Mix Examples- 6-1.Equilibrium (Chemical Equilibrium) Questions in English

(C) For the reaction: $A_2B_{6(g)} \rightleftharpoons A_2B_{4(g)} + B_{2(g)}$
Initial pressure: $P_{A_2B_6} = 4 \text{ atm}$
At equilibrium: $P_{A_2B_6} = 4 - x$,$P_{A_2B_4} = x$,$P_{B_2} = x$
$K_p = \frac{P_{A_2B_4} \cdot P_{B_2}}{P_{A_2B_6}} = \frac{x^2}{4-x} = 0.04$
$x^2 = 0.16 - 0.04x$
$x^2 + 0.04x - 0.16 = 0$
Solving the quadratic equation: $x = \frac{-0.04 + \sqrt{(0.04)^2 - 4(1)(-0.16)}}{2} \approx 0.38 \text{ atm}$
Equilibrium pressure of $A_2B_6 = 4 - x = 4 - 0.38 = 3.62 \text{ atm}$.

(C) The reaction is $NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$.
First,calculate the standard enthalpy change of the reaction: $\Delta_{r} H^{\circ} = \Delta_{f} H^{\circ}(NO_2) - \Delta_{f} H^{\circ}(NO) = 32.48 - 90.4 = -57.92 \ kJ \cdot mol^{-1} = -57920 \ J \cdot mol^{-1}$.
Next,calculate the standard Gibbs free energy change: $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} = -57920 - (298 \times -70.8) = -57920 + 21098.4 = -36821.6 \ J \cdot mol^{-1}$.
Finally,use the relation $\Delta G^{\circ} = -RT \ln K$ or $\Delta G^{\circ} = -2.303 RT \log K$.
$-36821.6 = -2.303 \times 8.314 \times 298 \times \log K$.
$\log K = \frac{36821.6}{5705.8} \approx 6.45$.
$K = 10^{6.45} \approx 2.8 \times 10^6$. Given the options,$3.162 \times 10^6$ is the intended answer.

(B) Let the initial concentration of $A$ be $a$ and $B$ be $1.5a$.
At equilibrium,let the concentration of $D$ be $x$.
According to the stoichiometry of the reaction $A + 2B \rightleftharpoons 2C + D$,the equilibrium concentrations are:
$[A] = a - x$
$[B] = 1.5a - 2x$
$[C] = 2x$
$[D] = x$
Given that $[A] = [C]$ at equilibrium:
$a - x = 2x \implies a = 3x \implies x = a/3$.
Substituting $x = a/3$ into the equilibrium concentrations:
$[A] = a - a/3 = 2a/3$
$[B] = 1.5a - 2(a/3) = 1.5a - 0.667a = 0.833a = 5a/6$
$[C] = 2(a/3) = 2a/3$
$[D] = a/3$
Now,calculate $K_C$:
$K_C = \frac{[C]^2 [D]}{[A] [B]^2} = \frac{(2a/3)^2 \times (a/3)}{(2a/3) \times (5a/6)^2} = \frac{(4a^2/9) \times (a/3)}{(2a/3) \times (25a^2/36)} = \frac{4a^3/27}{50a^3/108} = \frac{4}{27} \times \frac{108}{50} = \frac{4 \times 4}{50} = \frac{16}{50} = 0.32$.
Rounding to the nearest provided option,$K_C = 0.3$.

(B) Assuming the volume of the vessel is $1 \ L$.
The reaction is: $NO_2(g) + CO(g) \rightleftharpoons NO(g) + CO_2(g)$
Initial moles: $NO_2 = 1$,$CO = 2$,$NO = 0$,$CO_2 = 0$.
At equilibrium,$25 \%$ of $CO$ is consumed,so amount consumed $= 2 \times 0.25 = 0.5 \ mol$.
Equilibrium moles: $NO_2 = 1 - 0.5 = 0.5$,$CO = 2 - 0.5 = 1.5$,$NO = 0.5$,$CO_2 = 0.5$.
Since volume is $1 \ L$,molar concentrations are equal to the number of moles.
$K_c = \frac{[NO][CO_2]}{[NO_2][CO]} = \frac{0.5 \times 0.5}{0.5 \times 1.5} = \frac{0.5}{1.5} = 1/3$.
For this reaction,$\Delta n_g = (1+1) - (1+1) = 0$.
Since $\Delta n_g = 0$,$K_p = K_c(RT)^{\Delta n_g} = K_c(RT)^0 = K_c$.
Therefore,$K_p = 1/3$.

(B) The overall reaction is the sum of the two given equilibrium steps:
Step $1$: $Ag^{+} + NH_3 \rightleftharpoons [Ag(NH_3)]^{+}$,$K_1 = 1.6 \times 10^3$
Step $2$: $[Ag(NH_3)]^{+} + NH_3 \rightleftharpoons [Ag(NH_3)_2]^{+}$,$K_2 = 6.8 \times 10^3$
Adding these two equations gives the net reaction:
$Ag^{+} + 2 NH_3 \rightleftharpoons [Ag(NH_3)_2]^{+}$
The equilibrium constant $(K_{net})$ for the net reaction is the product of the equilibrium constants of the individual steps:
$K_{net} = K_1 \times K_2$
$K_{net} = (1.6 \times 10^3) \times (6.8 \times 10^3)$
$K_{net} = 10.88 \times 10^6 = 1.088 \times 10^7$

(A) The reaction is $SO_{2(g)} + NO_{2(g)} \rightleftharpoons SO_{3(g)} + NO_{(g)}$.
Initial concentrations are $[SO_2] = 1 \ M, [NO_2] = 1 \ M, [SO_3] = 1 \ M, [NO] = 1 \ M$.
Let $x$ be the change in concentration at equilibrium.
Equilibrium concentrations: $[SO_2] = 1-x, [NO_2] = 1-x, [SO_3] = 1+x, [NO] = 1+x$.
$K_c = \frac{[SO_3][NO]}{[SO_2][NO_2]} = \frac{(1+x)(1+x)}{(1-x)(1-x)} = 16$.
Taking the square root on both sides: $\frac{1+x}{1-x} = 4$.
$1+x = 4 - 4x \implies 5x = 3 \implies x = 0.6$.
Equilibrium concentrations:
$[SO_3] = 1 + 0.6 = 1.6 \ mol \ L^{-1}$.
$[SO_2] = 1 - 0.6 = 0.4 \ mol \ L^{-1}$.
Thus,the concentrations are $1.6$ and $0.4$ respectively.

(A) Let the number of moles of $SO_2$ at equilibrium be $x$. Then,the number of moles of $SO_3$ is $2x$.
Given volume $V = 10 \ L$ and $K_c = 100$.
The equilibrium expression is $K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]}$.
Substituting the concentrations: $[SO_3] = \frac{2x}{10}$,$[SO_2] = \frac{x}{10}$,and $[O_2] = \frac{n_{O_2}}{10}$.
$100 = \frac{(\frac{2x}{10})^2}{(\frac{x}{10})^2 \times (\frac{n_{O_2}}{10})}$
$100 = \frac{4x^2 / 100}{(x^2 / 100) \times (n_{O_2} / 10)}$
$100 = \frac{4}{n_{O_2} / 10} = \frac{40}{n_{O_2}}$
$n_{O_2} = \frac{40}{100} = 0.4 \ mol$.

(B) For the reaction: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
Degree of dissociation $\alpha = 0.5$
Total pressure $P = 1 \ atm$
Temperature $T = 60 + 273 = 333 \ K$
The equilibrium constant $K_p$ is given by the expression: $K_p = \frac{4\alpha^2 P}{1-\alpha^2}$
Substituting the values: $K_p = \frac{4(0.5)^2 \times 1}{1-(0.5)^2} = \frac{1}{0.75} = 1.333$
Standard free energy change $\Delta G^\circ = -RT \ln K_p$
$\Delta G^\circ = -8.314 \times 333 \times \ln(1.333) \approx -8.314 \times 333 \times 0.2874 \approx -796.8 \ J \ mol^{-1}$
Given the options,the closest value is $-763.8 \ J \ mol^{-1}$.

(C) The equilibrium constant $K_c$ for the reaction $CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$ is given by the expression:
$K_c = \frac{[CO][H_2O]}{[CO_2][H_2]}$
Given that the volume of the vessel is $1 \ L$,the molar concentrations are equal to the number of moles.
Substituting the given values:
$0.53 = \frac{0.25 \times x}{0.5 \times 0.6}$
$0.53 = \frac{0.25x}{0.3}$
$0.25x = 0.53 \times 0.3$
$0.25x = 0.159$
$x = \frac{0.159}{0.25} = 0.636$
Thus,the value of $x$ is $0.636$.

(A) The reaction is $AO_{2(g)} + BO_{2(g)} \rightleftharpoons AO_{3(g)} + BO_{(g)}$.
Given $K_c = 16$ and initial concentrations $[AO_2] = [BO_2] = [AO_3] = [BO] = 1.0 \ M$.
Let the change in concentration be $x$.
At equilibrium:
$[AO_2] = 1 - x$
$[BO_2] = 1 - x$
$[AO_3] = 1 + x$
$[BO] = 1 + x$
$K_c = \frac{[AO_3][BO]}{[AO_2][BO_2]} = \frac{(1+x)(1+x)}{(1-x)(1-x)} = 16$
Taking the square root of both sides:
$\frac{1+x}{1-x} = 4$
$1 + x = 4 - 4x$
$5x = 3 \implies x = 0.6$
Equilibrium concentration of $[BO] = 1 + x = 1 + 0.6 = 1.6 \ mol \ L^{-1}$.

(A) The dissociation reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
Given $K_c = 2 \times 10^{-2} \ mol \ L^{-1}$ and volume $V = 1.0 \ L$.
Let the initial moles of $PCl_5$ be $x$.

Reaction	$PCl_{5(g)}$	$PCl_{3(g)} + Cl_{2(g)}$
Initial moles	$x$	$0, 0$
Equilibrium moles	$x - 0.2$	$0.2, 0.2$

Since $V = 1.0 \ L$,the concentrations at equilibrium are $[PCl_5] = (x - 0.2) \ M$,$[PCl_3] = 0.2 \ M$,and $[Cl_2] = 0.2 \ M$.
The equilibrium constant expression is $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}$.
Substituting the values: $2 \times 10^{-2} = \frac{0.2 \times 0.2}{x - 0.2}$.
$0.02 = \frac{0.04}{x - 0.2} \implies x - 0.2 = \frac{0.04}{0.02} = 2$.
$x = 2 + 0.2 = 2.2 \ mol$.

(C) For the reaction $A_{(s)} \rightleftharpoons B_{(s)} + C_{(g)}$,the equilibrium constant $K_p = P_C$.
Statement $(a)$ is correct because the concentration of pure solids is taken as unity and does not appear in the equilibrium expression.
Statement $(b)$ is correct because $K_p$ is defined by the partial pressure of the gaseous product $C$ at equilibrium.
Statement $(c)$ is incorrect because $\Delta H$ (enthalpy of reaction) varies with temperature according to Kirchhoff's law.
Statement $(d)$ is correct because a catalyst provides an alternative pathway with lower activation energy but does not change the $\Delta H$ of the reaction.
Therefore,statements $(a), (b),$ and $(d)$ are correct.

(C) The given equilibrium reaction is $A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)}$.
At $t = 0$,moles are $1, 0, 0$ respectively.
At equilibrium,moles are $(1-\alpha), \alpha, \alpha$ respectively,where $\alpha = 0.33$.
Total moles at equilibrium $= 1 - \alpha + \alpha + \alpha = 1 + \alpha$.
Partial pressures are $P_{A} = \frac{1-\alpha}{1+\alpha} P$,$P_{B} = \frac{\alpha}{1+\alpha} P$,and $P_{C} = \frac{\alpha}{1+\alpha} P$.
$K_{p} = \frac{P_{B} \cdot P_{C}}{P_{A}} = \frac{(\frac{\alpha}{1+\alpha} P)(\frac{\alpha}{1+\alpha} P)}{(\frac{1-\alpha}{1+\alpha} P)} = \frac{\alpha^{2} P}{1-\alpha^{2}}$.
Given $\alpha = 0.33 \approx \frac{1}{3}$.
$K_{p} = \frac{(1/3)^{2} P}{1-(1/3)^{2}} = \frac{(1/9) P}{8/9} = \frac{P}{8}$.
Therefore,$P = 8 K_{p}$.

(B) The reaction quotient $(Q)$ is calculated using the formula $Q = \frac{[B][C]}{[A]^2}$.
Substituting the given concentrations: $Q = \frac{(3 \times 10^{-4})(3 \times 10^{-4})}{(3 \times 10^{-4})^2} = 1$.
Comparing $Q$ and $K_c$: $Q = 1$ and $K_c = 2 \times 10^{-3}$.
Since $Q > K_c$,the concentration of products is higher than the equilibrium concentration.
Therefore,the reaction will proceed in the backward direction (to the left) to reach equilibrium.

(B) The reaction is $2 A \rightleftharpoons B + C$.
The reaction quotient $Q$ is given by the expression: $Q = \frac{[B][C]}{[A]^2}$.
Substituting the given concentrations: $Q = \frac{(6 \times 10^{-5}) \times (6 \times 10^{-5})}{(6 \times 10^{-5})^2} = 1$.
Since $Q = 1$ and $K_{eq} = 2 \times 10^{-3}$,we observe that $Q > K_{eq}$.
When $Q > K_{eq}$,the reaction proceeds in the backward direction to reach equilibrium.

(A) The reaction is $0.5 C_{(s)} + 0.5 CO_{2(g)} \rightleftharpoons CO_{(g)}$.
Let the initial moles of $CO_{2(g)}$ be $1 \ mol$.
At equilibrium,$50 \%$ of $CO_{2(g)}$ is converted,so $0.5 \ mol$ of $CO_{2(g)}$ remains.
The amount of $CO_{(g)}$ formed is $1 \ mol$ (based on stoichiometry: $0.5 \ mol$ of $CO_2$ produces $1 \ mol$ of $CO$).
Total moles at equilibrium = $0.5 \ (CO_2) + 1 \ (CO) = 1.5 \ mol$.
Mole fraction of $CO_2$ $(x_{CO_2})$ = $0.5 / 1.5 = 1/3$.
Mole fraction of $CO$ $(x_{CO})$ = $1 / 1.5 = 2/3$.
Partial pressure $P_{CO_2} = (1/3) \times 12 \ atm = 4 \ atm$.
Partial pressure $P_{CO} = (2/3) \times 12 \ atm = 8 \ atm$.
$K_p = \frac{P_{CO}}{(P_{CO_2})^{0.5}} = \frac{8}{(4)^{0.5}} = \frac{8}{2} = 4 \ atm^{0.5}$.
Thus,the value of $K_p$ is $4$.

(A) The reaction is $A + B \rightleftharpoons C + D$.
Initial moles at $t = 0$: $A = 1 \ mol$,$B = 1 \ mol$,$C = 0 \ mol$,$D = 0 \ mol$.
At equilibrium,$40 \%$ of $B$ has reacted,so $0.4 \ mol$ of $B$ is consumed.
Moles at equilibrium: $A = (1 - 0.4) = 0.6 \ mol$,$B = (1 - 0.4) = 0.6 \ mol$,$C = 0.4 \ mol$,$D = 0.4 \ mol$.
Since the volume is $10 \ L$,the concentrations are $[A] = 0.6/10 = 0.06 \ M$,$[B] = 0.06 \ M$,$[C] = 0.04 \ M$,$[D] = 0.04 \ M$.
$K_C = \frac{[C][D]}{[A][B]} = \frac{0.04 \times 0.04}{0.06 \times 0.06} = \frac{0.0016}{0.0036} = \frac{16}{36} = \frac{4}{9} \approx 0.44$.

(A) The equilibrium constant expression for the reaction $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)}$ is $K_c = \frac{[Z]}{[X][Y]}$.
Given $K_c = 10^4 \ mol^{-1} \ L$.
At equilibrium,we are given $[X] = \frac{1}{2}[Y] = \frac{1}{2}[Z]$.
From this,we can express $[X]$ and $[Y]$ in terms of $[Z]$:
$[X] = \frac{1}{2}[Z]$
$[Y] = [Z]$
Substituting these into the $K_c$ expression:
$10^4 = \frac{[Z]}{(\frac{1}{2}[Z])([Z])} = \frac{[Z]}{\frac{1}{2}[Z]^2} = \frac{2}{[Z]}$
Therefore,$[Z] = \frac{2}{10^4} = 2 \times 10^{-4} \ mol \ L^{-1}$.

(C) The reaction is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Initially: $1 \ mol \ N_2, 3 \ mol \ H_2, 0 \ mol \ NH_3$.
At equilibrium: $(1-x) \ mol \ N_2, (3-3x) \ mol \ H_2, 2x \ mol \ NH_3$.
Given that at equilibrium,moles of $N_2 = 0.6$.
So,$1-x = 0.6 \implies x = 0.4$.
Now,calculate moles of each component at equilibrium:
Moles of $N_2 = 0.6 \ mol$.
Moles of $H_2 = 3 - 3(0.4) = 3 - 1.2 = 1.8 \ mol$.
Moles of $NH_3 = 2(0.4) = 0.8 \ mol$.
Total moles at equilibrium = $0.6 + 1.8 + 0.8 = 3.2 \ mol$.

(C) According to the van't Hoff equation,the relationship between the equilibrium constant $(K)$ and temperature $(T)$ is given by $\ln K = -\frac{\Delta H^0}{RT} + C$.
For the first reaction $(2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)})$,as temperature increases from $298 \ K$ to $700 \ K$,the value of $K_p$ decreases from $4.0 \times 10^{24}$ to $3.0 \times 10^4$. Since $K_p$ decreases with an increase in temperature,the reaction is exothermic,meaning $\Delta H_1^0 < 0$.
For the second reaction $(N_2O_{4(g)} \rightleftharpoons 2 NO_{2(g)})$,as temperature increases from $298 \ K$ to $500 \ K$,the value of $K_p$ increases from $0.98$ to $1700$. Since $K_p$ increases with an increase in temperature,the reaction is endothermic,meaning $\Delta H_2^0 > 0$.
Therefore,$\Delta H_1^0$ is negative and $\Delta H_2^0$ is positive.

(B) The relationship between the equilibrium constant $K$ and temperature $T$ is given by the equation: $\Delta G^{\circ} = -RT \ ln \ K$.
Substituting $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$,we get: $\Delta H^{\circ} - T \Delta S^{\circ} = -RT \ ln \ K$.
Dividing both sides by $-RT$,we obtain: $ln \ K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}$.
Comparing this with the linear equation $y = mx + c$,where $y = ln \ K$,$x = \frac{1}{T}$,$m = -\frac{\Delta H^{\circ}}{R}$,and $c = \frac{\Delta S^{\circ}}{R}$.
Thus,the intercept on the ordinate axis is $\frac{\Delta S^{\circ}}{R}$.

(B) The equilibrium reaction is $X_{2(g)} + Y_{2(g)} \rightleftharpoons 2Z_{(g)}$.
The equilibrium constant $K_{C}$ is calculated as:
$K_{C} = \frac{[Z]^2}{[X_2][Y_2]} = \frac{(9/1)^2}{(3/1) \times (3/1)} = \frac{81}{9} = 9$.
When $10 \ mol$ of $Z_{(g)}$ is added,the total moles of $Z$ become $9 + 10 = 19 \ mol$. The reaction shifts backward to re-establish equilibrium.
Let $x$ moles of $X_2$ and $Y_2$ be formed.
Equilibrium moles: $X_2 = 3+x$,$Y_2 = 3+x$,$Z = 19-2x$.
$K_{C} = \frac{(19-2x)^2}{(3+x)(3+x)} = 9$.
Taking the square root on both sides:
$\frac{19-2x}{3+x} = 3$.
$19-2x = 9+3x$.
$10 = 5x \Rightarrow x = 2$.
Moles of $Z$ at new equilibrium $= 19 - 2(2) = 15 \ mol$.

(B) The reaction is $A_{2(g)} \rightleftharpoons 2A_{(g)}$.
First,calculate the standard Gibbs free energy change for the reaction: $\Delta_r G^{\circ} = 2 \times \Delta G_f^{\circ}(A) - \Delta G_f^{\circ}(A_2) = 2 \times (-50.832) - (-100.00) = -101.664 + 100.00 = -1.664 \ kJ \ mol^{-1} = -1664 \ J \ mol^{-1}$.
Using the relation $\Delta_r G^{\circ} = -RT \ln K_p$:
$-1664 = -8.3 \times 300 \times \ln K_p$
$\ln K_p = \frac{1664}{2490} \approx 0.668 \approx 0.693$ (using the provided $\ln 2 = 0.693$ approximation).
Thus,$K_p = 2$.
For the dissociation $A_2 \rightleftharpoons 2A$,if $\alpha$ is the degree of dissociation,$K_p = \frac{4\alpha^2 P}{(1-\alpha^2)}$.
Given $P = 1 \ bar$,$2 = \frac{4\alpha^2}{1-\alpha^2} \implies 2 - 2\alpha^2 = 4\alpha^2 \implies 6\alpha^2 = 2 \implies \alpha^2 = \frac{1}{3} = 0.3333$.
$\alpha = (33.33 \times 10^{-2})^{1/2}$.
Comparing with $(x \times 10^{-2})^{1/2}$,we get $x = 33$.

(A) Reaction: $A(g) \rightleftharpoons B(g)$,$K_{c} = 4.0$.
Initial moles: $n_{A} = 1, n_{B} = 0, n_{He} = 1$.
Moles at equilibrium: $n_{A} = 1-x, n_{B} = x, n_{He} = 1$.
Total moles at equilibrium: $n_{total} = (1-x) + x + 1 = 2$.
Total pressure $P_{total} = \frac{n_{total}RT}{V} = \frac{2 \times 0.082 \times 400}{10} = 6.56 \text{ atm}$.
For the reaction $A(g) \rightleftharpoons B(g)$,$\Delta n = 1 - 1 = 0$,so $K_{p} = K_{c} = 4.0$.
$K_{p} = \frac{P_{B}}{P_{A}} = \frac{x_{B} \times P_{total}}{x_{A} \times P_{total}} = \frac{x / 2}{(1-x) / 2} = \frac{x}{1-x} = 4.0$.
$x = 4 - 4x \implies 5x = 4 \implies x = 0.8$.
Partial pressure of $He$: $P_{He} = \frac{n_{He}}{n_{total}} \times P_{total} = \frac{1}{2} \times 6.56 = 3.28 \text{ atm}$.
Partial pressure of $B(g)$: $P_{B} = \frac{n_{B}}{n_{total}} \times P_{total} = \frac{0.8}{2} \times 6.56 = 2.624 \text{ atm}$.
Thus,the partial pressures are $3.28 \text{ atm}$ and $2.624 \text{ atm}$ respectively.

(D) Let the initial pressure of $A$ be $P_0$. The reaction is $2A(g) \rightarrow 4B(g) + C(g)$.
At $t = 30$ min,let the decrease in pressure of $A$ be $2x$. Then the pressures are: $P_A = P_0 - 2x$,$P_B = 4x$,$P_C = x$.
The total pressure $P_t = (P_0 - 2x) + 4x + x = P_0 + 3x = 300$ mm Hg.
At $t = \infty$,the reaction is complete,so $P_A = 0$,which means $P_0 - 2x = 0 \Rightarrow P_0 = 2x$ or $x = P_0/2$.
The total pressure at $t = \infty$ is $P_\infty = P_0 + 3(P_0/2) = 2.5 P_0 = 600$ mm Hg.
Solving for $P_0$: $P_0 = 600 / 2.5 = 240$ mm Hg.
Substitute $P_0$ into the equation for $t = 30$ min: $240 + 3x = 300$.
$3x = 60 \Rightarrow x = 20$ mm Hg.
The pressure of $C(g)$ at $30$ minutes is $x = 20$ mm Hg.

(B) For the first equilibrium: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$,the equilibrium constant is $K_{p1} = P_{CO_2} = 0.08 \text{ atm}$.
For the second equilibrium: $C(s) + CO_2(g) \rightleftharpoons 2CO(g)$,the equilibrium constant is $K_{p2} = \frac{P_{CO}^2}{P_{CO_2}} = 2$.
Substituting the value of $P_{CO_2}$ from the first equilibrium into the second expression:
$P_{CO}^2 = K_{p2} \times P_{CO_2} = 2 \times 0.08 = 0.16 \text{ atm}^2$.
Taking the square root,$P_{CO} = \sqrt{0.16} = 0.4 \text{ atm}$.
Converting to the required format: $0.4 \text{ atm} = 4 \times 10^{-1} \text{ atm}$.
Thus,the partial pressure of $CO$ is $4 \times 10^{-1} \text{ atm}$.