Kp and Kc Relationship Questions in English

(D) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{\ln(2)}{k}$.
Since the half-life is independent of the concentration,both forward and reverse reactions are first-order reactions.
For the forward reaction: $k_f = \frac{\ln(2)}{t_{1/2, f}} = \frac{\ln(2)}{400}$.
For the reverse reaction: $k_b = \frac{\ln(2)}{t_{1/2, b}} = \frac{\ln(2)}{100}$.
The equilibrium constant $K_{eq}$ is defined as the ratio of the rate constant of the forward reaction to the rate constant of the reverse reaction: $K_{eq} = \frac{k_f}{k_b}$.
Substituting the values: $K_{eq} = \frac{\ln(2) / 400}{\ln(2) / 100} = \frac{100}{400} = 0.25$.

(B) For the reaction $2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2(g)}$,let the initial pressure of $NOBr$ be $P_0$ and at equilibrium,the pressure of $Br_2$ be $x = \frac{P}{9}$.
From the stoichiometry,the pressure of $NO$ is $2x = \frac{2P}{9}$ and the pressure of $NOBr$ is $P_0 - 2x$.
The total pressure $P = P_{NOBr} + P_{NO} + P_{Br_2} = (P_0 - 2x) + 2x + x = P_0 + x$.
Thus,$P_0 = P - x = P - \frac{P}{9} = \frac{8P}{9}$.
Now,$P_{NOBr} = P_0 - 2x = \frac{8P}{9} - \frac{2P}{9} = \frac{6P}{9} = \frac{2P}{3}$.
The equilibrium constant $K_p$ is given by $K_p = \frac{(P_{NO})^2 \times P_{Br_2}}{(P_{NOBr})^2} = \frac{(\frac{2P}{9})^2 \times \frac{P}{9}}{(\frac{2P}{3})^2} = \frac{\frac{4P^2}{81} \times \frac{P}{9}}{\frac{4P^2}{9}} = \frac{P}{81}$.
Therefore,$\frac{P}{K_p} = \frac{P}{P/81} = 81$.

(B) The dissociation reaction is: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$.
Initially,we have $1 \ mol$ of $PCl_5$ in $1 \ L$ volume.
At equilibrium,$20\%$ of $PCl_5$ is not dissociated,meaning $80\%$ is dissociated.
Amount of $PCl_5$ dissociated = $1 \times 0.8 = 0.8 \ mol$.
Amount of $PCl_5$ remaining at equilibrium = $1 - 0.8 = 0.2 \ mol$.
Amount of $PCl_3$ formed = $0.8 \ mol$.
Amount of $Cl_2$ formed = $0.8 \ mol$.
Since the volume is $1 \ L$,the concentrations are $[PCl_5] = 0.2 \ M$,$[PCl_3] = 0.8 \ M$,and $[Cl_2] = 0.8 \ M$.
The equilibrium constant $K_C$ is given by: $K_C = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{0.8 \times 0.8}{0.2} = \frac{0.64}{0.2} = 3.2$.

(B) For the reaction: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
At $t=0$,moles: $1$ for $N_2O_4$ and $0$ for $NO_2$.
At equilibrium,moles: $(1-\alpha)$ for $N_2O_4$ and $2\alpha$ for $NO_2$. Total moles $n_f = 1+\alpha$.
Mole fractions: $x_{N_2O_4} = \frac{1-\alpha}{1+\alpha}$ and $x_{NO_2} = \frac{2\alpha}{1+\alpha}$.
Partial pressures: $P_{N_2O_4} = \left(\frac{1-\alpha}{1+\alpha}\right)P$ and $P_{NO_2} = \left(\frac{2\alpha}{1+\alpha}\right)P$.
$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(\frac{2\alpha}{1+\alpha})^2 P^2}{(\frac{1-\alpha}{1+\alpha})P} = \frac{4\alpha^2 P}{(1+\alpha)(1-\alpha)} = \frac{4\alpha^2 P}{1-\alpha^2}$.
Rearranging for $\alpha^2$: $K_p(1-\alpha^2) = 4\alpha^2 P \implies K_p = \alpha^2(4P + K_p)$.
Thus,$\alpha^2 = \frac{K_p}{4P + K_p}$,which gives $\alpha = \sqrt{\frac{K_p}{4P + K_p}}$.

(A) The reaction is $CO + Cl_2 \rightleftharpoons COCl_2$.
Initially,$CO = 2\,mol$,$Cl_2 = 3\,mol$,$COCl_2 = 0\,mol$.
At equilibrium,$CO = 1\,mol$.
Change in $CO = 2 - 1 = 1\,mol$ reacted.
According to stoichiometry,$1\,mol$ of $CO$ reacts with $1\,mol$ of $Cl_2$ to produce $1\,mol$ of $COCl_2$.
At equilibrium: $[CO] = \frac{1\,mol}{5\,L} = 0.2\,M$,$[Cl_2] = \frac{3-1\,mol}{5\,L} = \frac{2\,mol}{5\,L} = 0.4\,M$,$[COCl_2] = \frac{1\,mol}{5\,L} = 0.2\,M$.
$K_c = \frac{[COCl_2]}{[CO][Cl_2]} = \frac{0.2}{0.2 \times 0.4} = \frac{1}{0.4} = 2.5$.

(A) The dissociation reaction of $HI$ is:
$2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$
Let the initial moles of $HI$ be $1$.
Degree of dissociation $\alpha = 50\% = 0.5$.
At equilibrium:
Moles of $HI = 1 - \alpha = 1 - 0.5 = 0.5$
Moles of $H_2 = \frac{\alpha}{2} = \frac{0.5}{2} = 0.25$
Moles of $I_2 = \frac{\alpha}{2} = \frac{0.5}{2} = 0.25$
The equilibrium constant $K_c$ is given by:
$K_c = \frac{[H_2][I_2]}{[HI]^2}$
Since the number of moles on both sides is the same,volume $V$ cancels out.
$K_c = \frac{0.25 \times 0.25}{(0.5)^2} = \frac{0.0625}{0.25} = 0.25$

(B) The decomposition reaction is $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$.
The equilibrium constant expression is $K_p = (P_{NH_3})^2 \times (P_{CO_2})$.
From the stoichiometry,$NH_3$ and $CO_2$ are produced in a $2:1$ molar ratio. If $P$ is the total pressure at equilibrium,then $P_{NH_3} = \frac{2P}{3}$ and $P_{CO_2} = \frac{P}{3}$.
Substituting these into the $K_p$ expression: $K_p = (\frac{2P}{3})^2 \times (\frac{P}{3}) = \frac{4P^3}{27}$.
Given $K_p = 2.9 \times 10^{-5}$,we have $2.9 \times 10^{-5} = \frac{4P^3}{27}$.
Solving for $P$: $P^3 = \frac{2.9 \times 10^{-5} \times 27}{4} = 19.575 \times 10^{-5} = 195.75 \times 10^{-6}$.
$P = (195.75)^{1/3} \times 10^{-2} \ atm \approx 5.82 \times 10^{-2} \ atm$.

(D) To obtain the target reaction $2NH_{3(g)} + \frac{5}{2}O_{2(g)} \rightleftharpoons 2NO_{(g)} + 3H_2O_{(g)}$,we manipulate the given equations:
Reverse equation $(1)$: $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$ with equilibrium constant $K_1^{-1}$.
Keep equation $(2)$ as is: $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ with equilibrium constant $K_2$.
Multiply equation $(3)$ by $3$: $3H_{2(g)} + \frac{3}{2}O_{2(g)} \rightleftharpoons 3H_2O_{(g)}$ with equilibrium constant $K_3^3$.
Adding these three equations:
$(2NH_{3(g)}) + (N_{2(g)} + O_{2(g)}) + (3H_{2(g)} + \frac{3}{2}O_{2(g)}) \rightleftharpoons (N_{2(g)} + 3H_{2(g)}) + (2NO_{(g)}) + (3H_2O_{(g)})$
Canceling common terms ($N_{2(g)}$ and $3H_{2(g)}$) gives:
$2NH_{3(g)} + \frac{5}{2}O_{2(g)} \rightleftharpoons 2NO_{(g)} + 3H_2O_{(g)}$
The equilibrium constant $K_4$ is given by $K_4 = K_1^{-1} \times K_2 \times K_3^3 = \frac{K_2 K_3^3}{K_1}$.

(C) For reaction $(1)$: $A_{2(g)} + B_{2(g)} \leftrightarrows 2AB_{(g)}$ with equilibrium constant $K_1$.
For reaction $(2)$: $6AB_{(g)} \leftrightarrows 3A_{2(g)} + 3B_{2(g)}$ with equilibrium constant $K_2$.
Reaction $(2)$ is obtained by reversing reaction $(1)$ and multiplying by $3$.
If a reaction is reversed,the equilibrium constant becomes $1/K$.
If a reaction is multiplied by a factor $n$,the equilibrium constant becomes $K^n$.
Therefore,$K_2 = (1/K_1)^3 = K_1^{-3}$.

(B) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$,which implies $K_p/K_c = (RT)^{\Delta n_g}$.
Given $RT = 24.62 \ dm^3 \ atm \ mol^{-1}$.
For reaction $(i): N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,$\Delta n_g = 2 - (1+1) = 0$. So,$K_p/K_c = (RT)^0 = 1$.
For reaction $(ii): N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,$\Delta n_g = 2 - 1 = 1$. So,$K_p/K_c = (RT)^1 = 24.62 \ dm^3 \ atm \ mol^{-1}$.
For reaction $(iii): N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,$\Delta n_g = 2 - (1+3) = -2$. So,$K_p/K_c = (RT)^{-2} = (24.62)^{-2} \approx 1.65 \times 10^{-3} \ dm^{-6} \ atm^{-2} \ mol^2$.

(D) The chemical equation for the decomposition is: $NH_4SH_{(s)} \leftrightarrow NH_{3(g)} + H_2S_{(g)}$
Initial moles of $NH_4SH = \frac{5.1 \ g}{51 \ g/mol} = 0.1 \ mol$.
Moles decomposed $= 30\% \times 0.1 \ mol = 0.03 \ mol$.
According to the stoichiometry,moles of $NH_3$ formed $= 0.03 \ mol$ and moles of $H_2S$ formed $= 0.03 \ mol$.
Temperature $T = 327 + 273 = 600 \ K$.
Partial pressure of $NH_3$ $(P_{NH_3})$ $= \frac{nRT}{V} = \frac{0.03 \times 0.082 \times 600}{3} = 0.492 \ atm$.
Partial pressure of $H_2S$ $(P_{H_2S})$ $= \frac{nRT}{V} = \frac{0.03 \times 0.082 \times 600}{3} = 0.492 \ atm$.
$K_p = P_{NH_3} \times P_{H_2S} = 0.492 \times 0.492 \approx 0.242 \ atm^2$.

(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.
If $\Delta n_g = 0$,then $K_p = K_c$.
If $\Delta n_g \neq 0$,then $K_p \neq K_c$.
Let us calculate $\Delta n_g$ for each option:
$(A)$ $2NO_{(g)} \rightleftharpoons N_{2(g)} + O_{2(g)}$: $\Delta n_g = (1 + 1) - 2 = 0$.
$(B)$ $2C_{(s)} + O_{2(g)} \rightleftharpoons 2CO_{(g)}$: $\Delta n_g = 2 - 1 = 1 \neq 0$.
$(C)$ $NO_{2(g)} + SO_{2(g)} \rightleftharpoons NO_{(g)} + SO_{3(g)}$: $\Delta n_g = (1 + 1) - (1 + 1) = 0$.
$(D)$ $2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$: $\Delta n_g = (1 + 1) - 2 = 0$.
Since $\Delta n_g \neq 0$ only in option $(B)$,$K_p \neq K_c$ for this reaction.

(B) The dissociation reaction is: $XY_{2(g)} \rightleftharpoons XY_{(g)} + Y_{(g)}$
Initial pressure: $P_0 = 600 \ mm \ Hg$,$P_{XY} = 0$,$P_Y = 0$
At equilibrium: $P_{XY_2} = 600 - x$,$P_{XY} = x$,$P_Y = x$
Total pressure at equilibrium: $P_{total} = (600 - x) + x + x = 600 + x$
Given $P_{total} = 800 \ mm \ Hg$,so $600 + x = 800$,which gives $x = 200 \ mm \ Hg$
Equilibrium partial pressures are: $P_{XY_2} = 600 - 200 = 400 \ mm \ Hg$,$P_{XY} = 200 \ mm \ Hg$,$P_Y = 200 \ mm \ Hg$
$K_p = \frac{P_{XY} \times P_Y}{P_{XY_2}} = \frac{200 \times 200}{400} = 100 \ mm \ Hg$

(C) We know the relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Given that the heat of reaction at constant volume $(Q_v = \Delta U)$ is $1500 \text{ cal}$ more than the heat at constant pressure $(Q_p = \Delta H)$,we have $\Delta U = \Delta H + 1500$.
Rearranging gives $\Delta H - \Delta U = -1500$.
Since $\Delta H - \Delta U = \Delta n_g RT$,we have $\Delta n_g RT = -1500$.
Given $T = 27^\circ\text{C} = 300 \text{ K}$ and $R \approx 2 \text{ cal/mol} \cdot \text{K}$,we get $\Delta n_g (2)(300) = -1500$.
$600 \Delta n_g = -1500$,which implies $\Delta n_g = -2.5$.
The relationship between $K_p$ and $K_c$ is $K_p = K_c(RT)^{\Delta n_g}$.
Since $\Delta n_g = -2.5$ (which is negative) and $RT > 1$,it follows that $K_p < K_c$.

(B) The chemical equation is $NH_4HS_{(s)} \rightleftharpoons NH_{3_{(g)}} + H_2S_{(g)}$.
Let the partial pressure of $NH_3$ be $P$ and the partial pressure of $H_2S$ be $P$ at equilibrium.
The total pressure at equilibrium is $P_{total} = P_{NH_3} + P_{H_2S} = P + P = 2P$.
Given that $P_{total} = 1.12 \ atm$,we have $2P = 1.12 \ atm$,which gives $P = 0.56 \ atm$.
The equilibrium constant $K_p$ is given by $K_p = P_{NH_3} \times P_{H_2S} = P \times P = P^2$.
Substituting the value of $P$,$K_p = (0.56)^2 = 0.3136 \ atm^2$.

(A) Let the total mass of the gaseous mixture be $100\, g$.
Mass of $CO_{(g)} = 84\, g$ and mass of $CO_{2_{(g)}} = (100 - 84)\, g = 16\, g$.
Moles of $CO_{(g)} = \frac{84}{28} = 3\, mol$.
Moles of $CO_{2_{(g)}} = \frac{16}{44} = 0.3636\, mol$.
Total moles = $3 + 0.3636 = 3.3636\, mol$.
Partial pressure of $CO$ $(P_{CO})$ = $\frac{3}{3.3636} \times 2\, atm = 1.784\, atm$.
Partial pressure of $CO_{2}$ $(P_{CO_{2}})$ = $\frac{0.3636}{3.3636} \times 2\, atm = 0.216\, atm$.
For the reaction $C_{(s)} + CO_{2_{(g)}} \rightleftharpoons 2CO_{(g)}$,the equilibrium constant $K_{p}$ is given by:
$K_{p} = \frac{(P_{CO})^{2}}{P_{CO_{2}}} = \frac{(1.784)^{2}}{0.216} \approx 14.73\, atm$.
Rounding to the nearest provided option,$K_{p} = 14.89\, atm$.

(D) The reaction is $FeO_{(s)} + CO_{(g)} \rightleftharpoons Fe_{(s)} + CO_{2_{(g)}}$.
$Q_p = \frac{P_{CO_2}}{P_{CO}} = \frac{0.8}{1.6} = 0.5$.
Since $Q_p > K_p$ $(0.5 > 0.25)$,the reaction proceeds in the backward direction.
For the backward reaction $Fe_{(s)} + CO_{2_{(g)}} \rightleftharpoons FeO_{(s)} + CO_{(g)}$,$K_p' = \frac{1}{K_p} = \frac{1}{0.25} = 4$.
Let $x$ be the change in pressure at equilibrium.
$t = 0$: $P_{CO_2} = 0.8 \ atm$,$P_{CO} = 1.6 \ atm$.
$t = t_{eq}$: $P_{CO_2} = (0.8 - x)$,$P_{CO} = (1.6 + x)$.
$K_p' = \frac{P_{CO}}{P_{CO_2}} \Rightarrow 4 = \frac{1.6 + x}{0.8 - x}$.
$3.2 - 4x = 1.6 + x$ $\Rightarrow 5x = 1.6$ $\Rightarrow x = 0.32$.
Equilibrium $P_{CO} = 1.6 + 0.32 = 1.92 \ atm$.
Equilibrium $P_{CO_2} = 0.8 - 0.32 = 0.48 \ atm$.

(D) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.
First,calculate the change in the number of moles of gaseous species,$\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - \frac{3}{2} = -0.5$.
The temperature $T$ in Kelvin is $191 + 273 = 464 \, K$.
Substituting the values into the equation: $1.24 = K_c(0.0821 \times 464)^{-0.5}$.
Rearranging for $K_c$: $K_c = 1.24 \times (0.0821 \times 464)^{0.5}$.
$K_c = 1.24 \times (38.0944)^{0.5} \approx 1.24 \times 6.172 \approx 7.65$.
Thus,the value of $K_c$ is approximately $7.6$.

(A) For the reaction: $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$
Initial concentrations: $[H_2] = 1 \ M$,$[CO_2] = 1 \ M$,$[CO] = 0$,$[H_2O] = 0$
At equilibrium: $[H_2] = 1 - x$,$[CO_2] = 1 - x$,$[CO] = x$,$[H_2O] = x$
$K_c = \frac{[CO][H_2O]}{[H_2][CO_2]} = \frac{x \cdot x}{(1 - x)(1 - x)} = \frac{x^2}{(1 - x)^2}$
Since $\Delta n_g = (1 + 1) - (1 + 1) = 0$,$K_p = K_c(RT)^{\Delta n_g} = K_c(RT)^0 = K_c$.
Therefore,$K_p = \frac{x^2}{(1 - x)^2}$.

(A) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.
Rearranging this,we get $\frac{K_p}{K_c} = (RT)^{\Delta n_g}$.
For the reaction $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - (1 + 1) = 1 - 2 = -1$.
Substituting $\Delta n_g = -1$ into the equation,we get $\frac{K_p}{K_c} = (RT)^{-1} = \frac{1}{RT}$.
Since $R$ is the gas constant $(0.0821 \ L \ atm \ K^{-1} \ mol^{-1})$ and $T$ is the absolute temperature in Kelvin,$RT$ is typically greater than $1$ at standard temperatures.
Therefore,$\frac{1}{RT} < 1$,which implies $\frac{K_p}{K_c} < 1$.

(D) The reaction is $2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2(g)}$.
Let the initial pressure of $NOBr$ be $P_0$.
At equilibrium,let $P_{Br_2} = x = \frac{P}{9}$.
From stoichiometry,$P_{NO} = 2x = \frac{2P}{9}$ and $P_{NOBr} = P_0 - 2x$.
The total pressure $P = P_{NOBr} + P_{NO} + P_{Br_2} = (P_0 - 2x) + 2x + x = P_0 + x$.
Thus,$P_0 = P - x = P - \frac{P}{9} = \frac{8P}{9}$.
So,$P_{NOBr} = \frac{8P}{9} - \frac{2P}{9} = \frac{6P}{9} = \frac{2P}{3}$.
$K_P = \frac{(P_{NO})^2 (P_{Br_2})}{(P_{NOBr})^2} = \frac{(\frac{2P}{9})^2 (\frac{P}{9})}{(\frac{2P}{3})^2} = \frac{\frac{4P^2}{81} \times \frac{P}{9}}{\frac{4P^2}{9}} = \frac{P}{81}$.
Therefore,$\frac{P}{K_P} = \frac{P}{P/81} = 81$.

(D) 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 $\Delta n = (1 + 1) - 1 = 1$.
Given that $K_P = 2K_C$,we have $\frac{K_P}{K_C} = 2$.
Substituting these values into the equation: $2 = (RT)^1$.
Using the gas constant $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,we get $2 = 0.0821 \times T$.
Solving for $T$: $T = \frac{2}{0.0821} \approx 24.36 \ K$.

(A) The given reaction is $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$.
The expression for the equilibrium constant $(K_P)$ is:
$K_P = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \times P_{O_2}}$
Given that at equilibrium,the number of moles of $SO_2$ and $SO_3$ are equal,their partial pressures will also be equal:
$P_{SO_3} = P_{SO_2}$
Substituting this into the $K_P$ expression:
$3.5 = \frac{(P_{SO_2})^2}{(P_{SO_2})^2 \times P_{O_2}}$
$3.5 = \frac{1}{P_{O_2}}$
$P_{O_2} = \frac{1}{3.5} \approx 0.2857 \ atm \approx 0.29 \ atm$.

(B) The chemical equation is $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$.
At equilibrium,let the partial pressure of both $NH_3$ and $H_2S$ be $P \ atm$.
The total pressure at equilibrium is $P_{total} = P_{NH_3} + P_{H_2S} = P + P = 2P$.
Given that $2P = 1.12 \ atm$,we find $P = \frac{1.12}{2} = 0.56 \ atm$.
The equilibrium constant $K_P$ is given by $K_P = P_{NH_3} \times P_{H_2S} = P \times P = P^2$.
Substituting the value of $P$,$K_P = (0.56)^2 = 0.3136 \ atm^2$.

(A) The relationship between the equilibrium constant $(K_{eq})$,the rate constant of the forward reaction $(K_f)$,and the rate constant of the backward reaction $(K_b)$ is given by the formula:
$K_{eq} = \frac{K_f}{K_b}$
Given that $K_{eq} = 20.0$ and $K_f = 10.0$,we substitute these values into the equation:
$20.0 = \frac{10.0}{K_b}$
Solving for $K_b$:
$K_b = \frac{10.0}{20.0} = 0.5$

(C) Given reactions:
$(i) 2NO + O_2 \rightleftharpoons 2NO_2$ with equilibrium constant $K_1$
$(ii) 4NO + 2Cl_2 \rightleftharpoons 4NOCl$ with equilibrium constant $K_2$
$(iii) NO_2 + \frac{1}{2}Cl_2 \rightleftharpoons NOCl + \frac{1}{2}O_2$ with equilibrium constant $K_3$
To obtain reaction $(iii)$,we manipulate reactions $(i)$ and $(ii)$:
Reverse reaction $(i)$ and divide by $2$:
$NO_2 \rightleftharpoons NO + \frac{1}{2}O_2$; $K' = \frac{1}{\sqrt{K_1}}$
Divide reaction $(ii)$ by $4$:
$NO + \frac{1}{2}Cl_2 \rightleftharpoons NOCl$; $K'' = (K_2)^{1/4}$
Adding these two reactions gives reaction $(iii)$:
$NO_2 + \frac{1}{2}Cl_2 \rightleftharpoons NOCl + \frac{1}{2}O_2$
Therefore,$K_3 = K' \times K'' = \frac{(K_2)^{1/4}}{\sqrt{K_1}}$
Squaring both sides:
$K_3^2 = \frac{(K_2)^{1/2}}{K_1} = \sqrt{\frac{K_2}{K_1^2}}$
Wait,checking the options provided,the standard relation is $K_3 = \frac{K_2^{1/4}}{K_1^{1/2}}$. Thus $K_3^2 = \frac{K_2^{1/2}}{K_1}$. Given the options,there is a discrepancy in the provided choices. Based on the derivation,$K_3^2 = \sqrt{K_2}/K_1$.

(A) The equilibrium constant $K_c$ for the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is given by the expression:
$K_c = \frac{[NO_2]^2}{[N_2O_4]}$
Substituting the given equilibrium concentrations:
$[NO_2] = 1.2 \times 10^{-2} \ mol/L$
$[N_2O_4] = 4.8 \times 10^{-2} \ mol/L$
$K_c = \frac{(1.2 \times 10^{-2})^2}{4.8 \times 10^{-2}}$
$K_c = \frac{1.44 \times 10^{-4}}{4.8 \times 10^{-2}}$
$K_c = 0.3 \times 10^{-2} = 3 \times 10^{-3} \ mol/L$
Thus,the correct option is $A$.

(B) For the reaction: $2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2(g)}$
Let the initial pressure of $NOBr$ be $P_0$.
At equilibrium,let the pressure of $Br_2$ be $x = \frac{P}{9}$.
According to stoichiometry,the pressure of $NO$ will be $2x = \frac{2P}{9}$ and the pressure of $NOBr$ will be $P_0 - 2x$.
The total pressure $P = P_{NOBr} + P_{NO} + P_{Br_2} = (P_0 - 2x) + 2x + x = P_0 + x$.
Given $x = \frac{P}{9}$,we have $P = P_0 + \frac{P}{9}$,so $P_0 = \frac{8P}{9}$.
Thus,$P_{NOBr} = \frac{8P}{9} - \frac{2P}{9} = \frac{6P}{9} = \frac{2P}{3}$.
$K_P = \frac{(P_{NO})^2 (P_{Br_2})}{(P_{NOBr})^2} = \frac{(\frac{2P}{9})^2 (\frac{P}{9})}{(\frac{2P}{3})^2} = \frac{(\frac{4P^2}{81}) (\frac{P}{9})}{\frac{4P^2}{9}} = \frac{P}{81}$.
Therefore,$\frac{K_P}{P} = \frac{1}{81}$.

(B) The reaction is $2A_{(g)} \rightleftharpoons B_{(g)} + 3C_{(g)}$.
At equilibrium,the number of moles are $n_A = 2 \ mol$,$n_B = 2 \ mol$,and $n_C = 2 \ mol$.
Let the volume of the flask be $V \ L$.
The concentrations are $[A] = \frac{2}{V}$,$[B] = \frac{2}{V}$,and $[C] = \frac{2}{V}$.
The equilibrium constant expression is $K_c = \frac{[B][C]^3}{[A]^2}$.
Substituting the values: $16 = \frac{(\frac{2}{V})(\frac{2}{V})^3}{(\frac{2}{V})^2}$.
Simplifying the expression: $16 = \frac{(\frac{2}{V}) \times \frac{8}{V^3}}{\frac{4}{V^2}} = \frac{16}{V^4} \times \frac{V^2}{4} = \frac{4}{V^2}$.
Thus,$V^2 = \frac{4}{16} = \frac{1}{4}$.
Therefore,$V = \sqrt{\frac{1}{4}} = \frac{1}{2} \ L$.

(D) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n_g}$.
For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta n_g = 2 - 1 = 1$.
Substituting this into the formula,we get $K_p = K_c(RT)^1$.
Rearranging to find the ratio $\frac{K_c}{K_p}$,we get $\frac{K_c}{K_p} = \frac{1}{(RT)^1} = (RT)^{-1}$.

(A) For the reaction,$C_{(s)} + CO_{2_{(g)}} \rightleftharpoons 2CO_{(g)}$,the equilibrium constant $K_p$ is defined as the ratio of the square of the partial pressure of the product to the partial pressure of the gaseous reactant.
$K_p = \frac{(P_{CO})^2}{P_{CO_2}}$
Given,$P_{CO} = 8 \ atm$ and $P_{CO_2} = 4 \ atm$.
Substituting these values into the expression:
$K_p = \frac{(8)^2}{4} = \frac{64}{4} = 16$
Thus,the $K_p$ of the reaction is $16$.
Therefore,the correct option is $A$.

(D) The standard Gibbs free energy change is given by $\Delta G^{\circ} = - RT \ln K_p$.
For the reaction $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$,the total equilibrium pressure is $P_{total} = 3X \ bar$.
Let the partial pressure of $CO_2$ be $p$ and the partial pressure of $NH_3$ be $2p$ based on the stoichiometry.
Thus,$p + 2p = 3X$,which implies $3p = 3X$,so $p = X$.
Therefore,$p_{CO_2} = X$ and $p_{NH_3} = 2X$.
The equilibrium constant $K_p$ is given by $K_p = (p_{NH_3})^2 (p_{CO_2}) = (2X)^2 (X) = 4X^3$.
Substituting this into the Gibbs free energy equation:
$\Delta G^{\circ} = - RT \ln(4X^3) = - RT (\ln 4 + 3 \ln X) = - RT \ln 4 - 3 RT \ln X$.

(A) The reaction is $SnO_{2(s)} + 2H_{2(g)} \rightleftharpoons 2H_2O_{(g)} + Sn_{(l)}$.
Since $SnO_2$ and $Sn$ are in solid and liquid states respectively,they do not appear in the $K_p$ expression.
Let the total moles of gaseous mixture at equilibrium be $n_T = 1$.
Given that the mixture contains $40\%$ $H_2$ by volume,the mole fraction of $H_2$ is $X_{H_2} = 0.4$.
Therefore,the mole fraction of $H_2O$ is $X_{H_2O} = 1 - 0.4 = 0.6$.
$K_p = \frac{(P_{H_2O})^2}{(P_{H_2})^2} = \frac{(X_{H_2O} \times P_T)^2}{(X_{H_2} \times P_T)^2} = \frac{(0.6)^2}{(0.4)^2}$.
$K_p = \frac{0.36}{0.16} = \frac{36}{16} = \frac{9}{4}$.

(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 $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta n_g = (2 + 1) - 2 = 1$.
Given values are $K_P = 8 \times 10^{12} \ atm$,$T = 500 \ K$,and $R = 0.08 \ L \ atm \ mol^{-1} \ K^{-1}$.
Substituting these values into the formula:
$8 \times 10^{12} = K_C \times (0.08 \times 500)^1$
$8 \times 10^{12} = K_C \times 40$
$K_C = \frac{8 \times 10^{12}}{40} = 0.2 \times 10^{12} = 2 \times 10^{11} \ mol \ L^{-1}$.
Thus,the correct option is $C$.

(C) The reaction is $2A_{(g)} \rightleftharpoons 2B_{(g)} + C_{(g)}$.
Since the reaction starts with pure $A$,the stoichiometry implies that at equilibrium,the moles (and thus volumes) of $B$ and $C$ will be in a $2:1$ ratio.
Given volume of $C = 100 \ mL$,therefore volume of $B = 200 \ mL$.
Total volume of mixture = $700 \ mL$.
Volume of $A = 700 - (200 + 100) = 400 \ mL$.
Since $P \propto n$ (or $V$ at constant $T$ and $P$),the partial pressure of each gas is $P_i = (V_i / V_{total}) \times P_{total}$.
$P_A = (400/700) \times 10 = 40/7 \ atm$.
$P_B = (200/700) \times 10 = 20/7 \ atm$.
$P_C = (100/700) \times 10 = 10/7 \ atm$.
$K_P = \frac{(P_B)^2 (P_C)}{(P_A)^2} = \frac{(20/7)^2 (10/7)}{(40/7)^2} = \frac{(400/49) \times (10/7)}{1600/49} = \frac{4000/343}{1600/49} = \frac{4000}{343} \times \frac{49}{1600} = \frac{40}{7 \times 4} = \frac{10}{28}$.

(A) For the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the equilibrium constant $K_p$ is given by $K_p = P_{CO_2}$.
According to the van't Hoff equation: $\ln K_p = -\frac{\Delta H^o}{RT} + C$.
Substituting $K_p = P_{CO_2}$,we get $\ln P_{CO_2} = -\frac{\Delta H^o}{RT} + C$.
Converting to base $10$ logarithm: $\log_{10} P_{CO_2} = -\frac{\Delta H^o}{2.303 R} \times \frac{1}{T} + C'$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log_{10} P_{CO_2}$ and $x = 1/T$,the slope $m = -\frac{\Delta H^o}{2.303 R}$.
Thus,a plot of $\log_{10} P_{CO_2}$ versus $1/T$ yields a straight line with a slope related to $\Delta H^o$,allowing its determination.

(C) The relationship between $K_P$ and $K_C$ is given by the formula $K_P = K_C (RT)^{\Delta n}$.
For the given equilibrium reaction $CO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons CO_{2(g)}$,the change in the number of moles of gaseous species is calculated as:
$\Delta n = n_p - n_r = 1 - (1 + 0.5) = -0.5$
Substituting the value of $\Delta n$ into the relationship:
$K_P = K_C (RT)^{-0.5}$
Therefore,the ratio $K_P/K_C$ is:
$\frac{K_P}{K_C} = (RT)^{-0.5} = (RT)^{-1/2}$

(D) The chemical equation for the reaction is $C_{(s)} + CO_{2(g)} \rightleftharpoons 2CO_{(g)}$.
The expression for the equilibrium constant $K_p$ is given by $K_p = \frac{(P_{CO})^2}{P_{CO_2}}$.
Given that $P_{CO} = 4.0 \ atm$ and $P_{CO_2} = 2.0 \ atm$ at equilibrium.
Substituting the values,we get $K_p = \frac{(4.0)^2}{2.0} = \frac{16}{2.0} = 8 \ atm$.

(D) The standard Gibbs free energy change for the reaction is calculated as: $\Delta G^{\circ} = 2 \times \Delta G^{\circ}_f(NO_2) - \Delta G^{\circ}_f(N_2O_4)$.
Substituting the given values: $\Delta G^{\circ} = 2 \times 12.39 - 23.49 = 1.29 \, kcal = 1290 \, cal$.
Using the relation $\Delta G^{\circ} = -RT \ln K_p$ or $\Delta G^{\circ} = -2.303 RT \log K_p$:
$1290 = -2.303 \times 1.987 \times 298 \times \log K_p$.
$\log K_p = -\frac{1290}{1364.2} \approx -0.9456$.
$K_p = 10^{-0.9456} \approx 0.1133 \, atm$.

(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 negative.
$\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$.
For option $C$: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$,$\Delta n = 2 - (2 + 1) = -1$.
Since $\Delta n = -1$,$K_p = K_c(RT)^{-1} = \frac{K_c}{RT}$,which means $K_p < K_c$.

(B) The equilibrium constant $K_c$ for a reversible reaction is given by the ratio of the rate constant of the forward reaction $(k_f)$ to the rate constant of the backward reaction $(k_b)$.
$K_c = \frac{k_f}{k_b}$
Given:
$k_f = 2.38 \times 10^{-4}$
$k_b = 8.15 \times 10^{-5}$
Substituting the values:
$K_c = \frac{2.38 \times 10^{-4}}{8.15 \times 10^{-5}}$
$K_c = \frac{2.38}{8.15} \times 10^{(-4 - (-5))}$
$K_c = 0.292 \times 10^1$
$K_c = 2.92$

(D) The equilibrium reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
Given volume $V = 1000\, cm^3 = 1\, L$.
The equilibrium concentrations are:
$[PCl_5] = 2\, mol / 1\, L = 2\, M$
$[PCl_3] = 2\, mol / 1\, L = 2\, M$
$[Cl_2] = 3\, mol / 1\, L = 3\, M$
Calculate $K_C$:
$K_C = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{2 \times 3}{2} = 3$
Calculate $K_P$ using the relation $K_P = K_C(RT)^{\Delta n_g}$:
Here,$\Delta n_g = (1 + 1) - 1 = 1$,$R = 0.0821\, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,and $T = 27 + 273 = 300\, K$.
$K_P = 3 \times (0.0821 \times 300)^1 = 3 \times 24.63 = 73.89 \approx 74\, atm$.

(C) The dissociation reaction is: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$.
Initial moles: $PCl_5 = 4$,$PCl_3 = 0$,$Cl_2 = 0$.
At equilibrium,$80\%$ of $PCl_5$ remains undissociated,so the amount of $PCl_5$ dissociated is $20\%$ of $4 \ moles = 0.2 \times 4 = 0.8 \ moles$.
Moles at equilibrium: $PCl_5 = 4 - 0.8 = 3.2 \ moles$,$PCl_3 = 0.8 \ moles$,$Cl_2 = 0.8 \ moles$.
Since the volume is $1 \ L$,the concentrations are: $[PCl_5] = 3.2 \ M$,$[PCl_3] = 0.8 \ M$,$[Cl_2] = 0.8 \ M$.
The equilibrium constant $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{0.8 \times 0.8}{3.2} = \frac{0.64}{3.2} = 0.2$.

(B) For a first-order reaction,the rate constant $k$ is related to the half-life $t_{1/2}$ by the formula $k = \frac{\ln(2)}{t_{1/2}}$.
For the forward reaction $(A \rightarrow B)$: $k_f = \frac{\ln(2)}{15 \ s}$.
For the backward reaction $(B \rightarrow A)$: $k_b = \frac{\ln(2)}{18 \ s}$.
The equilibrium constant $K_c$ for a reversible reaction is given by the ratio of the rate constants: $K_c = \frac{k_f}{k_b}$.
Substituting the values: $K_c = \frac{\ln(2) / 15}{\ln(2) / 18} = \frac{18}{15} = 1.2$.

(B) The chemical equation is: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$
Initial moles: $2 \ mol$ of $PCl_5$,$0 \ mol$ of $PCl_3$,$0 \ mol$ of $Cl_2$.
Degree of dissociation $\alpha = 0.40$.
At equilibrium:
Moles of $PCl_5 = 2(1 - 0.4) = 1.2 \ mol$
Moles of $PCl_3 = 2 \times 0.4 = 0.8 \ mol$
Moles of $Cl_2 = 2 \times 0.4 = 0.8 \ mol$
Concentrations at equilibrium (Volume = $2 \ L$):
$[PCl_5] = 1.2 / 2 = 0.6 \ M$
$[PCl_3] = 0.8 / 2 = 0.4 \ M$
$[Cl_2] = 0.8 / 2 = 0.4 \ M$
$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{0.4 \times 0.4}{0.6} = \frac{0.16}{0.6} = 0.266 \ mol/L$.

(A) The reaction is ${H_2} + {I_2} \rightleftharpoons 2HI$.
Initial moles: $[H_2] = 4.5 \, \text{mol}$,$[I_2] = 4.5 \, \text{mol}$,$[HI] = 0 \, \text{mol}$.
At equilibrium,$2x = 3 \, \text{mol}$ of $HI$ are formed,so $x = 1.5 \, \text{mol}$.
Equilibrium moles: $[H_2] = 4.5 - 1.5 = 3 \, \text{mol}$,$[I_2] = 4.5 - 1.5 = 3 \, \text{mol}$,$[HI] = 3 \, \text{mol}$.
Concentrations in $10 \, \text{L}$ volume: $[H_2] = 0.3 \, \text{M}$,$[I_2] = 0.3 \, \text{M}$,$[HI] = 0.3 \, \text{M}$.
$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.3)^2}{(0.3)(0.3)} = \frac{0.09}{0.09} = 1$.

(D) For a heterogeneous equilibrium reaction involving solids and gases,the concentration or partial pressure of pure solids is taken as unity $(1)$.
Given the reaction: $ZnCO_{3(s)} \rightleftharpoons ZnO_{(s)} + CO_{2(g)}$.
The equilibrium constant expression in terms of partial pressure is $K_p = \frac{P_{ZnO} \times P_{CO_2}}{P_{ZnCO_3}}$.
Since $ZnCO_{3(s)}$ and $ZnO_{(s)}$ are pure solids,their partial pressures are considered as $1$.
Therefore,$K_p = 1 \times P_{CO_2} / 1 = P_{CO_2}$.

(C) The reaction is $A + 2B \rightleftharpoons 2C + D$.
Initial moles: $A = 1.1, B = 2.2, C = 0, D = 0$.
At equilibrium,$0.2 \ mol$ of $C$ is formed. Since the stoichiometric coefficient of $C$ is $2$,the change in moles is:
$A: 1.1 - 0.1 = 1.0 \ mol$
$B: 2.2 - 0.2 = 2.0 \ mol$
$C: 0 + 0.2 = 0.2 \ mol$
$D: 0 + 0.1 = 0.1 \ mol$
Since the volume is $1 \ L$,the concentrations are equal to the number of moles.
$K_C = \frac{[C]^2 [D]}{[A] [B]^2} = \frac{(0.2)^2 \times (0.1)}{(1.0) \times (2.0)^2} = \frac{0.04 \times 0.1}{1.0 \times 4.0} = \frac{0.004}{4.0} = 0.001$.

(A) The reaction is $NH_4COONH_2(s) \rightleftharpoons 2NH_3(g) + CO_2(g)$.
Let the partial pressure of $CO_2$ at equilibrium be $x \ atm$. Then,the partial pressure of $NH_3$ is $2x \ atm$.
The total equilibrium pressure is $P_{total} = P_{NH_3} + P_{CO_2} = 2x + x = 3x$.
Given $P_{total} = 3 \ atm$,so $3x = 3 \implies x = 1 \ atm$.
Thus,$P_{CO_2} = 1 \ atm$ and $P_{NH_3} = 2 \ atm$.
The equilibrium constant $K_p$ is given by $K_p = (P_{NH_3})^2 \times (P_{CO_2})$.
$K_p = (2)^2 \times (1) = 4 \ atm^3$.

(A) The reaction is $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$.
Initial moles: $N_2O_4 = 0.8 \ mol$,$NO_2 = 0 \ mol$.
At equilibrium,let the amount of $N_2O_4$ reacted be $x \ mol$.
Then,$[N_2O_4] = (0.8 - x) \ M$ and $[NO_2] = 2x \ M$ (since volume is $1 \ L$).
$K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(2x)^2}{0.8 - x} = 0.00466$.
$4x^2 = 0.00466(0.8 - x) \implies 4x^2 + 0.00466x - 0.003728 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x \approx 0.03$.
Therefore,$[NO_2] = 2x = 2 \times 0.03 = 0.06 \ M$.