Chemical Equilibrium

Given reaction:

$\mathrm{X_2(g) + Y_2(g) \rightleftharpoons 2Z(g)}$

1) Find $K_c$ from the first equilibrium

At equilibrium in a $1\ \mathrm{L}$ flask:

$ [\mathrm{X_2}] = 3,\quad [\mathrm{Y_2}] = 3,\quad [\mathrm{Z}] = 9 $

$ K_c=\frac{[\mathrm{Z}]^2}{[\mathrm{X_2}][\mathrm{Y_2}]} =\frac{9^2}{3\times 3}=\frac{81}{9}=9 $

2) After adding $10$ mol of $Z$

New moles immediately (before shift):

$ \mathrm{X_2}=3,\quad \mathrm{Y_2}=3,\quad \mathrm{Z}=9+10=19 $

Since $Z$ is added, equilibrium shifts left. Let the backward reaction proceed by $x$ moles:

• $\mathrm{X_2}$ increases by $x$

• $\mathrm{Y_2}$ increases by $x$

• $\mathrm{Z}$ decreases by $2x$

So at new equilibrium:

$ [\mathrm{X_2}] = 3+x,\quad [\mathrm{Y_2}] = 3+x,\quad [\mathrm{Z}] = 19-2x $

3) Apply equilibrium constant again

$ K_c=\frac{(19-2x)^2}{(3+x)(3+x)}=9 $

$ (19-2x)^2=9(3+x)^2 $

Taking positive square root (concentrations are positive):

$ 19-2x = 3(3+x)=9+3x $

$ 19-9 = 3x+2x $

$ 10=5x \Rightarrow x=2 $

Hence,

$ \text{New moles of } Z = 19-2(2)=19-4=15 $

Nearest integer = $15$.

$\mathrm{k}_{\mathrm{P}}=\frac{\left(\frac{\alpha}{2}\right)^{1 / 2}\left(\frac{3 \alpha}{2}\right)^{3 / 2}}{(1-\alpha)}\left[\frac{\mathrm{P}_{\mathrm{T}}}{1+\alpha}\right]^1 \quad\left[\because \mathrm{P}_{\mathrm{T}}=\sqrt{3} \mathrm{~atm}\right]$

$\Rightarrow $ $9=\frac{\left(\frac{\alpha}{2}\right)^{1 / 2}\left(\frac{3 \alpha}{2}\right)^{3 / 2}}{(1-\alpha)} \times \frac{(3)^{1 / 2}}{1+\alpha}$

$\Rightarrow $ $9=\frac{9\left(\frac{\alpha}{2}\right)^2}{1-\alpha^2}$

$\Rightarrow $ $1-\alpha^2=\frac{\alpha^2}{4}$

$\Rightarrow $ $\frac{5 \alpha^2}{4}=1$

$\Rightarrow $ $\alpha^2=0.8$

$\Rightarrow $ $\alpha=(0.8)^{1 / 2}$

$\Rightarrow $ $\alpha=\left[\frac{1}{0.8}\right]^{-1 / 2}$

$\begin{aligned} & \alpha=\left[125 \times 10^{-2}\right]^{-1 / 2} \\\\ &\therefore x=125 .\end{aligned}$

$\begin{aligned} & -1.664 \times 10^3=-8.3 \times 300 \ln \mathrm{~K}_{\mathrm{p}} \\ & \ln \mathrm{K}_{\mathrm{p}}=0.693 \\ & \mathrm{~K}_{\mathrm{p}}=2 \\ & 2=\frac{4 \alpha^2 \mathrm{P}_0}{1-\alpha^2} \\ & \alpha=\frac{1}{\sqrt{3}} \\ & \alpha=\left(\frac{100}{3} \times 10^{-2}\right)^{1 / 2} \\ & =\left(33.33 \times 10^{-2}\right)^{1 / 2}\end{aligned}$

First, let's consider the first equilibrium:

$ \mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{~s})+\mathrm{CO}_2(\mathrm{~g}) \quad K_{p_1} = 0.08 \mathrm{~atm} $

As we know from our study of chemical equilibrium, for the equilibrium constant expression $K_p$, we only consider the partial pressures of the gaseous species. The concentrations of pure solids (like $\mathrm{CaCO}_3$ and $\mathrm{CaO}$) are considered to be unity (1).

So, for this reaction, the expression for $K_{p_1}$ is:

$ K_{p_1} = p_{CO_2} $

We are given that $K_{p_1} = 0.08 \mathrm{~atm}$.

Therefore, the partial pressure of $\mathrm{CO}_2$ at equilibrium is fixed by this reaction.

$ p_{CO_2} = 0.08 \mathrm{~atm} $

This is an important first step. The presence of both $\mathrm{CaCO}_3(\mathrm{~s})$ and $\mathrm{CaO}(\mathrm{~s})$ ensures that the partial pressure of $\mathrm{CO}_2$ is maintained at this value.

Now, let's look at the second equilibrium:

$ \mathrm{C}(\mathrm{~s})+\mathrm{CO}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{~g}) \quad K_{p_2} = 2 \mathrm{~atm} $

Again, we write the expression for $K_{p_2}$. We consider the solid carbon, $\mathrm{C}(\mathrm{s})$, to have an activity of 1.

$ K_{p_2} = \frac{(p_{CO})^2}{p_{CO_2}} $

Since both reactions are taking place in the same vessel and are at equilibrium, the partial pressure of the common species, $\mathrm{CO}_2$, must be the same for both.

We have already found that $p_{CO_2} = 0.08 \mathrm{~atm}$ from the first equilibrium. We also know that $K_{p_2} = 2 \mathrm{~atm}$. Let's substitute these values into the expression for $K_{p_2}$:

$ 2 = \frac{(p_{CO})^2}{0.08} $

$\Rightarrow $ $ (p_{CO})^2 = 2 \times 0.08 $

$\Rightarrow $ $ (p_{CO})^2 = 0.16 $

$\Rightarrow $ $ p_{CO} = \sqrt{0.16} $

$\Rightarrow $$ p_{CO} = 0.4 \mathrm{~atm} $

$\Rightarrow $ $p_{CO} = 4 \times 10^{-1} \mathrm{~atm}$.

Therefore, the value to be filled in the blank is 4.

$\begin{aligned} & \mathrm{P}^{\circ}(1+\alpha)=1 \\ & \alpha=\frac{3}{4} ; \mathrm{P}^{\circ}=\frac{4}{7} \\ & \mathrm{k}_{\mathrm{P}}=\frac{4 \mathrm{P}_{\mathrm{o}} \alpha^2}{1-\alpha^2}=\frac{36}{7} \\ & \Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ell \mathrm{nk}_{\mathrm{P}} \\ & \left|\Delta \mathrm{G}^{\circ}\right|=8.314 \times 600 \ln \left(\frac{36}{7}\right) \\ & =8169.5 \mathrm{~J} \mathrm{~mol} \\ & =8.169 \mathrm{~kJ} / \mathrm{mol}^{-1}\end{aligned}$

For the relation between equilibrium constant and temperature, we use the integrated van’t Hoff equation:

$ \log_{10} K_p = -\frac{\Delta H^\circ}{2.303R}\cdot \frac{1}{T} + \text{constant} $

This is in the form of a straight line:

$ y = mx + c $

where

• $y = \log_{10} K_p$

• $x = \dfrac{1}{T}$

• slope $m = -\dfrac{\Delta H^\circ}{2.303R}$

Now calculate the slope from the given data.

Using points $(0.05, 3.5)$ and $(0.06, 2.5)$:

$ m = \frac{2.5 - 3.5}{0.06 - 0.05} = \frac{-1}{0.01} = -100 $

So,

$ -\frac{\Delta H^\circ}{2.303R} = -100 $

Hence,

$ \frac{\Delta H^\circ}{R} = 100 \times 2.303 $

$ \frac{\Delta H^\circ}{R} = 230.3 $

Therefore, the magnitude of $\dfrac{\Delta H^\circ}{R}$ is

$ \boxed{230} $

Nearest integer: $\boxed{230}$

For the reaction

$ 2\mathrm{N}_2\mathrm{O}_5(g) \rightleftharpoons 2\mathrm{N}_2\mathrm{O}_4(g) + \mathrm{O}_2(g) $

we have to find the magnitude of standard free energy change, that is, $|\Delta G^\circ|$.

We use the relation

$ \Delta G^\circ = -RT \ln K_p $

So first, let us calculate $K_p$.

1. Take 1 mole of $\mathrm{N}_2\mathrm{O}_5$ initially

Given that $\mathrm{N}_2\mathrm{O}_5$ is $50\%$ dissociated.

So, initial moles of $\mathrm{N}_2\mathrm{O}_5 = 1$

Dissociated moles $= 0.5$

Now from the reaction,

$ 2\mathrm{N}_2\mathrm{O}_5 \rightarrow 2\mathrm{N}_2\mathrm{O}_4 + \mathrm{O}_2 $

If $2$ moles of $\mathrm{N}_2\mathrm{O}_5$ dissociate, then:

• $2$ moles of $\mathrm{N}_2\mathrm{O}_4$ are formed

• $1$ mole of $\mathrm{O}_2$ is formed

Therefore, if $0.5$ mole of $\mathrm{N}_2\mathrm{O}_5$ dissociates:

• $\mathrm{N}_2\mathrm{O}_4$ formed $= 0.5$ mole

• $\mathrm{O}_2$ formed $= 0.25$ mole

So at equilibrium:

$ n(\mathrm{N}_2\mathrm{O}_5)=1-0.5=0.5 $

$ n(\mathrm{N}_2\mathrm{O}_4)=0.5 $

$ n(\mathrm{O}_2)=0.25 $

Total moles at equilibrium:

$ n_\text{total}=0.5+0.5+0.25=1.25 $

2. Calculate mole fractions

$ y_{\mathrm{N}_2\mathrm{O}_5}=\frac{0.5}{1.25}=0.4 $

$ y_{\mathrm{N}_2\mathrm{O}_4}=\frac{0.5}{1.25}=0.4 $

$ y_{\mathrm{O}_2}=\frac{0.25}{1.25}=0.2 $

Given total pressure $P=2$ atm.

Hence partial pressures are:

$ P_{\mathrm{N}_2\mathrm{O}_5}=0.4 \times 2 = 0.8 \text{ atm} $

$ P_{\mathrm{N}_2\mathrm{O}_4}=0.4 \times 2 = 0.8 \text{ atm} $

$ P_{\mathrm{O}_2}=0.2 \times 2 = 0.4 \text{ atm} $

3. Expression for $K_p$

For the reaction

$ 2\mathrm{N}_2\mathrm{O}_5(g) \rightleftharpoons 2\mathrm{N}_2\mathrm{O}_4(g)+\mathrm{O}_2(g) $

$ K_p=\frac{(P_{\mathrm{N}_2\mathrm{O}_4})^2(P_{\mathrm{O}_2})}{(P_{\mathrm{N}_2\mathrm{O}_5})^2} $

Substitute the values:

$ K_p=\frac{(0.8)^2(0.4)}{(0.8)^2}=0.4 $

So,

$ K_p=0.4 $

4. Calculate $\Delta G^\circ$

Temperature:

$ T=50+273=323\text{ K} $

Now,

$ \Delta G^\circ = -RT \ln K_p $

$ \Delta G^\circ = -8.314 \times 323 \times \ln(0.4) $

Now,

$ 0.4=\frac{2}{5} $

So,

$ \ln(0.4)=\ln 2 - \ln 5 $

Also,

$ \log 5 = 1-\log 2 = 1-0.30=0.70 $

Hence,

$ \log(0.4)=\log 4 -1 = 2\log 2 -1 = 0.60-1=-0.40 $

Therefore,

$ \ln(0.4)=2.303 \times (-0.40)=-0.9212 $

Now,

$ \Delta G^\circ = -8.314 \times 323 \times (-0.9212) $

$ \Delta G^\circ \approx 2472 \text{ J mol}^{-1} $

So the magnitude is

$ x=2472 \text{ J mol}^{-1} $

Final Answer:

$ \boxed{2472} $

$\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}$

$\begin{aligned} & \mathrm{n}_{\mathrm{T}}=1+\frac{\alpha}{2} \simeq 1(\alpha \ll 1) \\ & \mathrm{k}_{\mathrm{P}}=\frac{\mathrm{P}_{\mathrm{H}_2} \cdot \mathrm{P}_{\mathrm{O}_2}^{1 / 2}}{\mathrm{P}_{\mathrm{H}_2 \mathrm{O}}}=\frac{(\alpha \cdot \mathrm{P})\left(\frac{\alpha}{2} \mathrm{P}\right)^{\frac{1}{2}}}{(1-\alpha) \mathrm{P}} \\ & \mathrm{P}=1 \\ & 8 \times 10^{-3}=\frac{\alpha^{3 / 2}}{\sqrt{2}} \\ & \alpha^{3 / 2}=8 \sqrt{2} \times 10^{-3} \\ & \alpha^3=128 \times 10^{-6} \\ & \alpha=\sqrt[3]{128} \times 10^{-2} \\ & =5.03 \times 10^{-2} \end{aligned}$

$\begin{aligned} & V=2 L \\ & T=500 K \\ & P_{\text {total }}=5 \text { bar } \\ & n_{\text {Total }}=0.25=\frac{1}{4} \mathrm{~mol} \\ & P_{\text {total }}=n_{\text {total }} \times \frac{R T}{V} \\ & \Rightarrow 5=(0.06+a-0.08+0.04) \times \frac{0.08 \times 500}{2} \\ & \Rightarrow 10=(0.02+a) \times 0.08 \times 500 \end{aligned}$

$\begin{aligned} &\Rightarrow \mathrm{a}=0.25-0.02=0.23 \mathrm{~mol}\\ &\begin{aligned} & \mathrm{K}_{\mathrm{P}}=\frac{\mathrm{X}_{\mathrm{CH}_3 \mathrm{OH}}}{\mathrm{X}_{\mathrm{CO}} \times \mathrm{X}_{\mathrm{H}_2}^2} \times \frac{1}{\left(\mathrm{P}_{\mathrm{T}}\right)^2}=\frac{0.04}{0.06 \times(0.15)^2} \times\left[\frac{1 / 4}{5}\right]^2 \\ & =\frac{4}{6 \times(0.15)^2 \times 16} \times \frac{1}{25} \\ & =\frac{100 \times 100}{24 \times 225 \times 25}=\frac{100 \times 100}{135000} \\ & =0.074=74 \times 10^{-3} \end{aligned} \end{aligned}$

$\begin{aligned} & \text { Initial pressure of } \mathrm{N}_2 \mathrm{O}_5 \\ & \qquad \\ & \qquad=\frac{\frac{37.8}{108} \times 0.082 \times 500}{1}=14.35 \mathrm{bar} \\ & 2 \mathrm{~N}_2 \mathrm{O}_5 \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_4+\mathrm{O}_2 \end{aligned}$

$\begin{aligned} & \mathrm{t}=0 \quad 14.35 \\ & \mathrm{t}=\mathrm{eq} \quad 14.35-2 \mathrm{P} \quad 2 \mathrm{P} \quad \mathrm{P} \\ & \mathrm{P}_{\text {Total }} \text { at eqb }=14.35+\mathrm{P}=18.65 \\ & \mathrm{P}=4.3 \\ & \mathrm{P}_{\mathrm{N}_2 \mathrm{O}_{\mathrm{s}}}=5.75 \mathrm{bar} \\ & \mathrm{P}_{\mathrm{N}_2 \mathrm{O}_4}=8.6 \mathrm{bar} \\ & \mathrm{P}_{\mathrm{O}_2}=4.3 \mathrm{bar} \\ & \mathrm{k}_{\mathrm{p}}=\frac{(8.6)^2 \times(4.3)}{(5.75)^2}=9.619=\mathrm{x} \times 10^{-2} \\ & \mathrm{x}=961.9 \approx 962 \end{aligned}$

To calculate the equilibrium constant $\mathrm{K}_C$ for the formation of $\mathrm{NH}_3$ from $\mathrm{N}_2$ and $\mathrm{H}_2$ at $500 \mathrm{~K}$, we use the formula:

$ \mathrm{K}_C = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} $

Substituting the given concentrations:

$ \mathrm{K}_C = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2}) \times (3 \times 10^{-2})^3} $

Calculating the above expression results in:

$ \mathrm{K}_C = 417 $

For the reaction:

$ \mathrm{N}_2 \mathrm{O}_{4(\mathrm{~g})} \rightleftarrows 2 \mathrm{NO}_{2(\mathrm{~g})} $

we need to find the equilibrium constant $\mathrm{K}_{\mathrm{c}}$ at 300 K given that $\mathrm{K}_{\mathrm{p}} = 0.492 \text{ atm}$.

The relationship between $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{c}}$ is given by the following equation:

$ \mathrm{K}_{\mathrm{p}} = \mathrm{K}_{\mathrm{c}} (RT)^{\Delta n} $

where:

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

$T = 300 \, \mathrm{K}$ is the temperature,

$\Delta n$ is the change in moles of gas from reactants to products.

For the reaction given:

The number of moles of gaseous products is 2 (for $2 \mathrm{NO}_2$).

The number of moles of gaseous reactants is 1 (for $\mathrm{N}_2 \mathrm{O}_4$).

Thus, $\Delta n = 2 - 1 = 1$.

Substitute the known values into the equation:

$ 0.492 = \mathrm{K}_{\mathrm{c}} (0.082 \times 300)^{1} $

Solve for $\mathrm{K}_{\mathrm{c}}$:

$ 0.492 = \mathrm{K}_{\mathrm{c}} \times 24.6 $

$ \mathrm{K}_{\mathrm{c}} = \frac{0.492}{24.6} $

Calculate $\mathrm{K}_{\mathrm{c}}$:

$ \mathrm{K}_{\mathrm{c}} = 0.02 $

Thus, $\mathrm{K}_{\mathrm{c}}$ for the reaction at 300 K is:

$ 2 \times 10^{-2} $

$ \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=\frac{(3)^2}{3 \times 3}=\frac{9}{9}=1 $

The given reaction is :

$\mathrm{CO(g)} + \mathrm{H_{2}O(g)} \rightleftharpoons \mathrm{CO_{2}(g)} + \mathrm{H_{2}(g)} $

We are given that $40\%$ of water by mass reacts with carbon monoxide.

The molecular weight of water (H₂O) is approximately $18 \, \mathrm{g/mol}$, so $1 \, \mathrm{mole}$ of water weighs $18 \, \mathrm{g}$. Therefore, the mass of the reacted water is $0.40 \times 18 \, \mathrm{g} = 7.2 \, \mathrm{g}$.

Since from the stoichiometry of the reaction we can see that $1 \, \mathrm{mole}$ of CO reacts with $1 \, \mathrm{mole}$ of H₂O to form $1 \, \mathrm{mole}$ of CO₂ and $1 \, \mathrm{mole}$ of H₂, this means that $7.2 \, \mathrm{g}$ of H₂O is equivalent to $7.2 \, \mathrm{g} / 18 \, \mathrm{g/mol} = 0.4 \, \mathrm{mole}$.

We start with $1 \, \mathrm{mole}$ of CO and $1 \, \mathrm{mole}$ of H2O. At equilibrium, we have :

- CO: $1 \, \mathrm{mole} - 0.4 \, \mathrm{mole} = 0.6 \, \mathrm{mole}$

- H₂O : $1 \, \mathrm{mole} - 0.4 \, \mathrm{mole} = 0.6 \, \mathrm{mole}$

- CO₂ : $0 \, \mathrm{mole} + 0.4 \, \mathrm{mole} = 0.4 \, \mathrm{mole}$

- H₂ : $0 \, \mathrm{mole} + 0.4 \, \mathrm{mole} = 0.4 \, \mathrm{mole}$

The volume of the container is $10 \, \mathrm{litres}$. Therefore, we can convert the moles to concentrations (in M = moles/L) as :

- [CO] = $0.6 \, \mathrm{M}$

- [H₂O] = $0.6 \, \mathrm{M}$

- [CO₂] = $0.4 \, \mathrm{M}$

- [H₂] = $0.4 \, \mathrm{M}$

The equilibrium constant $K_{c}$ for the reaction can be calculated as :

$K_{c} = \frac{[\mathrm{CO_{2}}][\mathrm{H_{2}}]}{[\mathrm{CO}][\mathrm{H_{2}O}]}$

So, substituting the values we get :

$K_{c} = \frac{(0.4 \times 0.4)}{(0.6 \times 0.6)} = 0.44$

As per the question, we need to calculate the value of $K_{c} \times 10^{2}$ :

$K_{c} \times 10^{2} = 0.44 \times 10^{2} = 44 \, (\text{rounded to the nearest integer})$

Therefore, $K_{c} \times 10^{2} = 44$.

Given the reaction:

$\mathrm{A}_{(\mathrm{g})} \rightleftharpoons 2 \mathrm{~B}_{(\mathrm{g})}+\mathrm{C}_{(\mathrm{g})}$

At the beginning of the reaction (at time = 0), the total pressure is solely due to A, hence it is 450 mmHg.

As the reaction progresses, let's denote 'x' as the pressure decrease due to the decomposition of A. Correspondingly, the pressure increases by '2x' and 'x' for B and C respectively, following the stoichiometry of the reaction.

At time 't', the total pressure (P(T)) is the sum of the pressures of A, B, and C, which is given to be 720 mmHg.

This gives us the equation:

$450 \text{ mmHg} - x \text{ mmHg (decrease in A's pressure)} + 2x \text{ mmHg (increase in B's pressure)} + x \text{ mmHg (increase in C's pressure)} = 720 \text{ mmHg (total pressure at time t)}$

Solving for 'x' gives:

$x = 135 \text{ mmHg}$

The fraction of A decomposed would then be this change in pressure divided by the initial pressure:

$\text{Fraction decomposed} = \frac{x}{\text{Initial pressure}} = \frac{135 \text{ mmHg}}{450 \text{ mmHg}} = 0.3 = 3 \times 10^{-1}$

(A) Equilibrium is possible only in a closed system at a given temperature.

• This statement is true. In a closed system, there is no exchange of matter with the surroundings. This allows the reaction to reach equilibrium at a constant temperature.

(B) Both the opposing processes occur at the same rate.

• This statement is true. At equilibrium, the forward and reverse reactions occur at the same rate, meaning the concentrations of reactants and products remain constant.

(C) When equilibrium is attained at a given temperature, the value of all its parameters became equal.

• This statement is false. When a system reaches equilibrium, it doesn't mean that the values of all parameters (such as the concentrations of reactants and products) are equal. It means that these values are constant, i.e., they are not changing with time because the rates of the forward and reverse reactions are equal.

(D) For dissolution of solids in liquids, the solubility is constant at a given temperature.

• This statement is true. The solubility of a solid in a liquid (in a saturated solution) is indeed constant at a given temperature and pressure.

Therefore, there are 3 correct statements: (A), (B), and (D).

The initial equilibrium concentrations are given as:

$[\mathrm{PCl}_{3}]_{\text{eq}} = 0.2 \, \mathrm{mol/L}$, $[\mathrm{Cl}_{2}]_{\text{eq}} = 0.1 \, \mathrm{mol/L}$, $[\mathrm{PCl}_{5}]_{\text{eq}} = 0.4 \, \mathrm{mol/L}$.

After adding $0.2 \, \mathrm{mol}$ of $\mathrm{Cl_2}$, the new concentration of $\mathrm{Cl_2}$ becomes $0.1 + 0.2 = 0.3 \, \mathrm{mol/L}$.

Let the change in concentration be $x$, so at the new equilibrium, the concentrations are:

$[\mathrm{PCl}_{3}] = 0.2 - x \, \mathrm{mol/L}$, $[\mathrm{Cl}_{2}] = 0.3 - x \, \mathrm{mol/L}$, $[\mathrm{PCl}_{5}] = 0.4 + x \, \mathrm{mol/L}$.

Using the equilibrium constant expression $K_c = 20$, we get:

$20 = \frac{0.4+x}{(0.2-x)(0.3-x)}$.

Solving for $x$, we find $x \approx 0.086$.

Therefore, the new equilibrium concentration of $\mathrm{PCl}_5$ is:

$[\mathrm{PCl}_5]_{\text{new eq}} = 0.4 + 0.086 = 0.486 \, \mathrm{mol/L} = 48.6 \times 10^{-2} \, \mathrm{mol/L}$.

$ \begin{aligned} & \mathrm{k}_{\mathrm{p}_1}=\frac{\left(\frac{\alpha}{1+\alpha} \times \mathrm{p}_1\right)^2}{\frac{1-\alpha}{1+\alpha} \mathrm{p}_1} \\\\ & 3=\frac{\alpha^2 \times \mathrm{p}_1}{1-\alpha^2} \end{aligned} $



$ \begin{aligned} & \mathrm{k}_{\mathrm{p}_2}=\frac{\left(\frac{2 \alpha}{1+\alpha} \times \mathrm{p}_2\right)^2}{\frac{1-\alpha}{1+\alpha} \times \mathrm{p}_2} \\\\ & 1=\frac{4 \alpha^2 \times \mathrm{p}_2}{1-\alpha^2} \\\\ & \frac{\mathrm{k}_{\mathrm{p}_1}}{\mathrm{k}_{\mathrm{p}_2}}=\frac{\mathrm{p}_1}{4 \mathrm{p}_2} \\\\ & \frac{3}{1}=\frac{\mathrm{p}_1}{4 \mathrm{p}_2} \\\\ & {\mathrm{p}_1} : {\mathrm{p}_2} = 12 : 1 \\\\ & \therefore \mathrm{x}=12 \end{aligned} $

$ \begin{aligned} & \mathrm{SO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{SO}_{3(\mathrm{~g})} \\\\ & \mathrm{K}_{\mathrm{P}}=2 \times 10^{12} \text { at } 300 \mathrm{~K} \\\\ & \mathrm{~K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}} \times(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} \\\\ & \Rightarrow 2 \times 10^{12}=\mathrm{K}_{\mathrm{C}} \times(0.082 \times 300)^{-1 / 2} \\\\ & \Rightarrow \mathrm{~K}_{\mathrm{C}}=9.92 \times 10^{12} \\\\ & \Rightarrow \mathrm{~K}_{\mathrm{C}}=0.992 \times 10^{13} \end{aligned} $

$\mathrm{2SO_2+O_2\rightleftharpoons 2SO_3~~\Delta H=-190~kJ}$

It is an exothermic reaction

$\therefore$ factor B, C, D will increase the amount of SO$_3$.

$\mathrm{2NH_3(g)\frac{5}{2}O_2(g)\rightleftharpoons 2NO(g)+3H_2O(g)}$

Clearly, $\mathrm{{K_{eq}} = {1 \over {{k_1}}} \times {k_2} \times k_3^3}$

$ = {{1.6 \times {{10}^{12}} \times {{10}^{ - 39}}} \over {4 \times {{10}^5}}}$

$ = 0.4 \times {10^{ - 32}}$

$ = 4 \times {10^{ - 33}}$

$\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}$

$ \begin{aligned} & \mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}_{\mathrm{T}}^{\frac{3}{2}}\left(1+\frac{\alpha}{2}\right) \alpha^{\frac{3}{2}}}{2^{\frac{1}{2}} \mathrm{P}_{\mathrm{T}}\left(1+\frac{\alpha}{2}\right)^{\frac{3}{2}}(1-\alpha)} \\\\ & 2 \times 10^{-3}=\frac{\alpha^{\frac{3}{2}}}{2^{\frac{1}{2}}}\left[\text { as } \alpha<<1 \text { and let } \mathrm{P}_{\mathrm{T}}=1\right] \end{aligned} $

$ \begin{aligned} & \alpha^{\frac{3}{2}}=2^{\frac{3}{2}} \times 10^{-3} \\\\ & \approx 2 \times 10^{-2} \end{aligned} $

$\therefore \%$ of water decomposition $=2 \%$

The Gibbs energy change for a reaction at standard conditions, $\Delta_r G^\theta$, can be calculated using the equation:

$ \Delta \mathrm{G}^\theta = -\mathrm{RT} \ln \mathrm{K}_{\mathrm{eq}} $

Where $\mathrm{K}_{\mathrm{eq}}$ is the equilibrium constant given by $\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}}$. For the reaction:

$ \mathrm{A + B} \rightleftharpoons \mathrm{C + D} $

The forward rate constant, $k_f$, is $10^3$, and the reverse rate constant, $k_r$, is $10^2$. Therefore,

$ \mathrm{K}_{\mathrm{eq}} = \frac{10^3}{10^2} = 10 $

Substitute the values into the Gibbs energy change formula:

$ \Delta \mathrm{G}^\theta = -\mathrm{RT} \ln 10 $

where $R = 8.3 \, \mathrm{J \, K^{-1} \, mol^{-1}}$ and $T = 300 \, \mathrm{K}$ (since the temperature is $27^\circ \mathrm{C}$ which converts to 300 K). The natural logarithm of 10 is approximately 2.3.

$ \Delta \mathrm{G}^\theta = -(8.3 \times 300 \times 2.3) $

Calculating gives:

$ \Delta \mathrm{G}^\theta = -5711 \, \mathrm{J \, mol^{-1}} $

Converting to kJ:

$ \Delta \mathrm{G}^\theta = -5.7 \, \mathrm{kJ \, mol^{-1}} $

Rounding to the nearest integer, the standard Gibb's energy change is approximately $-6 \, \mathrm{kJ \, mol^{-1}}$.

For reaction $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g})$

$\log \mathrm{k}_{\mathrm{f}}=\frac{-\mathrm{E}_{\mathrm{f}}}{2.303 \mathrm{RT}}+\log \mathrm{A}_{\mathrm{f}}$ [Arrhenius equation for forward reaction]

From plot when, $\frac{1}{\mathrm{~T}}=0.002, \log \mathrm{k}_{\mathrm{f}}=9$

$\begin{aligned} & \Rightarrow 9=\frac{-E_f}{2.303 R}(0.002)+\log \left(A_f\right) \\\\ & \text { Given }: A_f=10^{15} \mathrm{~s}^{-1} \\\\ & \Rightarrow 9=\frac{-E_f}{2.303 R}(0.002)+15 \\\\ & \Rightarrow \frac{E_f}{2.303 R}=\frac{6}{0.002}=3000\end{aligned}$

$\begin{aligned} \text { Now, } \mathrm{K} & =\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{b}}}=\frac{\mathrm{A}_{\mathrm{f}}}{\mathrm{A}_{\mathrm{b}}} \mathrm{e}^{-\left(\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{b}}\right) / \mathrm{RT}} \\\\ \log \mathrm{K} & =-\frac{1}{2.303} \frac{\left(\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{b}}\right)}{R T}+\log \left(\frac{10^{15}}{10^{11}}\right)\end{aligned}$

At $500 \mathrm{~K}$

$ \begin{aligned} & \Rightarrow 6=\frac{-\left(\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{b}}\right)}{500 \mathrm{R}(2.303)}+4 \\\\ & \Rightarrow(1000 \mathrm{R})(2.303)=\mathrm{E}_{\mathrm{b}}-\mathrm{E}_{\mathrm{f}} \\\\ & \Rightarrow(1000 \mathrm{R})(2.303)=\mathrm{E}_{\mathrm{b}}-3000(2.303 \mathrm{R}) \\\\ & \Rightarrow \mathrm{E}_{\mathrm{b}}=4000 \mathrm{R}(2.303) ..........(i) \end{aligned} $

Now $\mathrm{k}_{\mathrm{b}}=\mathrm{A}_{\mathrm{b}} \mathrm{e}^{-\mathrm{E}_{\mathrm{b}} / \mathrm{RT}}$

$\begin{aligned} & \Rightarrow \log \mathrm{k}_{\mathrm{b}}=\frac{-\mathrm{E}_{\mathrm{b}}}{2.303 \mathrm{RT}}+\log \mathrm{A}_{\mathrm{b}} \\\\ & \text { At } 250 \mathrm{~K}, \\\\ & \Rightarrow \log \mathrm{k}_{\mathrm{b}}=-\frac{4000}{250}+\log \left(10^{11}\right) \text { [From equation (1)] } \\\\ & \quad=-16+11=-5 \\\\ & \left|\log \mathrm{k}_{\mathrm{b}}\right|=5\end{aligned}$

Partial pressure of $\mathrm{NO}(\mathrm{g})=\frac{1.2}{2.6} \times 1$

Partial pressure of $\mathrm{O}_{2}(\mathrm{~g})=\frac{0.6}{2.6}$

Partial pressure of $\mathrm{NO}_{2}(\mathrm{~g})=\frac{0.8}{2.6}$

$ \begin{aligned} \mathrm{K}_{\mathrm{p}}=\frac{\left(\mathrm{P}_{\mathrm{NO}_{2}}\right)^{2}}{\left(\mathrm{P}_{\mathrm{NO}}\right)^{2}\left(\mathrm{P}_{\mathrm{O}_{2}}\right)} &=\frac{0.8 \times 0.8 \times 2.6}{1.2 \times 1.2 \times 0.6} \\\\ &=1.925 \\\\ & \approx 2 \end{aligned} $

$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$

$k=2 \times 10^{15}$

$\frac{1}{2} \mathrm{Cu}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{+2}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$

$\mathrm{K}^{\prime}=\frac{1}{(\mathrm{~K})^{1 / 2}}=\frac{1}{\left(2 \times 10^{15}\right)^{1 / 2}}$

$=2.23 \times 10^{-8}$

$x \simeq 2$

We know, 760 Torr = 1 atm

$\therefore$ 32 Torr = ${{32} \over {760}}$ atm

As all the liquid water evaporates so entire water is in gaseous state.

$\therefore$ Weight of water vapour = 0.9 g

$\therefore$ Moles of water vapour (n) = ${{0.9} \over {18}}$

Pressure (P) = ${{32} \over {760}}$ atm

Temperature (T) = (27 + 273) K = 300 K

R = 0.082 L atm K^$-$1 mol^$-$1

Given water vapour act as an ideal gas, so we can apply ideal gas equation.

From ideal gas equation,

PV = nRT

$ \Rightarrow {{32} \over {760}} \times v = {{0.9} \over {18}} \times 0.082 \times 300$

$ \Rightarrow v = 29$ L

Given that at equilibrium, concentration of NO = 0.4 mol/L

$\therefore$ 2x = 0.4

$\Rightarrow$ x = 0.2

$\therefore$ Concentration of NOCl at equilibrium,

[NOCl]_eq = 2 $-$ 2 $\times$ 0.2 = 1.6

and [NO]_eq = 0.4

and [Cl₂]_eq = 0.2

We know,

${K_C} = {{{{[NO]}^2}[C{l_2}]} \over {{{[NOCl]}^2}}}$

$ = {{{{[0.4]}^2}[0.2]} \over {{{[1.6]}^2}}}$

$ \Rightarrow {K_C} = 12.5 \times {10^{ - 3}}$

$ \Rightarrow {K_C} = 125 \times {10^{ - 4}}$

$\therefore$ $K = {{{{\left( {{\alpha \over 2}} \right)}^{1/2}} \times {{\left( {{\alpha \over 2}} \right)}^{1/2}}} \over {(1 - \alpha )}}$

Given $\alpha = {{40} \over {100}} = 0.4$

$\therefore$ $K = {{{{\left( {{{0.4} \over 2}} \right)}^{1/2}} \times {{\left( {{{0.4} \over 2}} \right)}^{1/2}}} \over {(1 - 0.4)}}$

$ = {1 \over 3}$

We know,

$\Delta G^\circ = - RT\ln K$

$ = - RT\ln \left( {{1 \over 3}} \right)$

$ = + RT\ln 3$

$ = + 8.314 \times 300 \times \ln 3$

= 2735 J/mol

$\mathrm{k}_{\mathrm{P}}=\frac{\left(\frac{1}{1.5} \times 1\right)^2}{\left(\frac{0.5}{1.5} \times 1\right)}=\frac{1}{0.75}=\frac{100}{75}$

$=1.33$

$\Delta \mathrm{G}^0=-\mathrm{RT} \ell \mathrm{nk}_{\mathrm{P}}$

$=-8.31 \times 300 \times \ln (1.33)=-710.45 \mathrm{~J} / \mathrm{mol}$

$=-710 \mathrm{~J} / \mathrm{mol}$

Here 2 moles of N₂ also present that is why 2 moles always have to add in total mole calculation.

At equilibrium,

Pressure (P) = 2.46 atm

Volume (V) = 200 L

Temperature (T) = 600 K

$\therefore$ Applying ideal gas equation,

PV = nRT

$\Rightarrow$ 2.46 $\times$ 200 = (7 + x) $\times$ 0.082 $\times$ 600

$\Rightarrow$ x = 3

Now,

${K_P} = {{{P_{PC{l_3}}} \times {P_{C{l_2}}}} \over {{P_{PC{l_5}}}}}$

$ = {{\left[ {{3 \over {7 + 3}} \times 2.46} \right]\left[ {{3 \over {7 + 3}} \times 2.46} \right]} \over {\left[ {{{5 - 3} \over {7 + 3}} \times 2.46} \right]}}$

$ = {{{3 \over {10}} \times {3 \over {10}} \times {{(2.46)}^2}} \over {{2 \over {10}} \times 2.46}}$

$ = {9 \over {20}} \times 2.46$

$ = 1107 \times {10^{ - 3}}$ atm

$\underset{1-x}{2 \mathrm{O}_3(\mathrm{~g})} \rightleftharpoons \underset{\frac{3 \mathrm{x}}{2}}{3 \mathrm{O}_2(\mathrm{~g})}$

Given, $x=0.5$

$\therefore \mathrm{k}_{\mathrm{p}}=\frac{[3(0.5)]^{3} \times 1}{[2]^{3} \times(0.5)^{2} \times 1.25}$

$\therefore \mathrm{k}_{\mathrm{p}}=\frac{27}{8} \times \frac{0.5}{1.25}=1.35$

$ \begin{aligned} \Delta \mathrm{G}^{\circ} &=-2.303 \mathrm{RT} \log \mathrm{k}_{\mathrm{p}} \\\\ &=-2.303 \times 8.3 \times 300 \log 1.35 \\\\ &=-8.3 \times 300 \ln (1.35) \\\\ &=-747 \mathrm{~J} \mathrm{~mol}^{-1} \end{aligned} $

moles of NH₄HS initially taken = ${{5.1g} \over {51g/mol}}$ = 0.1 mol

volume of vessel = 2 litre



$\Rightarrow$ partial pressure of each component

$P = {{nRT} \over V} = {{0.1 \times 0.2 \times 0.082 \times 300} \over 2}$

= 0.246 atm

$\Rightarrow$ k_P = P_{$N{H_3}$} $\times$ P_{${H_2}S$} = (0.246)² = 0.060516

= 6.05 $\times$ 10^$-$2

$ \therefore $ x = 6

Let moles added = a


${{0.8} \over {(5 \times {{10}^{ - 8}})\left( {{a \over 2} - 1.6} \right)}} = {10^8}$

$\Rightarrow$ ${{a \over 2}}$ $-$ 1.6 = 0.4 $\Rightarrow$ a = 4

$\therefore$ ${K_C} = {\left( {{{1 + x} \over {1 - x}}} \right)^2}$

$100 = {\left( {{{1 + x} \over {1 - x}}} \right)^2}$

${{1 + x} \over {1 - x}} = 10$

$x = {9 \over {11}}$

Moles of D = 1 + x

$ = 1 + {9 \over {11}} = {{20} \over {11}}$

$ = 1.818 = 181.8 \times {10^{ - 2}} = 181.8 \times {10^{ - 2}}$

$ \cong 182 \times {10^{ - 2}}$ M

The rate equation provided in your question describes the rates of the forward and reverse reactions for this chemical system. The equilibrium constant, $K_c$, is the ratio of the forward rate constant ($k_f$) to the reverse rate constant ($k_r$). At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, so the rate equation simplifies to:

$k_f[\text{{PtCl}}_4]^{2-} = k_r[\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$

Given the rate equation:

$-\frac{d[\text{{PtCl}}_4]^{2-}}{dt} = 4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} - 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$

At equilibrium, $-\frac{d[\text{{PtCl}}_4]^{2-}}{dt} = 0$, so:

$0 = 4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} - 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$

Rearranging terms, we find:

$4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} = 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$

Now, the equilibrium constant $K_c$ is defined as the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients. For the reaction in question, we have:

$K_c = \frac{[\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]}{[\text{{PtCl}}_4]^{2-}}$

Dividing both sides of our rate equation by $[\text{{PtCl}}_4]^{2-}$, we find that:

$K_c = \frac{4.8 \times 10^{-5}}{2.4 \times 10^{-3}} = \frac{1}{50}$ = 0.02

So, the equilibrium constant $K_c$ for this reaction is approximately 0, when rounded to the nearest integer.

$\left[ {NH_4^ + } \right]$ = 0.0504 & [NH₃] = 0.0210

So, ${K_b} = {{[NH_4^ + ][H{O^ - }]} \over {[N{H_3}]}}$

$[H{O^ - }] = {{{K_b} \times [N{H_3}]} \over {[NH_4^ + ]}} = 1.8 \times {10^{ - 5}} \times {2 \over 5} \times {{210} \over {504}} = 3 \times {10^{ - 6}}$

In the given equilibrium, the solid substances do not contribute to the equilibrium constant expression since their activities are considered to be 1.

For the reaction:

$ \text{A(s)} \rightleftharpoons \text{M(s)} + \frac{1}{2}\text{O}_2(\text{g}) $

The equilibrium constant, $ K_p $, is given by:

$ K_p = P_{\text{O}_2}^{n} $

where $ P_{\text{O}_2} $ is the partial pressure of $ \text{O}_2 $ and $ n $ represents the stoichiometric coefficient of $ \text{O}_2 $ in the balanced chemical reaction, which is $ \frac{1}{2} $. Therefore, the expression for $ K_p $ becomes:

$ K_p = \left( P_{\text{O}_2} \right)^{\frac{1}{2}} $

Given that $ K_p = 4 $, substituting into the equation gives:

$ 4 = \left( P_{\text{O}_2} \right)^{\frac{1}{2}} $

To find $ P_{\text{O}_2} $, square both sides of the equation:

$ 4^2 = P_{\text{O}_2} $

$ 16 = P_{\text{O}_2} $

Thus, the partial pressure of $ \text{O}_2 $ at equilibrium is $\boxed{16}$ atm.

PCl_5(g) $\rightleftharpoons$ PCl_3(g) + Cl_2(g) K₂ = 1.844

t = 0 3moles

t = $\infty$ x x

$ \Rightarrow {{[PC{l_3}][C{l_2}]} \over {[PC{l_5}]}} = {{{x^2}} \over {3 - x}} = 1.844$

$ \Rightarrow {x^2} + 1.844 - 5.532 = 0$

$ \Rightarrow x = {{ - 1.844 + \sqrt {{{(1.844)}^2} + 4 \times 5.532} } \over 2}$

$ \cong 1.604$

$\Rightarrow$ Moles of PCl₅ = 3 $-$ 1.604 $\cong$ 1.396

$Ba{(OH)_2} \to B{a^{ + 2}} + 2O{H^ - } \downarrow 2 \times 0.005 = 0.01 = {10^{ - 2}}$

At 298 K : in aq. solution $[{H_3}{O^ + }][O{H^ - }] = {10^{ - 14}}$

$[{H_3}{O^ + }] = {{{{10}^{ - 14}}} \over {{{10}^{ - 2}}}} = {10^{ - 12}}$

$K = {{[C]_{eq}^2} \over {{{[A]}_{eq}}{{[B]}_{eq}}}} = {{{{(1 + 2x)}^2}} \over {(1 - x)(1 - x)}}$

$100 = {\left( {{{1 + 2x} \over {1 - x}}} \right)^2}$

$\left( {{{1 + 2x} \over {1 - x}}} \right) = 10$

$x = {3 \over 4}$

$[C]{e_{q.}} = 1 + 2x$

$ = 1 + 2\left( {{3 \over 4}} \right)$

= 2.5 M

= 25 $\times$ 10^-1 M

For the equilibrium reaction

$ \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) $

the relationship between $ K_P $ and $ K_C $ is given by the equation:

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

where $ \Delta n $ is the change in the number of moles of gas, $ R $ is the ideal gas constant, and $ T $ is the temperature in Kelvin.

For the given reaction:

$ \Delta n = 2 - 1 = 1 $

Given:

$ K_P = 47.9 $

$ R = 0.083 \, \text{L bar K}^{-1} \text{mol}^{-1} $

$ T = 288 \, \text{K} $

Substitute these values into the equation:

$ 47.9 = K_C (0.083 \times 288)^{1} $

$ K_C = \frac{47.9}{0.083 \times 288} $

Calculate:

$ 0.083 \times 288 = 23.904 $

$ K_C = \frac{47.9}{23.904} $

$ K_C \approx 2.0033 $

Rounding to the nearest integer, the value of $ K_C $ is

$ \boxed{2} $

On reaction, 2SO₂(g) + O₂(g) $\rightarrow$ 2SO₃(g)

Given values are : p_SO3 = 45kPa, p_SO2 = 530 Pa = 0.53 kPa

p_SO2 = 43 kPa

Now, ${K_p} = {{{{[{p_{S{O_3}(g)}}]}^2}} \over {{{[{p_{S{O_2}(g)}}]}^2} \times [{p_{{O_2}}}]}}$

On putting given values, we get

$ \Rightarrow {K_p} = {{{{(43)}^2}} \over {{{(45)}^2} \times 0.53}}$

$ = {{1849} \over {2025 \times 0.53}} = {{1849} \over {1073.25}}$

$ = 1.7228$

$ = {{1.7228 \times {{10}^2}} \over {{{10}^2}}} = 172.28 \times {10^{ - 2}} = 172$

Hence, the equilibrium constant, K_p = 172.

Using formula,

$\Delta$G$^\circ$ = $-$ RTln(K_p)

$ \Rightarrow $ 25.2 $\times$ 10³ = $-$8.3 $\times$ 400 $\times$ 2.3 log (K_p)

$ \Rightarrow $ K_p = 10^$-$3.3

= 10^$-$3 $\times$ 0.501

= 5.01 $\times$ 10^$-$4 Bar^$-$1

Also,

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

$ \Rightarrow {{{K_p}} \over {{K_c}}} = {(RT)^{ - 1}}$

$ \Rightarrow {K_c} = {K_p}(RT)$

$ = 5.01 \times {10^{ - 4}} \times 8.3 \times 400$

$ = 1.66 \times {10^{ - 5}}$ m³/mole

$ = 1.66 \times {10^{ - 2}}$ L/mol

N₂O₄(g) $\rightleftharpoons$ 2NO₂(g)

$\Delta$n_g = 2 $-$ 1 = 1

K_P = K_C(RT)^$\Delta$n_g

600.1 = 20.4 (0.0831 $\times$ T)¹

$ \Rightarrow $ T = ${{600.1} \over {20.4 \times 0.0831}}$ = 354 K

Now,

${K_a} = {{[{H^ + }][{A^ - }]} \over {[HA]}}$

$ \Rightarrow 2 \times {10^{ - 6}} = {{(0.1)({{10}^{ - 2}}\alpha )} \over {{{10}^{ - 2}}}}$

$ \Rightarrow \alpha = 2 \times {10^{ - 5}}$

$\Delta$G^o = $-$RT ln Keq

$-$9.478 $\times$ 10³ = $-$495 $\times$ 8.314 ln Keq

ln Keq = 2.303 = ln 10

So, Keq = 10

Now, A(g) $\rightleftharpoons$ B(g)

$\matrix{ {t = 0} & {22} & 0 \cr {t = t} & {22 - x} & x \cr } $

$Keq = {{[B]} \over {[A]}} = {x \over {(22 - x)}} = 10$

x = 20

So, millimoles of B = 20

[p = Total pressure at equilibrium = 1.9 atm]

Now, at equilibrium pV = (1 + 2x)RT

$ \Rightarrow 1 + 2x = {{pV} \over {RT}} = {{1.9 \times 25} \over {0.082 \times 300}} = 1.93$

[V = 25 L, R = 0.082 L atm mol^$-$1 K^$-$1 T = 300 K]

$ \Rightarrow x = {{1.93 - 1} \over 2} = 0.465$

$ \Rightarrow {K_p} = {{{p_A} \times p_B^2} \over {{p_{A{B_2}}}}} \Rightarrow {{\left( {{x \over {1 + 2x}}p} \right) \times {{\left( {{{2x} \over {1 + 2x}}p} \right)}^2}} \over {\left( {{{1 - x} \over {1 + 2x}}p} \right)}}$

$ = {{4{x^3} \times {p^3}} \over {{{(1 + 2x)}^3}}} \times {{(1 + 2x)} \over {(1 - x) \times p}} = {{4{x^3} \times {p^2}} \over {{{(1 + 2x)}^2} \times (1 - x)}}$

$ = {{4 \times {{(0.465)}^3} \times {{(1.9)}^2}} \over {{{(1 + 2 \times 0.465)}^2} \times (1 - 0.465)}} = 0.7285$ atm

$ = 72.85 \times {10^{ - 2}}$ atm $ \simeq 73 \times {10^{ - 2}} = x \times {10^{ - 2}}$

$\therefore$ $x = 73$

Given, stability constant value,

K₁ = 10⁴

K₂ = 1.58 $\times$ 10³

K₃ = 5 $\times$ 10²

K₄ = 10²

Cu²⁺ + NH₃ $\buildrel {K_1} \over \rightleftharpoons $ [Cu(NH₃)]²⁺ .... (i)

[Cu(NH₃)]²⁺ + NH₃ $\buildrel {K_2} \over \rightleftharpoons $ [Cu(NH₃)₂]²⁺.... (ii)

[Cu(NH₃)₂]²⁺ + NH₃ $\buildrel {K_3} \over \rightleftharpoons $ [Cu(NH₃)₃]²⁺..... (iii)

[Cu(NH₃)₃]²⁺ + NH₃ $\buildrel {K_4} \over \rightleftharpoons $ [Cu(NH₃)₄]²⁺ ..... (iv)

On adding Eqs. (i), (ii), (iii) and (iv), we get

Cu²⁺ + 4NH₃ $\buildrel {K} \over \rightleftharpoons $ [Cu(NH₃)₄]²⁺

$\therefore$ The overall reaction constant (k) or equilibrium constant for formation of [Cu(NH₃)₄]²⁺ is

K = K₁ $\times$ K₂ $\times$ K₃ $\times$ K₄

$ \Rightarrow $ K = 10⁴ $\times$ 1.58 $\times$ 10³ $\times$ 5 $\times$ 10² $\times$ 10²

$ \Rightarrow $ K = 7.9 $\times$ 10¹¹

where, K = equilibrium constant for formation of [Cu(NH₃)₄]²⁺

So, equilibrium constant 'K' for dissociation of [Cu(NH₃)₄]²⁺ is ${1 \over K}$.

$K' = {1 \over K} = {1 \over {7.9 \times {{10}^{11}}}} = 1.26 \times {10^{ - 12}}$

Hence, K' = x $\times$ 10^$-$12

x = 1.26

Cl₂(g) $\rightleftharpoons$ 2Cl(g)

Let moles of both of Cl₂ and Cl molecule be x.

Partial pressure of Cl is, ${p_{Cl}} = {x \over {2x}} \times 1 = {1 \over 2}$

Partial pressure of Cl₂ is, ${p_{C{l_2}}} = {x \over {2x}} \times 1 = {1 \over 2}$

Now, ${K_p} = {{{{({p_{Cl}})}^2}} \over {{p_{C{l_2}}}}} $

$\Rightarrow {K_p} = {{{{(1/2)}^2}} \over {1/2}} = {1 \over 2} = 0.5$

= 5 $\times$ 10^$-$1

Hence, x $\times$ 10^$-$1

x = 5