Kp and Kc Relationship Questions in English

(C) The equilibrium reaction is $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$.
Let the partial pressures at equilibrium be $P_{NH_3} = P_{H_2S} = x$.
The total pressure $P_{total} = P_{NH_3} + P_{H_2S} = x + x = 2x$.
Given $P_{total} = 1.12 \ atm$,so $2x = 1.12$,which gives $x = 0.56 \ atm$.
The equilibrium constant $K_p$ is given by $K_p = P_{NH_3} \times P_{H_2S} = x \times x = x^2$.
$K_p = (0.56)^2 = 0.3136 \approx 0.31$.

(D) For an elementary reaction $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$,the equilibrium constant $K_c$ is the ratio of the forward rate constant $k_f$ to the reverse rate constant $k_r$.
Given $r_f = k_f [A_2][B_2] = 1.7 \times 10^{-18} [A_2][B_2]$,so $k_f = 1.7 \times 10^{-18}$.
Given $r_r = k_r [AB]^2 = 2.4 \times 10^{-21} [AB]^2$,so $k_r = 2.4 \times 10^{-21}$.
$K_c = \frac{k_f}{k_r} = \frac{1.7 \times 10^{-18}}{2.4 \times 10^{-21}} = \frac{1.7}{2.4} \times 10^3 \approx 0.708 \times 10^3 \approx 0.7 \times 10^3$.

(B) Initial moles: $A = 1 \ mol$,$B = 1.5 \ mol$,$C = 0 \ mol$.
Volume $V = 2 \ L$.
Initial concentrations: $[A]_0 = 1/2 = 0.5 \ M$,$[B]_0 = 1.5/2 = 0.75 \ M$.
At equilibrium,$[C] = 0.35 \ M$.
From the stoichiometry $A + 2B \rightleftharpoons 2C$,if $2x = 0.35$,then $x = 0.175 \ M$.
Equilibrium concentrations: $[A] = 0.5 - x = 0.5 - 0.175 = 0.325 \ M$.
$[B] = 0.75 - 2x = 0.75 - 0.35 = 0.4 \ M$.
$[C] = 0.35 \ M$.
$K_c = \frac{[C]^2}{[A][B]^2} = \frac{(0.35)^2}{(0.325)(0.4)^2} = \frac{0.1225}{0.325 \times 0.16} = \frac{0.1225}{0.052} \approx 2.3557 \approx 2.36 \ M^{-1}$.

(D) The given reaction is a heterogeneous equilibrium: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$.
For heterogeneous equilibria involving solids and gases,the equilibrium constant $K_p$ is defined only by the partial pressure of the gaseous species.
Thus,$K_p = P_{CO_2}$.
Given that $K_p = 1.16 \ atm$,the partial pressure of $CO_2$ at equilibrium is $P_{CO_2} = 1.16 \ atm$.

(C) The equilibrium constant expression for the reaction $C_{(s)} + CO_{2(g)} \rightleftharpoons 2CO_{(g)}$ is given by:
$K_p = \frac{(P_{CO})^2}{P_{CO_2}}$
Given that $P_{CO} = 4.0 \ atm$ and $P_{CO_2} = 2.0 \ atm$.
Substituting the values:
$K_p = \frac{(4.0)^2}{2.0} = \frac{16}{2.0} = 8$
Thus,the value of $K_p$ is $8$.

(A) The equilibrium constant $K_c$ is defined as the ratio of the forward rate constant $K_f$ to the backward rate constant $K_r$:
$K_c = \frac{K_f}{K_r}$
Given $K_c = 10$ and $K_f = 203$:
$10 = \frac{203}{K_r}$
$K_r = \frac{203}{10} = 20.3$

(C) The reaction is: $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$
Initial moles: $8, 0, 0$
At equilibrium,$n(A_2) = 2 \ mol$. Since the stoichiometry of $A_2$ is $1$,the amount of $AB_3$ reacted is $2 \times 2 = 4 \ mol$.
Remaining moles at equilibrium:
$n(AB_3) = 8 - 4 = 4 \ mol$
$n(A_2) = 2 \ mol$
$n(B_2) = 3 \times 2 = 6 \ mol$
Since the volume is $1.0 \ L$,the concentrations are $[AB_3] = 4 \ M, [A_2] = 2 \ M, [B_2] = 6 \ M$.
$K_c = \frac{[A_2][B_2]^3}{[AB_3]^2} = \frac{2 \times (6)^3}{(4)^2} = \frac{2 \times 216}{16} = \frac{432}{16} = 27$.

(A) The reaction is $PCl_5 \rightleftharpoons PCl_3 + Cl_2$.
At equilibrium,the number of moles of each species is $n(PCl_5) = 2$,$n(PCl_3) = 2$,and $n(Cl_2) = 2$.
Total moles $n_{total} = 2 + 2 + 2 = 6 \ mol$.
The mole fraction of each species is $x = \frac{2}{6} = \frac{1}{3}$.
The partial pressure of each species is $P_i = x_i \times P_{total} = \frac{1}{3} \times 3 \ atm = 1 \ atm$.
$K_p = \frac{P(PCl_3) \times P(Cl_2)}{P(PCl_5)} = \frac{1 \times 1}{1} = 1 \ atm$.

(A) Initial moles of $N_2O_4 = \frac{9.2 \ g}{92 \ g \ mol^{-1}} = 0.1 \ mol$.
Given that $50\%$ of $N_2O_4$ dissociates,the moles reacted $= 0.1 \times 0.5 = 0.05 \ mol$.
$\begin{array}{lccc} & N_2O_{4(g)} & \rightleftharpoons & 2NO_{2(g)} \\ \text{Initial moles} & 0.1 & & 0 \\ \text{Change} & -0.05 & & +2(0.05) \\ \text{Equilibrium moles} & 0.05 & & 0.1 \end{array}$
Since the volume is $1 \ L$,the concentrations are $[N_2O_4] = 0.05 \ mol \ L^{-1}$ and $[NO_2] = 0.1 \ mol \ L^{-1}$.
$K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.1)^2}{0.05} = \frac{0.01}{0.05} = 0.2 \ mol \ L^{-1}$.

(C) The reaction is $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$.
At equilibrium,the concentrations are:
$[H_2] = (1-x)$,$[CO_2] = (1-x)$,$[CO] = x$,$[H_2O] = x$.
Since the number of moles of gaseous reactants equals the number of moles of gaseous products $(\Delta n_g = 0)$,$K_p = K_c$.
$K_p = \frac{[CO][H_2O]}{[H_2][CO_2]} = \frac{x \times x}{(1-x) \times (1-x)} = \frac{x^2}{(1-x)^2}$.

(C) For the given reaction $2Ag_2O_{(s)} \rightleftharpoons 4Ag_{(s)} + O_{2(g)}$,the equilibrium constant expression $K_p$ is defined as the product of the partial pressures of the gaseous products raised to their stoichiometric coefficients,divided by the partial pressures of the gaseous reactants.
Since $Ag_2O$ and $Ag$ are in the solid state,their activities are taken as $1$.
Therefore,$K_p = P_{O_2}$.
Thus,the partial pressure of $O_2$ is equal to $K_p$.

(A) The given reaction is $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$.
Here,the change in the number of moles of gaseous species is $\Delta n_g = 1 - (1 + 1) = -1$.
The temperature $T = 250 + 273 = 523 \ K$.
The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.
Substituting the values: $K_p = 26 \times (0.0821 \times 523)^{-1}$.
$K_p = \frac{26}{0.0821 \times 523} \approx \frac{26}{42.94} \approx 0.605 \approx 0.61$.

(D) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.
For the reaction $A_{(g)} + 2B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = (3 + 1) - (1 + 2) = 4 - 3 = 1$.
Given $K_p = 0.05 \ atm$ and $T = 1000 \ K$.
Substituting the values: $0.05 = K_c(R \times 1000)^1$.
Therefore,$K_c = \frac{0.05}{1000 \ R} = \frac{5 \times 10^{-2}}{10^3 \ R} = 5 \times 10^{-5} / R$.

(C) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.
For $K_p = K_c$,the change in the number of moles of gaseous products and reactants must be zero,i.e.,$\Delta n_g = 0$.
In option $(C)$,the reaction is $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.
Here,$\Delta n_g = (2) - (1 + 1) = 0$.
Therefore,$K_p = K_c$ for this reaction.

(C) 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 positive $({\Delta n_g} > 0)$.
${\Delta n_g}$ is the difference between the number of moles of gaseous products and gaseous reactants.
For option $A$: ${\Delta n_g} = 2 - (1 + 3) = -2$.
For option $B$: ${\Delta n_g} = 2 - (1 + 1) = 0$.
For option $C$: ${\Delta n_g} = (1 + 1) - 1 = +1$.
For option $D$: ${\Delta n_g} = 2 - (2 + 1) = -1$.
Since ${\Delta n_g} > 0$ only for option $C$,the correct answer is $C$.

(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 $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$,the change in the number of moles of gaseous products and reactants is $\Delta n_g = (1 + 1) - 2 = 0$.
Substituting $\Delta n_g = 0$ into the formula,we get $K_p = K_c(RT)^0 = K_c \times 1 = K_c$.
Therefore,$K_p = K_c$.

(C) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$,which implies $\frac{K_p}{K_c} = (RT)^{\Delta n_g}$.
Since $RT = 300R$ is constant and greater than $1$,the ratio $\frac{K_p}{K_c}$ increases as $\Delta n_g$ increases.
Calculate $\Delta n_g$ (moles of gaseous products - moles of gaseous reactants) for each:
For $X: \Delta n_g = 2 - (2 + 1) = -1$
For $Y: \Delta n_g = (1 + 1) - 1 = 1$
For $Z: \Delta n_g = (1 + 1) - 2 = 0$
Comparing the values of $\Delta n_g$: $-1 < 0 < 1$.
Therefore,the increasing order of the ratio is $X < Z < Y$.

(B) The relationship between $K_p$ and $K_c$ is given by the formula $K_p = K_c(RT)^{\Delta n_g}$.
For the reaction ${N_{2(g)}} + {O_{2(g)}} \rightleftharpoons 2NO_{(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta n_g = 2 - (1 + 1) = 0$.
Since $\Delta n_g = 0$,we have $K_p = K_c(RT)^0 = K_c$.
Therefore,$K_p = 4.0 \times 10^{-6}$.

(B) The given reaction is $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$.
For this reaction,the change in the number of moles of gaseous species is $\Delta n_g = 1 - (1 + 1) = -1$.
The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$,which implies $K_c = K_p(RT)^{-\Delta n_g}$.
Given $K_p = 0.61 \ atm^{-1}$,$T = 250 + 273 = 523 \ K$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
$K_c = 0.61 \times (0.0821 \times 523)^{-( -1 )} = 0.61 \times (42.9383) = 26.19 \ mol \ L^{-1}$.

(C) The relationship between $K_p$ and $K_c$ is given by the formula $K_p = K_c(RT)^{\Delta n}$.
For the given reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta n = n_p - n_r = 2 - (1 + 3) = 2 - 4 = -2$.
Substituting $\Delta n = -2$ into the formula,we get $K_p = K_c(RT)^{-2}$.

(A) The first reaction is: $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ with equilibrium constant $K_1 = \frac{[NO]^2}{[N_2][O_2]}$.
The second reaction is: $NO_{(g)} \rightleftharpoons 1/2N_{2(g)} + 1/2O_{2(g)}$ with equilibrium constant $K_2 = \frac{[N_2]^{1/2}[O_2]^{1/2}}{[NO]}$.
Observe that the first reaction is the reverse of the second 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 $K^n$.
Therefore,$K_1 = (1/K_2)^2$.

(C) The first reaction is: $NO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons NO_{2(g)}$ with equilibrium constant $K_1 = 4 \times 10^{-3}$.
The second reaction is: $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$ with equilibrium constant $K_2$.
Notice that the second reaction is the reverse of the first reaction multiplied by $2$.
Therefore,the relationship between the equilibrium constants is $K_2 = (1/K_1)^2$.
Substituting the value of $K_1$: $K_2 = 1 / (4 \times 10^{-3})^2 = 1 / (16 \times 10^{-6}) = 10^6 / 16 = 0.0625 \times 10^6 = 6.25 \times 10^4$.

(A) Given reactions:
$(i) \, CoO_{2(g)} + H_{2(g)} \rightleftharpoons CoO_{(s)} + H_2O_{(g)} \quad K_1 = 67$
$(ii) \, CoO_{2(g)} + CO_{(g)} \rightleftharpoons CoO_{(s)} + CO_{2(g)} \quad K_2 = 490$
Target reaction:
$(iii) \, CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)} \quad K = ?$
To obtain reaction $(iii)$,subtract reaction $(ii)$ from reaction $(i)$:
$(iii) = (i) - (ii)$
Therefore,the equilibrium constant $K$ is given by:
$K = \frac{K_1}{K_2} = \frac{67}{490} \approx 0.137$

(C) For the reaction $AB_{(s)} \rightleftharpoons A_{(g)} + B_{(g)}$,the equilibrium constant expression is $K_c = [A][B]$.
Since $K_c$ is constant at a given temperature,if the concentration of $A$ is doubled (i.e.,$[A]' = 2[A]$),the new concentration of $B$ (let it be $[B]'$) must satisfy the same $K_c$ value.
$K_c = [A][B] = [A]'[B]'$
$K_c = [A][B] = (2[A]) \times [B]'$
$[B]' = \frac{[A][B]}{2[A]} = \frac{[B]}{2}$.
Therefore,the equilibrium concentration of $B$ will be halved.

(C) The equilibrium constant expression for the reaction $A_{(g)} \rightleftharpoons B_{(g)} + 2C_{(g)}$ is given by: $K_c = \frac{[B][C]^2}{[A]}$.
Assuming the temperature remains constant,$K_c$ remains constant.
Let the initial concentrations be $[A]_1, [B]_1, [C]_1$ and the new concentrations be $[A]_2, [B]_2, [C]_2$.
Since $K_c = \frac{[B]_1[C]_1^2}{[A]_1} = \frac{[B]_2[C]_2^2}{[A]_2}$,and given $[B]_2 = 2[B]_1$,we have $\frac{[B]_1[C]_1^2}{[A]_1} = \frac{2[B]_1[C]_2^2}{[A]_2}$.
Assuming the volume of the system is constant and the reaction reaches a new equilibrium,the relationship simplifies to $[C]_2^2 = \frac{[C]_1^2}{2}$ (assuming $[A]$ remains relatively constant or considering the ratio change).
Therefore,$[C]_2 = \frac{[C]_1}{\sqrt{2}}$.

(B) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.
For the reaction $CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - (1 + 0.5) = -0.5$ or $-\frac{1}{2}$.
Substituting this into the equation,we get $K_p = K_c(RT)^{-\frac{1}{2}}$.
Therefore,the ratio $\frac{K_p}{K_c} = (RT)^{-\frac{1}{2}}$.

(A) For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant expression is $K_C = \frac{[NH_3]^2}{[N_2][H_2]^3}$.
The units for concentration are $mol \, L^{-1}$.
Substituting these units into the expression: $K_C = \frac{(mol \, L^{-1})^2}{(mol \, L^{-1}) \times (mol \, L^{-1})^3} = \frac{mol^2 \, L^{-2}}{mol^4 \, L^{-4}} = mol^{-2} \, L^2$.
Thus,the unit of $K_C$ is $L^2 \, mol^{-2}$.
Both the Assertion and Reason are correct,and the Reason provides the correct explanation for the Assertion.

(N/A) The relationship between $K_{p}$ and $K_{c}$ is given by the equation: $K_{p} = K_{c}(RT)^{\Delta n}$.
For the given reaction: $2 \ NOCl \ (g) \rightleftharpoons 2 \ NO \ (g) + Cl_{2} \ (g)$.
The change in the number of moles of gaseous products and reactants is $\Delta n = (2 + 1) - 2 = 1$.
Given $K_{c} = 3.75 \times 10^{-6}$,$R = 0.0831 \ \text{L bar K}^{-1} \text{mol}^{-1}$,and $T = 1069 \ K$.
Substituting the values: $K_{p} = 3.75 \times 10^{-6} \times (0.0831 \times 1069)^{1}$.
$K_{p} = 3.75 \times 10^{-6} \times 88.8339$.
$K_{p} \approx 3.33 \times 10^{-4}$.

(A) For the reaction,let $x$ be the decrease in pressure of $CO_{2}$.
$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$
Initial pressure: $0.48 \ bar$ and $0 \ bar$.
At equilibrium: $(0.48 - x) \ bar$ and $2x \ bar$.
$K_{p} = \frac{p_{CO}^{2}}{p_{CO_{2}}} = 3.0$.
Substituting the values: $\frac{(2x)^{2}}{0.48 - x} = 3.0$.
$4x^{2} = 3(0.48 - x) \implies 4x^{2} + 3x - 1.44 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$:
$x = \frac{-3 \pm \sqrt{9 - 4(4)(-1.44)}}{8} = \frac{-3 \pm \sqrt{9 + 23.04}}{8} = \frac{-3 \pm 5.66}{8}$.
Taking the positive root: $x = \frac{2.66}{8} = 0.3325 \approx 0.33$.
Equilibrium partial pressures are:
$p_{CO} = 2x = 2 \times 0.33 = 0.66 \ bar$.
$p_{CO_{2}} = 0.48 - 0.33 = 0.15 \ bar$.

(N/A) The relation between $K_{p}$ and $K_{c}$ is given by the formula:
$K_{p} = K_{c}(RT)^{\Delta n}$
$(i)$ For the reaction $2 NOCl_{(g)} \longleftrightarrow 2 NO_{(g)} + Cl_{2(g)}$:
$\Delta n = (2 + 1) - 2 = 1$
$R = 0.0831 \ \text{bar L K}^{-1} \text{mol}^{-1}$
$T = 500 \ K$
$K_{p} = 1.8 \times 10^{-2}$
$1.8 \times 10^{-2} = K_{c}(0.0831 \times 500)^{1}$
$K_{c} = \frac{1.8 \times 10^{-2}}{41.55} = 4.33 \times 10^{-4}$
$(ii)$ For the reaction $CaCO_{3(s)} \longleftrightarrow CaO_{(s)} + CO_{2(g)}$:
$\Delta n = 1 - 0 = 1$
$R = 0.0831 \ \text{bar L K}^{-1} \text{mol}^{-1}$
$T = 1073 \ K$
$K_{p} = 167$
$167 = K_{c}(0.0831 \times 1073)^{1}$
$K_{c} = \frac{167}{89.1663} \approx 1.87$

(A) For the given reaction,$\Delta n = 2 - 3 = -1$.
Given: $T = 450 \, K$,$R = 0.0831 \, L \, bar \, K^{-1} \, mol^{-1}$,$K_{p} = 2.0 \times 10^{10} \, bar^{-1}$.
The relationship between $K_{p}$ and $K_{c}$ is $K_{p} = K_{c}(RT)^{\Delta n}$.
Substituting the values: $2.0 \times 10^{10} = K_{c} (0.0831 \times 450)^{-1}$.
$K_{c} = 2.0 \times 10^{10} \times (0.0831 \times 450) = 2.0 \times 10^{10} \times 37.395$.
$K_{c} = 74.79 \times 10^{10} = 7.48 \times 10^{11} \, M^{-1}$.

(4.0) The initial pressure of $HI$ is $0.2 \ atm$. At equilibrium,it has a partial pressure of $0.04 \ atm$. The decrease in the pressure of $HI$ is $0.2 - 0.04 = 0.16 \ atm$.
The reaction is: $2HI_{(g)} \longleftrightarrow H_{2_{(g)}} + I_{2_{(g)}}$
Initial pressure: $0.2 \ atm, 0, 0$
At equilibrium: $0.04 \ atm, \frac{0.16}{2} \ atm, \frac{0.16}{2} \ atm$
At equilibrium: $0.04 \ atm, 0.08 \ atm, 0.08 \ atm$
$K_{p} = \frac{p_{H_{2}} \times p_{I_{2}}}{p_{HI}^2}$
$K_{p} = \frac{0.08 \times 0.08}{(0.04)^{2}} = \frac{0.0064}{0.0016} = 4.0$
Hence,the value of $K_{p}$ is $4.0$.

(D) Let $p$ be the pressure exerted by ethene and hydrogen gas (each) at equilibrium.
According to the reaction:
$C_{2}H_{6(g)} \longleftrightarrow C_{2}H_{4(g)} + H_{2(g)}$
Initial pressure: $4.0 \ atm$,$0$,$0$
At equilibrium: $(4.0 - p) \ atm$,$p \ atm$,$p \ atm$
We can write the expression for $K_{p}$ as:
$K_{p} = \frac{p_{C_{2}H_{4}} \times p_{H_{2}}}{p_{C_{2}H_{6}}}$
Substituting the values:
$\frac{p \times p}{4.0 - p} = 0.04$
$p^{2} = 0.16 - 0.04p$
$p^{2} + 0.04p - 0.16 = 0$
Using the quadratic formula $p = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$:
$p = \frac{-0.04 \pm \sqrt{(0.04)^{2} - 4 \times 1 \times (-0.16)}}{2 \times 1}$
$p = \frac{-0.04 \pm \sqrt{0.0016 + 0.64}}{2}$
$p = \frac{-0.04 \pm \sqrt{0.6416}}{2}$
$p \approx \frac{-0.04 + 0.801}{2} \approx 0.3805 \ atm$
Hence,at equilibrium,the pressure of $C_{2}H_{6}$ is:
$p_{C_{2}H_{6}} = 4.0 - 0.3805 = 3.6195 \ atm \approx 3.62 \ atm$

(N/A) For the given reaction,
$FeO(s) + CO(g) \longleftrightarrow Fe(s) + CO_{2}(g)$
Initially: $p_{CO} = 1.4 \, atm$,$p_{CO_{2}} = 0.80 \, atm$
$Q_{p} = \frac{p_{CO_{2}}}{p_{CO}} = \frac{0.80}{1.4} \approx 0.571$
Since $Q_{p} > K_{p}$ $(0.571 > 0.265)$,the reaction proceeds in the backward direction.
Let $x$ be the change in pressure at equilibrium.
$K_{p} = \frac{p_{CO_{2}}}{p_{CO}} = \frac{0.80 - x}{1.4 + x} = 0.265$
$0.80 - x = 0.265(1.4 + x)$
$0.80 - x = 0.371 + 0.265x$
$1.265x = 0.429$
$x = 0.339 \, atm$
Equilibrium partial pressures:
$p_{CO_{2}} = 0.80 - 0.339 = 0.461 \, atm$
$p_{CO} = 1.4 + 0.339 = 1.739 \, atm$

(N/A) Let the total mass of the gaseous mixture be $100 \, g$.
Mass of $CO = 90.55 \, g$
Mass of $CO_2 = 100 - 90.55 = 9.45 \, g$
Moles of $CO$,$n_{CO} = \frac{90.55}{28} = 3.234 \, mol$
Moles of $CO_2$,$n_{CO_2} = \frac{9.45}{44} = 0.215 \, mol$
Total moles = $3.234 + 0.215 = 3.449 \, mol$
Partial pressure of $CO$,$p_{CO} = \frac{3.234}{3.449} \times 1 = 0.938 \, atm$
Partial pressure of $CO_2$,$p_{CO_2} = \frac{0.215}{3.449} \times 1 = 0.062 \, atm$
$K_P = \frac{(p_{CO})^2}{p_{CO_2}} = \frac{(0.938)^2}{0.062} = 14.19$
For the reaction,$\Delta n = 2 - 1 = 1$
Using $K_P = K_C(RT)^{\Delta n}$:
$14.19 = K_C(0.0821 \times 1127)^1$
$K_C = \frac{14.19}{92.5257} \approx 0.153$.

(D) Let the partial pressure of both carbon dioxide and hydrogen gas be $p$. The given reaction is:
$CO_{(g)} + H_2O_{(g)} \longleftrightarrow CO_{2(g)} + H_{2(g)}$
Initial: $4.0 \ bar \quad 4.0 \ bar \quad 0 \quad 0$
At equilibrium: $(4.0 - p) \ bar \quad (4.0 - p) \ bar \quad p \ bar \quad p \ bar$
It is given that $K_P = 10.1$
Now,
$\frac{P_{CO_2} \times P_{H_2}}{P_{CO} \times P_{H_2O}} = K_P$
$\frac{p \times p}{(4.0 - p)(4.0 - p)} = 10.1$
$\frac{p}{4.0 - p} = \sqrt{10.1} = 3.178$
$p = 12.712 - 3.178p$
$4.178p = 12.712$
$p = 3.04 \ bar$
Hence,at equilibrium,the partial pressure of $H_2$ will be $3.04 \ bar$.

(N/A) Consider the synthesis of $HI$ as the forward reaction:
$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)} \quad (Eq.-i)$
According to the law of chemical equilibrium,the equilibrium constant $K$ is:
$K = \frac{[HI]^2}{[H_2][I_2]} = X \quad (Eq.-ii)$
For the reverse reaction (decomposition of $HI$):
$2 HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)} \quad (Eq.-iii)$
The equilibrium constant $K'$ for this reverse reaction is:
$K' = \frac{[H_2][I_2]}{[HI]^2} = \frac{1}{X} \quad (Eq.-iv)$
Comparing $(Eq.-ii)$ and $(Eq.-iv)$,we get:
$K = \frac{1}{K'} \quad \text{or} \quad K \times K' = 1$
Thus,the equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward reaction.

(A) The reaction is $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.
Concentrations at equilibrium are:
$[H_2] = \frac{0.8 \ mol}{10 \ L} = 0.08 \ M$
$[I_2] = \frac{0.8 \ mol}{10 \ L} = 0.08 \ M$
$[HI] = \frac{2.4 \ mol}{10 \ L} = 0.24 \ M$
The equilibrium constant $K_c$ is given by:
$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.24)^2}{(0.08)(0.08)} = \frac{0.0576}{0.0064} = 9$.
For the reverse reaction $2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$,the equilibrium constant $K_c'$ is:
$K_c' = \frac{1}{K_c} = \frac{1}{9} \approx 0.111$.

(B) For reaction $(i)$: $K_c(1) = \frac{[NH_3]^2}{[N_2][H_2]^3}$
For reaction $(ii)$: $K_c(2) = \frac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}$
Squaring the expression for $K_c(2)$: $(K_c(2))^2 = \left( \frac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}} \right)^2 = \frac{[NH_3]^2}{[N_2][H_2]^3} = K_c(1)$
Therefore,the relation is $K_c(1) = (K_c(2))^2$ or $K_c(2) = \sqrt{K_c(1)}$.

(N/A) For reactions involving gases,it is often convenient to express the equilibrium constant in terms of partial pressure.
The ideal gas equation is given by $pV = nRT$ (Eq. $i$).
Rearranging for pressure,we get $p = (n/V)RT$ (Eq. $ii$).
Since concentration $c$ is defined as $n/V$ (in $mol \ L^{-1}$ or $mol \ dm^{-3}$),we can substitute this into the equation:
$p = cRT$ (Eq. $iii$).
Here,$R$ is the universal gas constant,$T$ is the temperature in Kelvin,and $c$ is the molar concentration of the gas.
At a constant temperature $(T)$,the product $RT$ is constant.
Therefore,the partial pressure $p$ of the gas is directly proportional to its concentration $c$,expressed as $p \propto c$ or $p \propto [\text{gas}]$.

(A) For the gaseous reaction $aA_{(g)} + bB_{(g)} \rightleftharpoons cC_{(g)} + dD_{(g)}$,the partial pressure of each component is given by $P_i = [i]RT$.
The equilibrium constant $K_p$ is defined as:
$K_p = \frac{P_C^c \cdot P_D^d}{P_A^a \cdot P_B^b}$
Substituting $P_i = [i]RT$ into the expression:
$K_p = \frac{([C]RT)^c \cdot ([D]RT)^d}{([A]RT)^a \cdot ([B]RT)^b}$
$K_p = \frac{[C]^c \cdot [D]^d}{[A]^a \cdot [B]^b} \cdot (RT)^{(c+d)-(a+b)}$
Since $K_c = \frac{[C]^c \cdot [D]^d}{[A]^a \cdot [B]^b}$ and $\Delta n = (c+d) - (a+b)$,we get:
$K_p = K_c(RT)^{\Delta n}$

The relationship is given by the equation $K_{p} = K_{c}(RT)^{\Delta n}$.
$1$. If $\Delta n = 0$,then $K_{p} = K_{c}(RT)^{0} = K_{c}$.
Example: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,where $\Delta n = 2 - (1+1) = 0$.
$2$. If $\Delta n > 0$,then $K_{p} = K_{c}(RT)^{\text{positive value}}$,which implies $K_{p} > K_{c}$.
Example: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,where $\Delta n = (1+1) - 1 = +1$.
$3$. If $\Delta n < 0$,then $K_{p} = K_{c}(RT)^{\text{negative value}}$,which implies $K_{p} < K_{c}$.
Example: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,where $\Delta n = 2 - (1+3) = -2$.

(A) The reaction is $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.
Initial moles: $H_2 = 0.4$,$I_2 = 0.4$,$HI = 0$.
At equilibrium,$n(HI) = 0.5 \ mol$.
From stoichiometry,$2 \ mol$ of $HI$ are formed from $1 \ mol$ of $H_2$ and $1 \ mol$ of $I_2$.
So,$0.5 \ mol$ of $HI$ is formed from $0.25 \ mol$ of $H_2$ and $0.25 \ mol$ of $I_2$.
Equilibrium moles: $n(H_2) = 0.4 - 0.25 = 0.15 \ mol$,$n(I_2) = 0.4 - 0.25 = 0.15 \ mol$,$n(HI) = 0.5 \ mol$.
Concentrations in $2 \ L$ vessel: $[H_2] = 0.15/2 = 0.075 \ M$,$[I_2] = 0.075 \ M$,$[HI] = 0.5/2 = 0.25 \ M$.
$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.25)^2}{(0.075)(0.075)} = \frac{0.0625}{0.005625} = 11.11$.
Since $\Delta n_g = 2 - (1+1) = 0$,$K_p = K_c(RT)^{\Delta n_g} = K_c = 11.11$.

(A) The reaction is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Initial moles: $N_2 = 1, H_2 = 3, NH_3 = 0$
At equilibrium: $N_2 = 1 - x, H_2 = 3 - 3x, NH_3 = 2x$
Given $2x = 0.5 \ mol$,so $x = 0.25 \ mol$.
Equilibrium moles: $N_2 = 0.75, H_2 = 2.25, NH_3 = 0.5$.
Total moles at equilibrium = $0.75 + 2.25 + 0.5 = 3.5 \ mol$.
Partial pressures: $P_{N_2} = (0.75/3.5) \times 100 = 21.43 \ atm$,$P_{H_2} = (2.25/3.5) \times 100 = 64.29 \ atm$,$P_{NH_3} = (0.5/3.5) \times 100 = 14.28 \ atm$.
$K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} = \frac{(14.28)^2}{(21.43)(64.29)^3} \approx 7.5 \times 10^{-6} \ atm^{-2}$.

(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 $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = n_p - n_r = 2 - (1 + 3) = 2 - 4 = -2$.
Given $K_c = 0.50$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 673 \ K$.
Substituting these values into the formula: $K_p = 0.50 \times (0.082 \times 673)^{-2}$.
$K_p = 0.50 \times (55.186)^{-2}$.
$K_p = 0.50 \times \frac{1}{(55.186)^2} = 0.50 \times \frac{1}{3045.49} \approx 1.64 \times 10^{-4}$.

(A) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n_g}$.
For the reaction $N_2O_{4_{(g)}} \rightleftharpoons 2NO_{2_{(g)}}$,the change in the number of moles of gaseous products and reactants is $\Delta n_g = 2 - 1 = 1$.
Given $K_p = 0.14 \ atm$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$.
Substituting these values into the equation: $0.14 = K_c \times (0.082 \times 298)^1$.
$K_c = \frac{0.14}{0.082 \times 298}$.
$K_c = \frac{0.14}{24.436} \approx 5.73 \times 10^{-3} \ M$.

(A) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.
For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,$\Delta n_g = 2 - (1 + 3) = -2$.
Given $K_p = 41$,$T = 400 \ K$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
$41 = K_c(0.0821 \times 400)^{-2} \implies K_c = 41 \times (32.84)^2 \approx 44216.7$.
$(a)$ For $2N_{2(g)} + 6H_{2(g)} \rightleftharpoons 4NH_{3(g)}$,the reaction is multiplied by $2$. Thus,$K_c' = (K_c)^2 = (44216.7)^2 \approx 1.95 \times 10^9$.
$(b)$ For $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$,the reaction is reversed and multiplied by $2$. Thus,$K_c'' = (1/K_c)^2 = (1/44216.7)^2 \approx 5.11 \times 10^{-10}$.
$(c)$ For $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$,the reaction is multiplied by $1/2$. Thus,$K_c''' = (K_c)^{1/2} = \sqrt{44216.7} \approx 210.28$.

(C) Given the reaction: $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ with equilibrium constant $K_{c1} = 7 \times 10^{25}$.
For the target reaction: $SO_{3(g)} \rightleftharpoons SO_{2(g)} + \frac{1}{2}O_{2(g)}$,we observe that the reaction is reversed and multiplied by a factor of $\frac{1}{2}$.
Therefore,the new equilibrium constant $K_{c2}$ is given by $K_{c2} = (\frac{1}{K_{c1}})^{1/2}$.
$K_{c2} = (\frac{1}{7 \times 10^{25}})^{1/2} = \frac{1}{\sqrt{7} \times 10^{12.5}} = \frac{1}{2.6457 \times 10^{12.5}} \approx 0.378 \times 10^{-12.5} = 3.78 \times 10^{-13}$.

(A) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n}$.
For the reaction $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$,the change in the number of moles of gaseous species is $\Delta n = (2 + 1) - 2 = 1$.
Given $K_p = 0.033 \ bar$,$T = 1060 \ K$,and $R = 0.08314 \ L \ bar \ K^{-1} \ mol^{-1}$.
Substituting the values: $0.033 = K_c \times (0.08314 \times 1060)^1$.
$K_c = \frac{0.033}{0.08314 \times 1060} = \frac{0.033}{88.1284} \approx 3.74 \times 10^{-4} \ mol \ L^{-1}$.