,$NH_3 = 0 \ mol$
Given that $50\%$ of $N_2$ reacts,the amount of $N_2$ reacted is $2 \times 0.5 = 1 \ mol$. Thus,$x = 1$.
At equilibrium:
$[N_2] = (2 - 1) = 1 \ mol/L$
$[H_2] = (6 - 3 \times 1) = 3 \ mol/L$
$[NH_3] = (2 \times 1) = 2 \ mol/L$
$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(2)^2}{(1)(3)^3} = \frac{4}{27}$

(A) The balanced chemical equation is: $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$
Initial concentrations: $[H_2] = 1 \ M$,$[CO_2] = 1 \ M$,$[CO] = 0 \ M$,$[H_2O] = 0 \ M$.
At equilibrium,if $x \ mol/L$ of $H_2$ is consumed:
$[H_2] = (1 - x) \ M$
$[CO_2] = (1 - x) \ M$
$[CO] = x \ M$
$[H_2O] = x \ M$
The equilibrium constant $K_c$ is given by:
$K_c = \frac{[CO][H_2O]}{[H_2][CO_2]} = \frac{(x)(x)}{(1 - x)(1 - x)} = \frac{x^2}{(1 - x)^2}$

(A) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.
For $K_p$ to be equal to $K_c$,the term $(RT)^{\Delta n_g}$ must be equal to $1$,which occurs when $\Delta n_g = 0$.
$\Delta n_g$ is defined as the difference between the sum of stoichiometric coefficients of gaseous products and gaseous reactants: $\Delta n_g = \sum n_p - \sum n_r$.
For option $A$: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$,$\Delta n_g = 2 - (1 + 1) = 0$.
Since $\Delta n_g = 0$,$K_p = K_c$ for this reaction.

(D) For the first equilibrium: $NO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons NO_{2(g)}$,the equilibrium constant is $K_1 = \frac{[NO_2]}{[NO][O_2]^{1/2}}$.
For the second equilibrium: $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,the equilibrium constant is $K_2 = \frac{[NO]^2[O_2]}{[NO_2]^2}$.
Observe that the second reaction is the reverse of the first reaction multiplied by $2$.
If a reaction is reversed,the equilibrium constant becomes $1/K$. If a reaction is multiplied by a factor $n$,the equilibrium constant becomes $K^n$.
Here,the first reaction is reversed and multiplied by $2$,so $K_2 = (1/K_1)^2 = 1/K_1^2$.

(A) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n_g}$.
For the reaction $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$,the change in the number of gaseous moles is $\Delta n_g = 1 - (1 + 1) = -1$.
The temperature $T = 250 \ ^\circ C = 250 + 273 = 523 \ K$.
Given $K_c = 26$ and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $K_p = 26 \times (0.0821 \times 523)^{-1} = 26 / 42.9383 \approx 0.6055$.
Rounding to two decimal places,we get $K_p = 0.61$.

(B) For a general gaseous reaction at equilibrium: $aA_{(g)} + bB_{(g)} \rightleftharpoons cC_{(g)} + dD_{(g)}$
$K_c$ is defined as the ratio of the product of concentrations of products to that of reactants,each raised to the power of their stoichiometric coefficients: $K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \ldots (1)$
$K_p$ is defined using partial pressures: $K_p = \frac{p_C^c p_D^d}{p_A^a p_B^b} \ldots (2)$
From the ideal gas equation,$PV = nRT$,we have $P = \frac{n}{V} RT = [Concentration]RT$.
Substituting $p_i = [i]RT$ into equation $(2)$:
$K_p = \frac{([C]RT)^c ([D]RT)^d}{([A]RT)^a ([B]RT)^b} = \frac{[C]^c [D]^d}{[A]^a [B]^b} (RT)^{(c+d)-(a+b)}$
Since $\Delta n = (c+d) - (a+b)$,we get $K_p = K_c (RT)^{\Delta n}$.

(C) The relationship between $K_p$ and $K_c$ is given by the formula $K_p = K_c(RT)^{\Delta n_g}$.
For the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$,the change in the number of moles of gaseous products and reactants is $\Delta n_g = n_p - n_r = 2 - (1 + 3) = -2$.
Substituting this into the formula,we get $K_p = K_c(RT)^{-2}$.

(A) For the reaction,$CaCO_{3_{(s)}} \rightleftharpoons CaO_{(s)} + CO_{2_{(g)}}$,the equilibrium constant $K_p$ is given by $K_p = pCO_2 / p^{\circ}$.
According to the van't Hoff equation,$\ln K_p = -\frac{\Delta H_r^{\circ}}{RT} + C$.
Converting to base $10$ logarithm,we get $\log_{10} K_p = -\frac{\Delta H_r^{\circ}}{2.303 RT} + \text{constant}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log_{10} (pCO_2 / p^{\circ})$ and $x = 1 / T$,the slope $m = -\frac{\Delta H_r^{\circ}}{2.303 R}$.
Thus,$\Delta H_r^{\circ}$ can be determined from the slope of the plot of $\log_{10} (pCO_2 / p^{\circ})$ versus $1 / T$.

(B) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n}$.
For $K_p < K_c$,the value of $\Delta n$ must be negative.
$\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}$.
For option $B$: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$,$\Delta n = 2 - (1 + 3) = -2$.
Since $\Delta n = -2$ is negative,$K_p < K_c$.

(C) For the first equilibrium: $SO_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}$,the equilibrium constant is $K_1 = \frac{[SO_3]}{[SO_2][O_2]^{1/2}}$.
For the second equilibrium: $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$,the equilibrium constant is $K_2 = \frac{[SO_2]^2 [O_2]}{[SO_3]^2}$.
Comparing the two expressions,we see that $K_2 = \left( \frac{1}{K_1} \right)^2 = \frac{1}{K_1^2}$.
Therefore,the correct relationship is $K_2 = \frac{1}{K_1^2}$.

(C) The relationship between $K_p$ and $K_c$ is given by the equation $K_p = K_c(RT)^{\Delta n}$,where $\Delta n$ is the change in the number of moles of gaseous products and reactants.
For $K_p < K_c$,the value of $\Delta n$ must be negative.
$(A)$ $N_2O_4 \rightleftharpoons 2NO_2$: $\Delta n = 2 - 1 = 1$
$(B)$ $2HI \rightleftharpoons H_2 + I_2$: $\Delta n = (1 + 1) - 2 = 0$
$(C)$ $2SO_2 + O_2 \rightleftharpoons 2SO_3$: $\Delta n = 2 - (2 + 1) = -1$
$(D)$ $N_2 + O_2 \rightleftharpoons 2NO$: $\Delta n = 2 - (1 + 1) = 0$
Since $\Delta n = -1$ for reaction $(C)$,$K_p = K_c(RT)^{-1} = K_c / RT$,which means $K_p < K_c$.

(C) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n}$.
For $K_p = K_c$,the value of $\Delta n$ must be $0$.
$\Delta n$ is defined as the difference between the sum of the stoichiometric coefficients of gaseous products and gaseous reactants.
For option $(C)$: $H_{2_{(g)}} + Cl_{2_{(g)}} \rightleftharpoons 2HCl_{(g)}$,$\Delta n = 2 - (1 + 1) = 0$.
Therefore,for this reaction,$K_p = K_c$.

(C) For the given reaction: ${H_2}_{(g)} + {I_2}_{(g)} \rightleftharpoons 2HI_{(g)}$
The relationship between ${K_p}$ and ${K_c}$ is given by the formula: ${K_p} = {K_c}(RT)^{\Delta n}$
Here,$\Delta n$ is the change in the number of moles of gaseous products and reactants:
$\Delta n = (n_{products}) - (n_{reactants}) = 2 - (1 + 1) = 0$
Since $\Delta n = 0$,the expression becomes: ${K_p} = {K_c}(RT)^0 = {K_c} \times 1 = {K_c}$
Given that ${K_c} = 50$,therefore ${K_p} = 50$.

(D) The relationship between ${K_p}$ and ${K_c}$ is given by the equation: ${K_p} = {K_c}{(RT)}^{\Delta n}$.
For ${K_p} > {K_c}$,the value of $\Delta n$ must be positive $(\Delta n > 0)$.
$\Delta n$ is calculated as the difference between the number of moles of gaseous products and gaseous reactants.
$(A)$ ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$: $\Delta n = 2 - (1 + 3) = -2$
$(B)$ ${H_2} + {I_2} \rightleftharpoons 2HI$: $\Delta n = 2 - (1 + 1) = 0$
$(C)$ $PC{l_3} + C{l_2} \rightleftharpoons PC{l_5}$: $\Delta n = 1 - (1 + 1) = -1$
$(D)$ $2S{O_3} \rightleftharpoons {O_2} + 2S{O_2}$: $\Delta n = (1 + 2) - 2 = +1$
Since $\Delta n = +1$ for reaction $(D)$,${K_p} > {K_c}$ holds true for this reaction.

(C) The relationship between $K_p$ and $K_c$ is given by the formula $K_p = K_c(RT)^{\Delta n}$.
For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta n = (1 + 1) - 1 = 1$.
Substituting $\Delta n = 1$ into the formula,we get $K_p = K_c(RT)^1 = K_c(RT)$.

(C) The relationship between $K_p$ and $K_c$ is given by the equation $K_p = K_c(RT)^{\Delta n}$.
For the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$,the change in the number of moles of gaseous products and reactants is $\Delta n = n_p - n_r = 2 - (1 + 1) = 0$.
Substituting $\Delta n = 0$ into the equation,we get $K_p = K_c(RT)^0 = K_c \times 1 = K_c$.
Therefore,$K_p = K_c$.

(A) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n}$.
For the reaction $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,the change in the number of moles of gas is $\Delta n = (2 + 1) - 2 = 1$.
Since $\Delta n = 1$,the equation becomes $K_p = K_c(RT)^1$.
Because $R$ and $T$ are positive values,$(RT)^1 > 1$,which implies $K_p > K_c$.

(D) The given reaction is $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$.
The change in the number of moles of gaseous products and reactants is $\Delta n = (2 + 1) - 2 = 1$.
The relationship between $K_P$ and $K_C$ is given by $K_P = K_C(RT)^{\Delta n}$.
Given $K_C = 3 \times 10^{-6}$,$R = 0.0821 \ L\ atm\ K^{-1}\ mol^{-1}$,and $T = 427 + 273 = 700 \ K$.
Substituting the values: $K_P = 3 \times 10^{-6} \times (0.0821 \times 700)^1$.
$K_P = 3 \times 10^{-6} \times 57.47 = 172.41 \times 10^{-6} = 1.7241 \times 10^{-4}$.
Rounding to the nearest value,$K_P \approx 1.75 \times 10^{-4}$.

(B) The relationship between $K_p$ and $K_c$ is given by the equation $K_p = K_c(RT)^{\Delta n_g}$,where $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
For $K_p = K_c$,the exponent $\Delta n_g$ must be equal to $0$.
Calculating $\Delta n_g$ for each reaction:
$A$: $\Delta n_g = 2 - (1 + 3) = -2$
$B$: $\Delta n_g = 2 - (1 + 1) = 0$
$C$: $\Delta n_g = (1 + 1) - 1 = 1$
$D$: $\Delta n_g = (2 + 1) - 2 = 1$
Since $\Delta n_g = 0$ for reaction $B$,$K_p = K_c$ for this reaction.

(B) For the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$,the equilibrium constant is $K = \frac{[NH_3]^2}{[N_2][H_2]^3}$.
For the reaction $\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightleftharpoons NH_3$,the equilibrium constant is $K' = \frac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}$.
Comparing the two expressions,we see that $K' = \sqrt{\frac{[NH_3]^2}{[N_2][H_2]^3}} = \sqrt{K}$.
Therefore,$K' = \sqrt{K}$.

(C) Given reaction $(i)$: $2NO_2 \rightleftharpoons 2NO + O_2$ with equilibrium constant $K = \frac{[NO]^2 [O_2]}{[NO_2]^2} = 1.6 \times 10^{-12}$.
Target reaction $(ii)$: $NO + \frac{1}{2}O_2 \rightleftharpoons NO_2$ with equilibrium constant $K' = \frac{[NO_2]}{[NO] [O_2]^{1/2}}$.
Comparing the two reactions,we observe that reaction $(ii)$ is the reverse of reaction $(i)$ multiplied by a factor of $\frac{1}{2}$.
Mathematically,$K' = (\frac{1}{K})^{1/2} = \frac{1}{\sqrt{K}}$.
Therefore,the correct option is $C$.

(B) The relationship between $K_p$ and $K_c$ is given by the equation $K_p = K_c(RT)^{\Delta n}$.
For the reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta n = (2 + 1) - 2 = 1$.
Substituting $\Delta n = 1$ into the equation,we get $K_p = K_c(RT)^1$,which implies $K_c = K_p / (RT)$.
Since the temperature $T = 106.5 + 273.15 = 379.65 K$ is greater than $1$,the value of $(RT)$ is much greater than $1$.
Therefore,$K_c = K_p / (RT)$ will be less than $K_p$.
Thus,$K_c < 1.2 \times 10^{-2}$.

(D) For the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$,the equilibrium constant is $K = \frac{[NH_3]^2}{[N_2][H_2]^3}$.
For the reaction $NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2$,the equilibrium constant $K'$ is given by $K' = \frac{[N_2]^{1/2}[H_2]^{3/2}}{[NH_3]}$.
Comparing the two expressions,we see that $K' = \sqrt{\frac{1}{K}} = \frac{1}{\sqrt{K}}$.

(A) The balanced chemical equation is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.
The change in the number of moles of gaseous products and reactants is $\Delta n = 2 - (1 + 3) = -2$.
The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n}$.
Given $K_c = 6 \times 10^{-2}$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 500 + 273 = 773 \ K$.
Substituting the values: $K_p = 6 \times 10^{-2} \times (0.0821 \times 773)^{-2}$.
$K_p = \frac{6 \times 10^{-2}}{(63.4633)^2} \approx \frac{0.06}{4027.59} \approx 1.49 \times 10^{-5} \approx 1.5 \times 10^{-5}$.

(D) The given reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
The change in the number of moles of gaseous products and reactants is $\Delta n = 2 - (1 + 3) = -2$.
The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n}$.
Substituting the values,$K_p = K_c(RT)^{-2}$.
Therefore,$K_c = \frac{K_p}{(RT)^{-2}}$.
Given $T = 500 \ ^oC = 500 + 273 = 773 \ K$ and $R = 0.082 \ \text{L atm K}^{-1} \text{mol}^{-1}$.
Thus,$K_c = \frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}$.

(B) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.
For $K_p < K_c$,the value of $\Delta n_g$ must be negative $(\Delta n_g < 0)$.
$\Delta n_g$ is calculated as (sum of stoichiometric coefficients of gaseous products) - (sum of stoichiometric coefficients of gaseous reactants).
For option $A$: $\Delta n_g = 2 - (1 + 1) = 0$,so $K_p = K_c$.
For option $B$: $\Delta n_g = 2 - (2 + 1) = -1$,so $K_p = K_c(RT)^{-1} = K_c/RT$,which means $K_p < K_c$.
For option $C$: $\Delta n_g = 2 - (1 + 1) = 0$,so $K_p = K_c$.
For option $D$: $\Delta n_g = (1 + 1) - 1 = 1$,so $K_p = K_c(RT)^1$,which means $K_p > K_c$.
Therefore,the correct option is $B$.

(D) The formation reaction of $NH_3$ is: $\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightleftharpoons NH_3(g)$ with equilibrium constant ${K_c}$.
The dissociation reaction of $NH_3$ is the reverse of the formation reaction: $NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)$.
When a chemical reaction is reversed,the new equilibrium constant is the reciprocal of the original equilibrium constant.
Therefore,the dissociation constant is $1/{K_c}$.

(A) The equilibrium expression is $K_p = p_{NH_3} \times p_{H_2S}$.
Let $p$ be the partial pressure of $H_2S$ produced at equilibrium.
Then,$p_{H_2S} = p$ and $p_{NH_3} = 0.5 + p$.
Substituting into the $K_p$ expression: $0.11 = (0.5 + p)(p)$.
This simplifies to the quadratic equation: $p^2 + 0.5p - 0.11 = 0$.
Using the quadratic formula $p = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$:
$p = \frac{-0.5 + \sqrt{0.5^2 - 4(1)(-0.11)}}{2(1)} = \frac{-0.5 + \sqrt{0.25 + 0.44}}{2} = \frac{-0.5 + \sqrt{0.69}}{2} \approx \frac{-0.5 + 0.83}{2} = 0.165 \ atm$.
Thus,$p_{H_2S} = 0.165 \ atm$ and $p_{NH_3} = 0.5 + 0.165 = 0.665 \ atm$.

(A) The change in the number of moles of gaseous species,$\Delta n_g$,is calculated as the difference between the sum of the stoichiometric coefficients of gaseous products and the sum of the stoichiometric coefficients of gaseous reactants.
For the reaction: $C_{12}H_{22}O_{11(s)} + 12O_{2(g)} \rightleftharpoons 12CO_{2(g)} + 11H_2O_{(l)}$
Number of moles of gaseous products = $12$ (from $CO_2$)
Number of moles of gaseous reactants = $12$ (from $O_2$)
$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$
$\Delta n_g = 12 - 12 = 0$

(A) For the given heterogeneous equilibrium reaction: $MgCO_{3(s)} \rightleftharpoons MgO_{(s)} + CO_{2(g)}$
The equilibrium constant $K_p$ is defined as the product of the partial pressures of the gaseous products divided by the partial pressures of the gaseous reactants,each raised to the power of their stoichiometric coefficients.
Pure solids and liquids are assigned an activity of $1$ and are not included in the expression for $K_p$.
Therefore,$K_p = P_{CO_2}$.