Chemical Equilibrium

$ \begin{gathered} \mathrm{A} \rightleftharpoons \mathrm{~B} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.5 \mathrm{M} \quad 0.375 \mathrm{M} \,\,\,\,\,\,\,\,\,\,\,\,\,\text { (At equilibrium) }\\ \mathrm{~K}_{\mathrm{eq}}=\frac{[\mathrm{B}]_{\mathrm{eq}}}{[\mathrm{~A}]_{\mathrm{eq}}}=\frac{0.375}{0.5}=0.75 \end{gathered} $

Now 0.1 mole of A is added so reaction will move in forward direction.

$ \begin{aligned} & \mathrm{A} \rightleftharpoons \mathrm{~B} \\ & 0.6-\mathrm{x} \quad 0.375+\mathrm{x} \\ & \mathrm{~K}_{\mathrm{eq}}=0.75=\frac{0.375+\mathrm{x}}{0.6-\mathrm{x}} \\ & 0.45-0.75 \mathrm{x}=0.375+\mathrm{x} \\ & 1.75 \mathrm{x}=0.075 \\ & \mathrm{X}=\frac{0.075}{1.75}=\frac{3}{70}=0.043 \end{aligned} $

Moles of $\mathrm{A}=0.043=0.557$

Moles of $\mathrm{B}=0.418$

For the equilibrium

$ \mathrm{P_2(g)}+\mathrm{Q_2(g)}\rightleftharpoons 2\,\mathrm{PQ(g)} $

$ K_c=\frac{[\mathrm{PQ}]^2}{[\mathrm{P_2}][\mathrm{Q_2}]} $

Since $[X]=n_X/V$, the volume cancels:

$ K_c=\frac{(n_{\mathrm{PQ}}/V)^2}{(n_{\mathrm{P_2}}/V)(n_{\mathrm{Q_2}}/V)} =\frac{n_{\mathrm{PQ}}^2}{n_{\mathrm{P_2}}\,n_{\mathrm{Q_2}}} $

1) Find $K_c$ from the given equilibrium

Initially at equilibrium: $n_{\mathrm{P_2}}=2,\; n_{\mathrm{Q_2}}=2,\; n_{\mathrm{PQ}}=2$

$ K_c=\frac{2^2}{2\cdot 2}=1 $

2) After adding 1 mol each of $ \mathrm{P_2}, \mathrm{Q_2}$

Immediately after addition: $n_{\mathrm{P_2}}=3,\; n_{\mathrm{Q_2}}=3,\; n_{\mathrm{PQ}}=2$

Let the reaction proceed forward by extent $x$:

$ n_{\mathrm{P_2}}=3-x,\quad n_{\mathrm{Q_2}}=3-x,\quad n_{\mathrm{PQ}}=2+2x $

Apply $K_c=1$:

$ 1=\frac{(2+2x)^2}{(3-x)(3-x)}=\frac{(2+2x)^2}{(3-x)^2} $

$ (2+2x)^2=(3-x)^2 \Rightarrow 2+2x=3-x \Rightarrow 3x=1 \Rightarrow x=\frac13 $

So,

$ n_{\mathrm{P_2}}=3-\frac13=\frac83=2.67 $

$ n_{\mathrm{Q_2}}=\frac83=2.67 $

$ n_{\mathrm{PQ}}=2+\frac{2}{3}=\frac83=2.67 $

Answer: Option B $(2.67,\;2.67,\;2.67)$.

For any gaseous reaction, the relationship between equilibrium constants is

$ K_p = K_c (RT)^{\Delta n_g} $

where $\Delta n_g =$ (moles of gaseous products) $-$ (moles of gaseous reactants).

For the given reaction $ x \mathrm{~A}(g) \rightleftharpoons y \mathrm{~B}(g), $ we have

$ \Delta n_g = y - x. $

For reaction (i) : $\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{C}}$

Since $K_p > K_c$, using $K_p = K_c (RT)^{\Delta n_g}$, $\Delta n_g$ must be positive. So the number of moles of gaseous products is more than that of reactants.

$ \begin{aligned} & \Delta n_g>0 \\ & y-x>0 \\ & 85.87=2.586(0.0821 \times 400)^{y-x} \end{aligned} $

Here $R = 0.0821\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}$ and $T = 400\ \mathrm{K}$, so $RT = 0.0821 \times 400$.

From the equation $ 85.87 = 2.586 (0.0821 \times 400)^{y-x}, $ we get

Solving $\mathrm{y}-\mathrm{x} \simeq 1$

This means the products have one mole more than the reactants. So

$ y - x = 1 \Rightarrow y = x + 1. $

Taking the simplest whole numbers, we can choose $x = 1$ and $y = 2$ for case (i).

For reaction (ii) : $\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{c}}$

Since $K_p < K_c$, using $K_p = K_c (RT)^{\Delta n_g}$, $\Delta n_g$ must be negative. So the number of moles of gaseous products is less than that of reactants.

$ y-x<0 . $

Using the same relation, $ K_p = K_c (RT)^{y-x}, $ we write for reaction (ii)

$ 0.862 = 28.62 (0.0821 \times 400)^{y-x} $

Solving this, we find

$ y - x \simeq -1 $

So the reactants have one mole more than the products:

$ y - x = -1 \Rightarrow x = y + 1. $

Taking the simplest whole numbers, we can choose $x = 2$ and $y = 1$ for case (ii).

When xenon gas, an inert gas, is added to the equilibrium system $ \mathrm{PCl}_5(\mathrm{g}) \leftrightharpoons \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) $ at a constant temperature and pressure, it affects the equilibrium according to Le Chatelier's principle. The addition of an inert gas at constant pressure effectively increases the volume of the system, reducing the concentration of the gases involved.

Since the system is at constant pressure, adding an inert gas increases the total volume, which favors the side of the reaction with more moles of gas. In this reaction, there are two moles of gas on the right side ($\mathrm{PCl}_3$ and $\mathrm{Cl}_2$) compared to one mole on the left ($\mathrm{PCl}_5$). Thus, the equilibrium will shift toward the right to increase the number of moles and counteract the change, decreasing the concentration of $\mathrm{PCl}_5$ and increasing the concentrations of $\mathrm{PCl}_3$ and $\mathrm{Cl}_2$.

However, due to the increase in overall volume, the individual concentrations of all species $\left([\mathrm{PCl}_5], [\mathrm{PCl}_3], [\mathrm{Cl}_2]\right)$ will decrease as a result of more space being available for the same amount of gas particles.

Let's analyze both statements:

Statement I: "A catalyst cannot alter the equilibrium constant ($K_c$) of the reaction, temperature remaining constant."

A catalyst speeds up both the forward and reverse reactions equally, allowing the system to reach equilibrium faster.

However, it does not change the energy difference between reactants and products (i.e., the Gibbs free energy change), which directly determines the equilibrium constant.

Therefore, this statement is true.

Statement II: "A homogenous catalyst can change the equilibrium composition of a system, temperature remaining constant."

A homogeneous catalyst acts in the same phase as the reactants and, like any catalyst, it only helps the reaction reach equilibrium more quickly.

It does not alter the equilibrium position, meaning the relative concentrations of reactants and products at equilibrium remain unchanged if the temperature is constant.

Thus, this statement is false.

In summary:

Statement I is true.

Statement II is false.

The correct answer is: Option A.

a moles of $A(g)$ taken initially and at time

Now moles fraction of $\mathrm{A}(\mathrm{g}), \mathrm{B}(\mathrm{g})$ and $\mathrm{C}(\mathrm{g})$ are

$\begin{aligned} & \mathrm{X}_{\mathrm{A}}=\frac{\mathrm{a}-\mathrm{a} \alpha}{\mathrm{a}+\mathrm{a} \alpha}=\frac{1-\alpha}{1+\alpha} \\ & \mathrm{X}_{\mathrm{B}}=\frac{\mathrm{a} \alpha}{\mathrm{a}+\mathrm{a} \alpha}=\frac{\alpha}{1+\alpha} \\ & \mathrm{X}_{\mathrm{C}}=\frac{\mathrm{a} \alpha}{\mathrm{a}+\mathrm{a} \alpha}=\frac{\alpha}{1+\alpha} \end{aligned}$

Now if P is total pressure then partial pressure of $\mathrm{A}(\mathrm{g}), \mathrm{B}(\mathrm{g})$ and $\mathrm{C}(\mathrm{g})$ are

$\begin{aligned} & \mathrm{P}_{\mathrm{A}}=\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P} \\ & \mathrm{P}_{\mathrm{B}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \\ & \mathrm{P}_{\mathrm{C}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \end{aligned}$

$\begin{aligned} & \mathrm{K}_{\mathrm{P}}=\frac{\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}}{\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P}} \\ & \mathrm{~K}_{\mathrm{P}}=\frac{\alpha^2 \mathrm{P}}{1-\alpha^2} \end{aligned}$

As $K_P$ is only function of temperature.

So as $\mathrm{P} \uparrow \quad \alpha \downarrow$

Given equilibrium

$\mathrm{\mathop {C{O_{(g)}} + 3{H_2}(g)}\limits_{(n = 4)}} \rightleftharpoons \mathrm{\mathop {C{H_4}_{(g)} + {H_2}O(g)}\limits_{(n = 2)}}$

Temperature $\to$ Constant

Pressure $\to$ Increases by two fold.

It is based on Le Chatelier's principle. The principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

When pressure increases in a chemical equilibrium, the equilibrium shifts towards the side of the reaction with fewer moles of gas molecules. According to the principle, the reaction will favor the side that provides fewer gas molecules to counteract the increased pressure.

In this equilibrium, fewer gas molecules are on the produce side and on increase in pressure, equilibrium shifts towards products - forwarded reaction occurs. As a result, concentration of the products increases.

A) Concentration of reactants and products increases.

$-$ Correct

With increase in pressure, at constant temperature, the concentration of both reactants and products will increase because the same amount of gas is now in a smaller volume.

(B) Equilibrium will shift in forward direction

Correct

Increase in pressure causes product formation as the equilibrium shift is in the forward direction.

(C) Equilibrium constant increases since concentration of products increases.

Not correct.

The expression for equilibrium constant (K$_c$) for the given equilibrium can be written as

${K_c} = {{[C{H_4}][{H_2}O]} \over {[CO]{{[{H_2}]}^3}}}$

Equilibrium constant is not changed with increase in pressure and increase in concentration of reactants or products.

So, the equilibrium constant value is does not affected by change in concentration. If the change increase in concentration of product takes place, the equilibrium position will be shifted to counteract the change. So, equilibrium const value does not increase. Statement is incorrect.

(D) Equilibrium constant remains unchanged as concentration of reactants and products remain same.

State is incorrect.

In the reaction, increase in pressure causes changes in the concentration of rectants and products but the equilibrium constant will not change.

Statements (A) and (B) are correct.

$A B_2(g) \rightleftharpoons A B_{(g)}+\frac{1}{2} B_2({g})$

Degree of dissociation $\rightarrow x$ (small compared to unity)

Equilibrium constant interms of pressure $\rightarrow k_p$

Partial pressure of each gas $\rightarrow P_{A B_2}, P_{A B}$ and $P_{B_2}$

Total pressure $\rightarrow P$

ICE Table:

$\begin{aligned} & \text { partial pressure } \\ & \text { at equilibrium }\end{aligned}=\frac{\text { moles at equilibrium }}{\begin{array}{l}\text { total moles at } \\ \text { equilibrium }\end{array}} \times$ total pressure

Partial pressure of $A B_2(g)$

$P_{A B_2}=\left(\frac{1-x}{1+\frac{x}{2}}\right) P$

Partial pressure of $A B(g)$

$P_{A B}=\left(\frac{x}{1+\frac{x}{2}}\right) P$

Partial pressure of $B_2(g)$

$P_{B_2}=\left(\frac{\frac{x}{2}}{1+\frac{x}{2}}\right)^P$

$K_p$ for the equation $A_2(g) \rightleftharpoons A B_{(g)}+\frac{1}{2} B_2(g)$ can be written as

$k_p=\frac{P_{A B} P_{B_2}^{1 / 2}}{P_{A B_2}}$

$\begin{aligned} & =\frac{\frac{x}{1+\frac{x}{2}} P 1 \times\left(\frac{\frac{x}{2}}{1+\frac{x}{2}} P\right)^{1 / 2}}{\frac{1-x}{1+\frac{x}{2}} P} \\ & =\frac{x \times\left(\frac{x}{1+\frac{x}{2}} P\right)^{1 / 2}}{1-x}\end{aligned}$

$\begin{aligned} &\text { given, } x \lll 1\\ &\begin{aligned} & \text { So, } 1-x \approx 1 \\ & K_p=\frac{x \times\left(\frac{\frac{x}{2}}{1+\frac{x}{2}} p\right)^{1 / 2}}{1} \end{aligned} \end{aligned}$

$=x \times \frac{\left(\frac{x}{2}\right)^{1 / 2}}{\left(\frac{2+x}{2}\right)^{1 / 2}} p^{1 / 2}$

$=\frac{x \times \frac{x^{1 / 2}}{2^{1 / 2}} \times p^{1 / 2}}{\frac{(2+x)^{1 / 2}}{2^{1 / 2}}}$

$=\frac{x^{3 / 2} \times p^{1 / 2}}{2^{1 / 2}} \times \frac{2^{1 / 2}}{(2+x)^{1 / 2}}$

$\begin{gathered} x \lll 1 \\ 2+x \approx 2 \end{gathered}$

$\begin{aligned} & =\frac{x^{3 / 2} \times p^{1 / 2}}{2^{1 / 2}} \\ & =\left(\frac{x^3 p}{2}\right)^{1 / 2} \end{aligned}$

$\begin{aligned} &k_p=\left(\frac{x^3 p}{2}\right)^{1 / 2}\\ &\text { Square on both sides, } \end{aligned}$

$\begin{aligned} k_p^2 & =\frac{x^3 p}{2} \\ x^3 p & =2 k_p^2 \\ x^3 & =\frac{2 k_p^2}{p} \\ x & =\sqrt[3]{\frac{2 k_p^2}{p}} \end{aligned}$

Correct answer option (2) $\sqrt[3]{\frac{2 K_p^2}{p}}$

$\mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}$

Concentration of $\mathrm{H}_2$ and $\mathrm{I}_2$ decreases untill equilibrium condition and concentration of HI increases till equilibrium condition and after equilibrium concentration of all the reactant and products remain constant. Correct option (2)

To determine the relationship between the degree of dissociation ($x$) of $\mathrm{X}_2 \mathrm{Y}(\mathrm{g})$ and its equilibrium constant $K_p$, let's analyze the reaction:

$ \mathrm{X}_2 \mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{X}_2(\mathrm{g}) + \frac{1}{2} \mathrm{Y}_2(\mathrm{g}) $

Initial and Change in Moles

Initially, we start with 1 mole of $\mathrm{X}_2 \mathrm{Y}$.

At equilibrium:

$\mathrm{X}_2 \mathrm{Y}$ is $(1-x)$ moles.

$\mathrm{X}_2$ is $x$ moles.

$\frac{1}{2} \mathrm{Y}_2$ is $\frac{x}{2}$ moles.

Partial Pressures

The total pressure $\mathrm{P}_\text{total}$ is given by:

$\mathrm{P}_{\mathrm{X}_2 \mathrm{Y}} = \frac{1-x}{1+\frac{x}{2}} \times \mathrm{P}$

$\mathrm{P}_{\mathrm{X}_2} = \frac{x}{1+\frac{x}{2}} \times \mathrm{P}$

$\mathrm{P}_{\mathrm{Y}_2} = \frac{x/2}{1+\frac{x}{2}} \times \mathrm{P}$

Equilibrium Constant Expression

The equilibrium constant $K_p$ can be written in terms of partial pressures:

$ K_p = \frac{\left(\frac{x}{1+\frac{x}{2}} \cdot \mathrm{P}\right) \cdot \left(\frac{\frac{x}{2}}{1+\frac{x}{2}} \cdot \mathrm{P}\right)^{1/2}}{\frac{1-x}{1+\frac{x}{2}} \cdot \mathrm{P}} $

Simplifying the expression, considering $x$ to be very small, results in:

$ K_p = \left(\frac{x}{1-x}\right)\left(\frac{x}{2\left(1+\frac{x}{2}\right)}\right)^{1/2} \cdot \mathrm{P}^{\frac{1}{2}} $

For small $x$, $(1-x) \approx 1$, thus:

$ K_p = \frac{x^{3/2}}{\frac{1}{3}} \cdot \mathrm{P}^{\frac{1}{2}} $

$ x^{3/2} = \frac{K_p \cdot 2^{1/2}}{\mathrm{P}^{1/2}} $

Finally, by cubing both sides:

$ x^3 = \frac{2 \cdot K_p^2}{\mathrm{P}} $

$ x = \left(\frac{2 \cdot K_p^2}{\mathrm{P}}\right)^{1/3} $

Thus, the degree of dissociation $x$ relates to $K_p$ by the equation:

$ x = \left(\frac{2 \cdot K_p^2}{\mathrm{P}}\right)^{1/3} $

The reaction is:

$\mathrm{CO}_2(g) + \mathrm{C}(s) \rightleftharpoons 2\mathrm{CO}(g)$

Initially, the pressure of $\mathrm{CO}_2$ is 0.5 atm, and the pressure of $\mathrm{CO}$ is 0 atm. Let 'x' be the change in pressure of $\mathrm{CO}_2$ at equilibrium. Since the stoichiometric coefficient of CO is twice that of $\mathrm{CO}_2$, the pressure of CO formed at equilibrium is 2x.

At equilibrium:

Pressure of $\mathrm{CO}_2 = 0.5 - x$

Pressure of $\mathrm{CO} = 2x$

The total pressure at equilibrium is given as 0.8 atm. Therefore:

$(0.5 - x) + 2x = 0.8$

$0.5 + x = 0.8$

$x = 0.8 - 0.5 = 0.3$

So, at equilibrium:

Pressure of $\mathrm{CO}_2 = 0.5 - 0.3 = 0.2 \text{ atm}$

Pressure of $\mathrm{CO} = 2(0.3) = 0.6 \text{ atm}$

The equilibrium constant $K_p$ is given by:

$K_p = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}}$

$K_p = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8 \text{ atm}$

Therefore, the value of $K_p$ is 1.8 atm.

Given,

$ \begin{gathered} 2 \mathrm{AO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{AO}_3(g) ; \\ K_p=4 \times 10^{10} \end{gathered} $

Divide above reaction by 2

$ \begin{aligned} A \mathrm{O}_2+\frac{1}{2} \mathrm{O}_2 \rightleftharpoons A \mathrm{O}_3(g) & \\ K_p^{\prime}=\sqrt{K_p}=2 \times 1.0^5 & \end{aligned} $

$ \begin{aligned} &\text { Multiply above by } 3 \text {, }\\ &\begin{aligned} & 3 \mathrm{AO}_2+\frac{3}{2} \mathrm{O}_2 \rightleftharpoons 3 \mathrm{AO}_3 ; \\ & K_p^{\prime}=\left(2 \times 10^5\right)^3=8 \times 10^{15} \end{aligned} \end{aligned} $

Given reaction,

$ \mathrm{CO}_2(g)+\mathrm{H}_2(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g) ; K=0.53 $

At equilibrium, $[\mathrm{CO}]=0.25 \mathrm{~mol} / \mathrm{L}$,

$ \begin{aligned} {\left[\mathrm{CO}_2\right] } & =0.5 \mathrm{~mol} / \mathrm{L} \\ {\left[\mathrm{H}_2\right] } & =0.6 \mathrm{~mol} / \mathrm{L},\left[\mathrm{H}_2 \mathrm{O}\right]=x \end{aligned} $

Rate reaction; $K=\frac{[\mathrm{CO}]\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{CO}_2\right]\left[\mathrm{H}_2\right]}=\frac{0.25 \times x}{0.5 \times 0.6}$

$ x=0.636 $

$ \text { Given, reaction is } $

Partial pressure is given by

$ \begin{aligned} p_{\mathrm{N}_2 \mathrm{O}_4} & =\left(\frac{1-\alpha}{1+\alpha}\right) p ; \quad p_{\mathrm{NO}_2}=\frac{2 \alpha}{1+\alpha} \cdot p \\ K_p & =\frac{\left(p_{\mathrm{NO}_2}\right)^2}{p_{\mathrm{N}_2 \mathrm{O}_4}} \end{aligned} $

Substituting the values,

$ \alpha=\left(\frac{\frac{K_p}{p}}{4+\frac{K_p}{p}}\right)^{1 / 2} . $

The given reaction is:

$A \mathrm{O}_2(g)+B \mathrm{O}_2(g) \rightleftharpoons A \mathrm{O}_3(g)+B O(g)$

The equilibrium constant $K_c = 16$.

The volume of the flask is 1 L.

Initial moles of each species are 1 mole.

1. Calculate the initial concentrations:

   Since the volume is 1 L, the initial concentrations are numerically equal to the initial moles:

   $[AO_2]_0 = 1 \text{ M}$

   $[BO_2]_0 = 1 \text{ M}$

   $[AO_3]_0 = 1 \text{ M}$

   $[BO]_0 = 1 \text{ M}$

2. Calculate the reaction quotient $Q_c$ to determine the direction of the reaction:

   $Q_c = \frac{[AO_3]_0[BO]_0}{[AO_2]_0[BO_2]_0} = \frac{(1)(1)}{(1)(1)} = 1$

3. Compare $Q_c$ with $K_c$:

   Since $Q_c = 1$ and $K_c = 16$, we have $Q_c < K_c$. This means the reaction will proceed in the forward direction (to the right) to reach equilibrium.

4. Set up an ICE table (Initial, Change, Equilibrium) for moles:

   Let $x$ be the change in moles. Since the reaction proceeds to the right, reactants decrease by $x$ and products increase by $x$.

   Reaction: $A \mathrm{O}_2(g)+B \mathrm{O}_2(g) \rightleftharpoons A \mathrm{O}_3(g)+B O(g)$

   Initial (mol): 1 1 1 1

   Change (mol): $-x$ $-x$ $+x$ $+x$

   Equilibrium (mol): $(1-x)$ $(1-x)$ $(1+x)$ $(1+x)$

   Since the volume of the flask is 1 L, the equilibrium concentrations are numerically equal to the equilibrium moles.

5. Use the $K_c$ expression to solve for $x$:

   $K_c = \frac{[AO_3]_{eq}[BO]_{eq}}{[AO_2]_{eq}[BO_2]_{eq}}$

   $16 = \frac{(1+x)(1+x)}{(1-x)(1-x)}$

   $16 = \left(\frac{1+x}{1-x}\right)^2$

   Take the square root of both sides:

   $\sqrt{16} = \pm \frac{1+x}{1-x}$

   $4 = \pm \frac{1+x}{1-x}$

   Since the reaction proceeds in the forward direction, $x > 0$. Also, for the amounts to be positive, $1-x > 0$, so $x < 1$. Thus, both $(1+x)$ and $(1-x)$ must be positive. We take the positive root.

   $4 = \frac{1+x}{1-x}$

   $4(1-x) = 1+x$

   $4 - 4x = 1+x$

   $3 = 5x$

   $x = \frac{3}{5} = 0.6$

6. Calculate the equilibrium moles of each species:

   $n_{AO_2} = 1 - x = 1 - 0.6 = 0.4 \text{ mol}$

   $n_{BO_2} = 1 - x = 1 - 0.6 = 0.4 \text{ mol}$

   $n_{AO_3} = 1 + x = 1 + 0.6 = 1.6 \text{ mol}$

   $n_{BO} = 1 + x = 1 + 0.6 = 1.6 \text{ mol}$

7. Evaluate each statement:

   I. Total number of moles at equilibrium is 4.

   Total moles at equilibrium = $n_{AO_2} + n_{BO_2} + n_{AO_3} + n_{BO}$

   Total moles = $0.4 + 0.4 + 1.6 + 1.6 = 4.0 \text{ mol}$.

   This statement is correct. (Note: For this reaction, the number of moles of gaseous reactants equals the number of moles of gaseous products ($\Delta n_g = 0$), so the total number of moles remains constant throughout the reaction).

   II. At equilibrium, the ratio of moles of $A \mathrm{O}_2$ and $A \mathrm{O}_3$ is $1: 4$.

   Ratio = $n_{AO_2} : n_{AO_3} = 0.4 : 1.6$

   To simplify, divide both sides by 0.4:

   Ratio = $1 : 4$.

   This statement is correct.

   III. Total number of moles of $A \mathrm{O}_2$ and $B \mathrm{O}_2$ at equilibrium is 0.8.

   Total moles of $AO_2$ and $BO_2 = n_{AO_2} + n_{BO_2}$

   Total moles = $0.4 + 0.4 = 0.8 \text{ mol}$.

   This statement is correct.

All three statements (I, II, and III) are correct.

The final answer is $\boxed{D}$

At equilibrium $[A]=[B]$

$ \begin{array}{r} a-x=1.5 a-2 x \\ x=0.5 a \end{array} $

Equilibrium $[A]=a-0.5 a=0.5 a$

$ \begin{aligned} & {[B]=0.5 a,[C]=a,[D]=0.5 a} \\ & K_C=\frac{[C]^2[D]}{[A][B]^2} \Rightarrow \frac{(a)^2(0.5 a)}{(0.5 a)(0.5 a)^2} \\ & K_C=4 \end{aligned} $

The equilibrium reaction is

$ \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) $

$\Delta n=$ moles of gaseous product - moles of gaseous reactant

$ \begin{aligned} & \Delta n=2-(1+3)=-2 \\ & \frac{K_p}{K_c}=(R T)^{\Delta n} \end{aligned} $

or  $ \frac{K_c}{K_p}=(R T)^{-\Delta t} $

Substituting the values,

$ \begin{aligned} & 1076=(0.082 \times T)^{+2} \\ & T=400 \mathrm{~K} \end{aligned} $

The given reaction is,

$ \mathrm{N}_2 \mathrm{O}_5 \longrightarrow 2 \mathrm{NO}_2+\frac{1}{2} \mathrm{O}_2 $

Let $p_0=$ initial pressure of $\mathrm{N}_2 \mathrm{O}_5 x=$ pressure of $\mathrm{N}_2 \mathrm{O}_5$ decomposed. Initial state $=\mathrm{N}_2 \mathrm{O}_5$ has pressure $p_0$

Equilibrium state $=\mathrm{N}_2 \mathrm{O}_5$ has pressure

$ =p_0-x $

$\mathrm{NO}_2=2 x$ pressure, $\mathrm{O}_2=\frac{1}{2} x$ pressure

$ \begin{aligned} p_{\text {equilibrium }} & =\left(p_0-x\right)+2 x+\frac{1}{2} x \\ 480 & =p_0+\frac{3}{2} x \\ \Rightarrow \quad p_0 & =300 \mathrm{mmHg} \\ 180 & =\frac{3}{2} x, x=120 \mathrm{mmHg} \end{aligned} $

Fraction decomposed

$ \begin{aligned} & =\frac{\text { pressure of } \mathrm{N}_2 \mathrm{O}_5 \text { decomposed }}{\text { initial pressure of } \mathrm{N}_2 \mathrm{O}_5} \\ & =\frac{x}{p_0}=\frac{120}{300}=\frac{2}{5}=0.4 \end{aligned} $

$ \begin{aligned} &\text { For reaction, }\\ &\begin{aligned} & \mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \\ & K_p=\frac{\left(p_{\mathrm{NO}_2}\right)^2}{p_{\mathrm{N}_2 \mathrm{O}_4}} \Rightarrow 0.113=\frac{\left(p_{\mathrm{NO}_2}\right)^2}{0.2} \\ & p_{\mathrm{NO}_2}=0.15 \mathrm{~atm} \end{aligned} \end{aligned} $

I. $\frac{1}{2} \mathrm{~N}_2+\frac{3}{2} \mathrm{H}_2 \rightleftharpoons \mathrm{NH}_3$

Reverse and double the equation I

$ \begin{array}{rlr} & 2 \mathrm{NH}_3 \rightleftharpoons \mathrm{~N}_2+3 \mathrm{H}_2 \\ \Rightarrow & K_1^{\prime}=\frac{1}{K_1^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

II. $\quad 2 \mathrm{NO} \rightleftharpoons \mathrm{N}_2+\mathrm{O}_2$

Reverse the equation II

$ \begin{array}{ll} & \mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} \\ \Rightarrow & K_2^{\prime}=\frac{1}{K_2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Multiply by equation III

$ \begin{aligned} & & 3 \mathrm{H}_2+\frac{3}{2} \mathrm{O}_2 & \rightleftharpoons 3 \mathrm{H}_2 \mathrm{O} \\ \Rightarrow & & K_3^{\prime}=K_3^3 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Required equation is

$ \begin{aligned} & 2 \mathrm{NH}_3+\frac{5}{2} \mathrm{O}_2 \rightarrow 2 \mathrm{NO}+3 \mathrm{H}_2 \mathrm{O} \\ & K=\frac{1}{K_1^2} \cdot \frac{1}{K_2} \times K_3^2=\frac{K_3^3}{K_1^2 K_2} \end{aligned} $

Given reaction:

$ W + X \rightleftharpoons Y + Z $

Initial concentrations:

W starts at $2x$, X starts at $x$, and both Y and Z start at 0.

At equilibrium:

Let the amount of Y and Z formed be $y$. So, at equilibrium:

• W = $2x - y$
• X = $x - y$
• Y = $y$
• Z = $y$

Relationship between Y and X at equilibrium:

At equilibrium, the amount of Y is four times the amount of X. This gives us:

$[Y]_{eq} = 4[X]_{eq}$

So, $ y = 4(x - y) $

Solve for $y$:

$ y = 4x - 4y $

$ y + 4y = 4x $

$ 5y = 4x $

$ y = \frac{4}{5}x $

Write $K_c$ expression and substitute values:

The equilibrium constant is:

$ K_c = \frac{[Y][Z]}{[W][X]} $

Plug in the equilibrium values:

$ K_c = \frac{\left(\frac{4}{5}x\right)\left(\frac{4}{5}x\right)}{(2x - \frac{4}{5}x)(x - \frac{4}{5}x)} $

Simplify:

Numerator: $\left(\frac{4}{5}x\right)^2 = \frac{16}{25}x^2$

Denominator:

• $2x - \frac{4}{5}x = \frac{10x - 4x}{5} = \frac{6x}{5}$
• $x - \frac{4}{5}x = \frac{5x - 4x}{5} = \frac{x}{5}$

So denominator: $\frac{6x}{5} \cdot \frac{x}{5} = \frac{6x^2}{25}$

Therefore, $ K_c = \frac{16x^2/25}{6x^2/25} = \frac{16}{6} = 2.666 $

We are given these starting amounts:

$ \left[AO_2\right] = 1~\mathrm{mol/L} \\ \left[BO_2\right] = 1~\mathrm{mol/L} \\ \left[AO_3\right] = 1~\mathrm{mol/L} \\ \left[BO\right] = 1~\mathrm{mol/L} \\ K_c = 16 $

We need to find the equilibrium concentration of $AO_3$.

Step 1: Write the reaction equation.

$ AO_2~(g) + BO_2~(g) \rightleftharpoons AO_3~(g) + BO~(g) $

Step 2: Set up an ICE (Initial, Change, Equilibrium) table.

At the start (initial):
$[AO_2] = 1$, $[BO_2] = 1$, $[AO_3] = 1$, $[BO] = 1$

Let $x$ be the amount that reacts.

Change in concentrations: $\begin{array}{cccc} AO_2 & BO_2 & AO_3 & BO \\ - x & - x & + x & + x \\ \end{array}$

At equilibrium:
$[AO_2] = 1 - x$
$[BO_2] = 1 - x$
$[AO_3] = 1 + x$
$[BO] = 1 + x$

Step 3: Write the equilibrium expression and solve for $x$.

$ K_c = \frac{[AO_3][BO]}{[AO_2][BO_2]} = \frac{(1 + x)(1 + x)}{(1 - x)(1 - x)} = 16 $

Take the square root of both sides: $ \frac{1 + x}{1 - x} = 4 $

Solve for $x$:

$1 + x = 4(1 - x)$

$1 + x = 4 - 4x$

$1 + x + 4x = 4$

$1 + 5x = 4$

$5x = 3$

$x = 0.6$

Step 4: Find the equilibrium concentration of $AO_3$.

$[AO_3] = 1 + x = 1 + 0.6 = 1.6~\mathrm{mol/L}$

Given reaction:

$ \mathrm{A_2O_4(g)} \rightleftharpoons 2\mathrm{AO_2(g)} $

At 298 K, the equilibrium constant in concentration units is $ K_c = x\ \mathrm{mol\,L^{-1}} $.

We need to find the approximate value of $ K_P $.

Relation between $ K_P $ and $ K_C $

For a gaseous reaction:

$ K_P = K_C (RT)^{\Delta n} $

where

$\Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)}$

Step 1: Determine $\Delta n$

$ \begin{align*} \text{Moles of gaseous products} &= 2 \\ \text{Moles of gaseous reactants} &= 1 \\ \Rightarrow \Delta n &= 2 - 1 = +1 \end{align*} $

Step 2: Substitute into the relation

$ K_P = K_C (RT)^{1} = K_C (RT) $

Step 3: Plug in the given values

At $T = 298\,\text{K}$,

$R = 0.082\, \mathrm{L \, atm \, mol^{-1} \, K^{-1}}$

$ RT = 0.082 \times 298 = 24.4\,\mathrm{L \, atm \, mol^{-1}} $

Step 4: Therefore,

$ K_P = K_C \times 24.4 $

$ \boxed{K_P = 24.4x} $

✅ Final Answer: Option A — $ 24.4x $

Gibbs free energy is given by

$ \begin{aligned} & \Delta G^{\circ}=-R T \ln K \\ \Rightarrow \quad & \ln K=-\frac{\Delta G^{\circ}}{R T} \\ \ln K= & \frac{-165.469}{8.3 \times 298} \approx-66.8 \\ & K=e^{-66.8} \approx 9.75 \times 10^{-30} \approx 10^{-29} \end{aligned} $

So, $K \cong 10^{-29}$

Volume occupied by $B$ atom $=20 \%$ of total volume

$ \begin{aligned} =20 \% \text { of } 22.7 \mathrm{~L} & =4.54 \mathrm{~L} \\ \text { Remaining volume } & =(22.7-4.54) \mathrm{L} \\ & =18.16 \mathrm{~L} \end{aligned} $

Mole fraction of $B=\chi_B=\frac{4.54}{22.7}=0.2$

Mole fraction of $B_2=\chi_{B_2}=\frac{18.16}{22.7}=0.8$

Partial pressure of $B=p_B=0.2 \times 1=0.2$ bar

Partial pressure of

$ \begin{aligned} B_2=p_{B_2} & =0.8 \times 1=0.8 \text { bar } \\ & K_p=\frac{p_B^2}{p_{B_2}}=\frac{0.2^2}{0.8}=\frac{0.04}{0.8}=0.05 \end{aligned} $

To find the equilibrium constant for the overall reaction $\mathrm{X} \rightleftharpoons \mathrm{W}$, we need to combine the equilibrium constants for the individual reactions given. Let's analyze this step-by-step:

The given reactions and their equilibrium constants are:

$\begin{aligned} & \mathrm{X} \rightleftharpoons \mathrm{Y} \quad K_1 = 1.0 \\ & \mathrm{Y} \rightleftharpoons \mathrm{Z} \quad K_2 = 2.0 \\ & \mathrm{Z} \rightleftharpoons \mathrm{W} \quad K_3 = 4.0 \end{aligned}$

The equilibrium constant for the overall reaction $\mathrm{X} \rightleftharpoons \mathrm{W}$ is the product of the equilibrium constants for the individual steps. This is because the equilibrium constant for a composite reaction is the product of the equilibrium constants for the sequential reactions that lead to the composite reaction.

So, we have:

$K_{\text{overall}} = K_1 \cdot K_2 \cdot K_3$

Now, substituting the given values:

$K_{\text{overall}} = 1.0 \cdot 2.0 \cdot 4.0 = 8.0$

Therefore, the equilibrium constant for the reaction $\mathrm{X} \rightleftharpoons \mathrm{W}$ is 8.0, which corresponds to Option B.

To solve this problem, we need to understand the relationship between the equilibrium constant in terms of pressure, $K_P$, and the equilibrium constant in terms of concentration, $K_C$. The relationship between these two constants for a general reaction is given by:

$ K_P = K_C (RT)^{\Delta n} $

where:

$\Delta n$ is the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants),

$R$ is the ideal gas constant, and

$T$ is the temperature in Kelvin.

For the given reaction:

$ \mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})} $

we need to calculate $\Delta n$.

On the reactants side, we have:

• 1 mole of $\mathrm{CO}$ (g)
• 0.5 moles of $\mathrm{O}_2$ (g)

So, the total number of moles of reactants is:

$ 1 + \frac{1}{2} = \frac{3}{2} = 1.5 $

On the products side, we have:

• 1 mole of $\mathrm{CO}_2$ (g)

The total number of moles of products is:

$ 1 $

Therefore, $\Delta n$ is:

$ \Delta n = \text{moles of products} - \text{moles of reactants} = 1 - 1.5 = -0.5 $

Now, we can use the relationship between $K_P$ and $K_C$:

$ K_P = K_C (RT)^{\Delta n} $

Substituting $\Delta n = -0.5$, we get:

$ K_P = K_C (RT)^{-0.5} $

or equivalently:

$ \frac{K_P}{K_C} = (RT)^{-0.5} $

Therefore, the correct option is:

Option D: $\frac{1}{\sqrt{R T}}$

At $-20^{\circ} \mathrm{C}$ and a pressure of $1 \mathrm{~atm}$, a cylinder contains equal quantities of $\mathrm{H}_2, \mathrm{I}_2,$ and $\mathrm{HI}$ molecules. The equilibrium constant for the reaction

$\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$

denoted as $\mathrm{K}_{\mathrm{p}}$, is given by the expression $x \times 10^{-1}$.

To solve for $x$, use the provided equilibrium expression:

$\mathrm{K_{P} = \mathrm{K_C}} =\frac{\left[\mathrm{HI}\right]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=1 = 10 \times 10^{-1}$

Therefore, $x = 10$.

To determine the correctness of the statements, we'll analyze each one individually by applying principles of solubility, common ion effect, and chemical equilibria.

Statement I:

On passing $\mathrm{HCl}_{(g)}$ through a saturated solution of $\mathrm{BaCl}_2$ at room temperature, white turbidity appears.

Analysis:

Dissolution of HCl Gas:

When $\mathrm{HCl}_{(g)}$ is bubbled through water, it dissolves and dissociates completely:

$ \mathrm{HCl}_{(aq)} \rightarrow \mathrm{H}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} $

This increases the concentration of $\mathrm{Cl}^-$ ions in the solution.

Effect on BaCl₂ Solubility:

The solubility equilibrium of $\mathrm{BaCl}_2$ in water is:

$ \mathrm{BaCl}_2 \leftrightarrow \mathrm{Ba}^{2+}_{(aq)} + 2\mathrm{Cl}^-_{(aq)} $

Adding more $\mathrm{Cl}^-$ shifts the equilibrium to the left (Le Chatelier's Principle), causing $\mathrm{BaCl}_2$ to precipitate.

The precipitation of $\mathrm{BaCl}_2$ manifests as a white turbidity.

Conclusion:

Statement I is correct.

Statement II:

When $\mathrm{HCl}$ gas is passed through a saturated solution of $\mathrm{NaCl}$, sodium chloride is precipitated due to common ion effect.

Analysis:

Dissolution of HCl Gas:

Similar to before, $\mathrm{HCl}$ increases $\mathrm{Cl}^-$ ion concentration.

Effect on NaCl Solubility:

The solubility equilibrium of $\mathrm{NaCl}$ is:

$ \mathrm{NaCl} \leftrightarrow \mathrm{Na}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} $

However, $\mathrm{NaCl}$ is highly soluble in water, and its solubility is not significantly affected by the common ion effect from $\mathrm{Cl}^-$.

The solubility product ($K_{sp}$) of $\mathrm{NaCl}$ is large, and the addition of $\mathrm{Cl}^-$ ions does not cause $\mathrm{NaCl}$ to precipitate under normal conditions.

No precipitation occurs; the solution remains clear.

Conclusion:

Statement II is incorrect.

Final Answer:

Statement I is correct but Statement II is incorrect.

For the reaction :

$\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}+3 \mathrm{CO}_{(\mathrm{g})} \rightleftharpoons \mathrm{Fe}_{(\mathrm{l})}+3 \mathrm{CO}_{2(\mathrm{~g})}$

Addition or removal of $\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{s})}$ and/or $\mathrm{Fe}_{\text {(l) }}$ will not affect the equilibrium quotient and the equilibrium.

The reaction

$2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3$

can be formed from the given reaction by reverting it and multiplying coefficients by 2.

Thus,

$\begin{aligned} \mathrm{K}_c^{\prime} & =\mathrm{K}_{\mathrm{c}}^{-2}=\frac{1}{\mathrm{~K}_{\mathrm{c}}^2}=\left(\frac{1}{4.9 \times 10^{-2}}\right)^2 \\ & =416 \end{aligned}$

$\begin{aligned} & \mathrm{A}_{(\mathrm{g})} \rightleftharpoons \mathrm{B}_{(\mathrm{g})}+\frac{\mathrm{C}}{2}(\mathrm{~g}) \\ & \mathrm{t}=\mathrm{t}_{\mathrm{eq}} \quad(1-\alpha) \quad \alpha \quad \frac{\alpha}{2} \end{aligned}$

$\mathrm{P}_{\mathrm{B}}=\frac{\alpha}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P}, \mathrm{P}_{\mathrm{A}}=\frac{(1-\alpha)}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P}, \mathrm{P}_{\mathrm{C}}=\frac{\frac{\alpha}{2}}{\left(1+\frac{\alpha}{2}\right)} . \mathrm{P}$

$\mathrm{K_P=\frac{P_B \cdot P_C^{\frac{1}{2}}}{P_A}}$

$\mathrm{=\frac{(\alpha)^{\frac{3}{2}}(P)^{\frac{1}{2}}}{(1-\alpha)(2+\alpha)^{\frac{1}{2}}}}$

$\begin{aligned} & \mathrm{K}_{\mathrm{C}}=\frac{\text { Products ion conc. }}{\text { Reactants ion conc. }} \\ & \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]} \end{aligned}$

Given the equilibrium constants for the reactions:

$2A(g) \rightleftharpoons B(g) + C(g)$ with $K_1 = 16$

$2B(g) + C(g) \rightleftharpoons 2D(g)$ with $K_2 = 25$

We need to find the equilibrium constant $K$ for the reaction:

$ A(g) + \frac{1}{2}B(g) \rightleftharpoons D(g) $

To determine this, we first manipulate the given reactions as follows:

Divide both sides of the first reaction by 2:

$ A(g) \rightleftharpoons \frac{1}{2}B(g) + \frac{1}{2}C(g) $

When you perform this division, the equilibrium constant becomes the square root of the original. Therefore, the equilibrium constant for this modified reaction is $\sqrt{K_1} = \sqrt{16} = 4$.

Next, divide both sides of the second reaction by 2:

$ B(g) + \frac{1}{2}C(g) \rightleftharpoons D(g) $

Similarly, the equilibrium constant for this modified reaction is $\sqrt{K_2} = \sqrt{25} = 5$.

By adding these two modified reactions, we obtain the target reaction:

$ A(g) + \frac{1}{2}B(g) \rightleftharpoons D(g) $

The equilibrium constant for the resulting reaction is the product of the equilibrium constants of the individual modified reactions:

$ K = \sqrt{K_1} \cdot \sqrt{K_2} = 4 \cdot 5 = 20 $

Thus, the equilibrium constant $K$ for the reaction $A(g) + \frac{1}{2}B(g) \rightleftharpoons D(g)$ at temperature $T(K)$ is 20.

The given reaction is,

$ \begin{array}{ccc} & \mathrm{PCl}_5 & \rightleftharpoons \mathrm{PCl}_3(g) & +\mathrm{Cl}_2(g) \\\\ \text { Moles at initial } & x & 0 & 0 \\\\ \text { Moles at equilibrium } & x-0.2 & 0.2 & 0.2 \end{array} $

Equilibrium constant of above reaction is,

$ \begin{aligned} K_C&=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]} \\\\ 2 \times 10^{-2}&=\frac{[0.2][0.2]}{[x-0.2]} \\\\ \Rightarrow \quad x&=2.2 \mathrm{~mol} \end{aligned} $

To determine the equilibrium concentration of $ BO(g) $, we need to consider the reaction and the given equilibrium constant:

$ AO_2(g) + BO_2(g) \rightleftharpoons AO_3(g) + BO(g) $

The reaction quotient, $ Q_c $, is given by:

$ Q_c = \frac{[AO_3][BO]}{[AO_2][BO_2]} $

Given $ K_c = 16 $ at equilibrium, the system satisfies:

$ K_c = \frac{[AO_3][BO]}{[AO_2][BO_2]} = 16 $

Let $ x $ be the change in concentration of products and reactants at equilibrium. The initial concentration for all species is $ 1 \, \text{mol/L} $, thus:

Initial concentrations: $[AO_2] = [BO_2] = [AO_3] = [BO] = 1 \, \text{mol/L}$

Change in concentrations:

$[AO_2] = [BO_2] = 1 - x $

$[AO_3] = [BO] = 1 + x $

At equilibrium:

$ K_c = \frac{(1 + x)(1 + x)}{(1 - x)(1 - x)} = 16 $

$ \frac{(1 + x)^2}{(1 - x)^2} = 16 $

Taking the square root of both sides:

$ \frac{1 + x}{1 - x} = 4 $

Solving for $ x $:

$ 1 + x = 4 - 4x $

$ 1 + 5x = 4 $

$ 5x = 3 $

$ x = 0.6 $

Thus, the equilibrium concentration of $ BO $ is:

$ [BO] = 1 + x = 1 + 0.6 = 1.6 \, \text{mol/L} $

Therefore, the correct option is:

Option A: 1.6

To determine $ K_p $ for the reaction $ A_2B_2(g) \rightleftharpoons A_2(g) + B_2(g) $ at 300 K, we are provided with the following information:

Equilibrium constant $ K_C = 100 \, \text{mol L}^{-1} $

Temperature $ T = 300 \, \text{K} $

Gas constant $ R = 0.082 \, \text{L atm mol}^{-1} \text{K}^{-1} $

The relationship between $ K_C $ and $ K_p $ is given by the equation:

$ K_p = K_C (RT)^{\Delta n} $

Where $\Delta n$ is the change in the number of moles of gas, calculated as the difference between the total moles of gaseous products and reactants:

$ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 1 = 1 $

Now, substituting these values into the formula:

$ K_p = 100 \times (0.082 \times 300)^1 $

$ K_p = 100 \times 24.6 = 2460 \, \text{atm} $

Thus, the value of $ K_p $ for the reaction at 300 K is $ 2460 \, \text{atm} $.

To find the equilibrium constant $ K_p $ for the decomposition of $ AB $ at 800 K, follow these steps:

Given Equilibrium Reaction:

$ A_2(g) + B_2(g) \rightleftharpoons 2AB(g) $

Concentrations at Equilibrium:

$[A_2] = 1.5 \times 10^{-3} \, \text{M}$

$[B_2] = 2.1 \times 10^{-3} \, \text{M}$

$[AB] = 1.4 \times 10^{-3} \, \text{M}$

Decomposition Reaction:

$ 2AB(g) \rightleftharpoons A_2(g) + B_2(g) $

Expression for $ K_p $:

$ K_p = \frac{[A_2][B_2]}{[AB]^2} $

Calculation:

Substitute the concentration values into the expression for $ K_p $:

$ K_p = \frac{(1.5 \times 10^{-3})(2.1 \times 10^{-3})}{(1.4 \times 10^{-3})^2} $

Calculate:

$ K_p = \frac{(1.5 \times 2.1) \times 10^{-6}}{1.96 \times 10^{-6}} $

$ K_p = \frac{3.15 \times 10^{-6}}{1.96 \times 10^{-6}} $

$ K_p = 1.6 $

Thus, the value of $ K_p $ for the decomposition of $ AB $ at 800 K is 1.6.

Given the reaction:

$ \mathrm{H}_2 + \mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI} $

Initial Conditions:

At $ t = 0 $, the initial concentrations of the reactants are:

$\mathrm{H}_2 = 15 \text{ moles}$

$\mathrm{I}_2 = 5.2 \text{ moles}$

At Equilibrium:

It is given that the number of moles of $\mathrm{HI}$ at equilibrium is 10.

Since the balanced equation shows that 2 moles of $\mathrm{HI}$ are formed per mole of reactants, we can set:

$2x = 10$, thus $x = 5 \text{ moles}$

Therefore, at equilibrium:

$\mathrm{H}_2 = 15 - 5 = 10 \text{ moles}$

$\mathrm{I}_2 = 5.2 - 5 = 0.2 \text{ moles}$

Calculating the Equilibrium Constant:

The equilibrium constant $(K_C)$ for the dissociation of $\mathrm{HI}$ is calculated by:

$ K_C = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} $

Plugging in the equilibrium concentrations:

$ K_C = \frac{(10)^2}{10 \times 0.2} = 50 $

Therefore, the equilibrium constant for the dissociation of $\mathrm{HI}$ is 50.

$ \begin{gathered} K_C=\frac{\alpha}{1-\alpha}=39 \\ 39-39 \alpha=\alpha \\ {\left[B_2\right]=\alpha=\frac{39}{40}=0.975} \\ {\left[A_2\right]=1-\alpha=0.025} \end{gathered} $

The reaction given is:

$ \mathrm{H}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{HBr}(g) $

The equilibrium constant $ K_C $ is $ 1.6 \times 10^5 $.

When 10 bar of HBr is introduced into a sealed vessel, the equilibrium of the decomposition reaction can be expressed as:

$ 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_2(g) + \mathrm{Br}_2(g) $

For this reverse reaction, the equilibrium constant $ K_C $ becomes:

$ K_C = \frac{1}{1.6 \times 10^5} $

Since $\Delta n_g = 0$, we have:

$ K_p = K_C = \frac{1}{1.6 \times 10^5} $

For the decomposition:

Initial pressure of HBr = 10 bar

At equilibrium, pressure of HBr = $10 - 2p$

The expression for the equilibrium constant in terms of pressure is:

$ K_p = \frac{p^2}{(10 - 2p)} $

Equating and solving:

$ \frac{1}{1.6 \times 10^5} = \frac{p^2}{(10 - 2p)} $

Solving this gives $ p = 0.025 \, \text{bar} $.

Thus, the equilibrium pressure of HBr is:

$ p_{\mathrm{HBr}} = 10 - 2(0.025) = 9.95 \, \text{bar} $

The given reaction is:

$ A_2(g) \stackrel{T(K)}{\rightleftharpoons} B_2(g) $

with an equilibrium constant $ K_C = 99 $.

We are interested in finding the equilibrium concentration of $ B_2(g) $ when starting with 2 moles of $ B_2(g) $ in a 1 L flask. This is effectively the reverse of the initial reaction:

$ B_2(g) \stackrel{T(K)}{\rightleftharpoons} A_2(g) $

For this reverse reaction, the equilibrium constant is:

$ K_C' = \frac{1}{K_C} = \frac{1}{99} $

Initially, at $ t = 0 $, the concentration of $ B_2 $ is:

$ [B_2] = 2 \, \text{mol/L} $

At equilibrium, let $ \alpha $ be the degree of dissociation of $ B_2 $ into $ A_2 $. Thus, the concentrations become:

$ [B_2] = 2 - 2\alpha $

$ [A_2] = 2\alpha $

Using the expression for the reverse reaction:

$ K_C' = \frac{[A_2]}{[B_2]} = \frac{2\alpha}{2 - 2\alpha} = \frac{1}{99} $

Solving for $ \alpha $:

$ \frac{2\alpha}{2 - 2\alpha} = \frac{1}{99} $

$ \Rightarrow 2\alpha \times 99 = 2 - 2\alpha $

$ \Rightarrow 198\alpha = 2 - 2\alpha $

$ \Rightarrow 200\alpha = 2 $

$ \Rightarrow \alpha = \frac{2}{200} = 0.01 $

Therefore, the equilibrium concentration of $ B_2 $ is:

$ [B_2]_{\text{eq}} = 2 - 2 \times 0.01 = 1.98 \, \text{mol/L} $