Chemical Equilibrium

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} $

1. Initial Equilibrium

We start with 2 moles of $A_2$ in a 1 L flask, so the initial concentration of $A_2$ is 2 M.

Let 'x' be the change in concentration of $A_2$ as it converts to $B_2$ at equilibrium.

$ \begin{array}{ccc} & A_2(g) \rightleftharpoons & B_2(g) \\ \text{Initial} & 2 & 0 \\ \text{Change} & -x & +x \\ \text{Equilibrium} & 2-x & x \end{array} $

The equilibrium constant $K_c$ is given by:

$K_c = \frac{[B_2]}{[A_2]} = \frac{x}{2-x} = 99.0$

Solving for x:

$x = 99(2 - x)$

$x = 198 - 99x$

$100x = 198$

$x = 1.98$

So, at the first equilibrium:

$C_1(A_2) = 2 - x = 2 - 1.98 = 0.02 \, \text{M}$

$C_2(B_2) = x = 1.98 \, \text{M}$

2. Adding More A₂ and Re-establishing Equilibrium

Now, we add 1 mole of $A_2$ to the flask. The concentration of $A_2$ becomes $0.02 + 1 = 1.02 \, \text{M}$. The concentration of $B_2$ remains $1.98 \, \text{M}$.

Let 'y' be the change in concentration of $A_2$ as it converts to $B_2$ to reach the new equilibrium.

$ \begin{array}{ccc} & A_2(g) \rightleftharpoons & B_2(g) \\ \text{Initial} & 1.02 & 1.98 \\ \text{Change} & -y & +y \\ \text{Equilibrium} & 1.02-y & 1.98+y \end{array} $

The equilibrium constant $K_c$ remains the same:

$K_c = \frac{[B_2]}{[A_2]} = \frac{1.98 + y}{1.02 - y} = 99.0$

Solving for y:

$1.98 + y = 99(1.02 - y)$

$1.98 + y = 100.98 - 99y$

$100y = 99$

$y = 0.99$

So, at the second equilibrium:

$C_3(A_2) = 1.02 - y = 1.02 - 0.99 = 0.03 \, \text{M}$

$C_4(B_2) = 1.98 + y = 1.98 + 0.99 = 2.97 \, \text{M}$

Therefore, the concentration of $A_2$ at the new equilibrium, $C_3(A_2)$, is 0.03 M.

Answer:

Option C is the correct answer: 0.03

To analyze the equilibrium concentrations for the reaction $ A_2(g) \rightleftharpoons B_2(g) $, where the equilibrium constant $ K_c $ at temperature $ T(\text{K}) $ is 99.0, follow these steps:

Given:

$ K_c = 99.0 $

Initial moles of $ A_2 $: 2 moles

Volume of the flask: 1 L

Calculation:

Initial Conditions and Definition:

Since the reaction starts with 2 moles of $ A_2(s) $ which sublimes to $ A_2(g) $, the initial concentration of $ A_2 $ is 2 mol/L (as the flask is 1 L in volume).

Setting up the Equilibrium Expression:

For the equilibrium constant expression:

$ K_c = \frac{[B_2]}{[A_2]} $

Let the concentration of $ B_2 $ at equilibrium be $ x $ mol/L. Therefore, the concentration of $ A_2 $ becomes $ (2 - x) $ mol/L at equilibrium.

Solving the Equilibrium Equation:

Substitute the equilibrium values into the $ K_c $ expression:

$ 99 = \frac{x}{2 - x} $

Solving for $ x $:

$ 99(2 - x) = x $

$ 198 - 99x = x $

$ 198 = 100x $

$ x = 1.98 $

Thus, the concentration of $ B_2 $ is $ [B_2] = 1.98 $ mol/L.

Finding the Concentration of $ A_2 $:

Calculate $ [A_2] $ using the expression $ 2 - x $:

$ [A_2] = 2 - 1.98 = 0.02 \text{ mol/L} $

Conclusion:

The concentrations at equilibrium are:

$ [A_2] = 0.02 \, \text{mol/L} $

$ [B_2] = 1.98 \, \text{mol/L} $

Let's work through the problem step by step.

The dissociation of the weak acid is given by:

$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-.$

The initial concentration of HA is 0.5 M. With a degree of dissociation, $\alpha = 0.01$ (or 1%), the equilibrium concentrations are:

[HA] = 0.5(1 - 0.01) = 0.5 × 0.99 = 0.495 M

[H⁺] = 0.5 × 0.01 = 0.005 M

[A⁻] = 0.005 M

The acid dissociation constant, $K_a$, is defined as:

$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}.$

Substitute the equilibrium concentrations:

$K_a = \frac{(0.005)(0.005)}{0.495} = \frac{0.000025}{0.495}.$

Simplifying the expression:

$K_a \approx 5 \times 10^{-5}.$

Thus, the approximate value of $K_a$ (or $K_e$, as referenced) is $5 \times 10^{-5}$.

The correct option is Option B.

The $K_C$ for the reaction, $\frac{1}{3} \mathrm{~N}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \frac{2}{3} \mathrm{NH}_3(\mathrm{~g})$ is 50 (at $T(\mathrm{~K}))$.

$ \therefore \quad K_C=\frac{\left[\mathrm{NH}_3\right]^{2 / 3}}{\left[\mathrm{~N}_2\right]^{1 / 3}\left[\mathrm{H}_2\right]}=50 $

Thus, for reaction,

$ \begin{aligned} 2 \mathrm{NH}_3(g) & \rightleftharpoons \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \\ K_{C_1} & =\frac{\left[\mathrm{H}_2\right]^3\left[\mathrm{~N}_2\right]}{\left[\mathrm{NH}_3\right]^2} \\ K_{C_1} & =\left[\frac{1}{K_C}\right]^3=\left(\frac{1}{50}\right)^3 \\ & =8 \times 10^{-6} \end{aligned} $

Note ' $A$ ' will be formed in same amount as ' $Z$ '.

$ \begin{aligned} & \therefore \text { Equilibrium constant, } K_C=\frac{[Z][A]}{[X][Y]} \\ & \quad=\frac{0.5 \times 0.5}{(1-0.5)(1-0.5)}=1 \end{aligned} $

At 750 K and 10 atm , the equilibrium constant of reaction $2 A(g) \rightleftharpoons B(g)+C(g)$ is 3.52 . It will be same at 750 K and 7.04 atm pressure as equilibrium constant doesn't depend on pressure, catalyst and concentration. It just depends on absolute temperature.

Here 4 moles of inert gas argon also present.

$\therefore$ Total moles of mixture present at equilibrium,

n_T = 5 + x + 4

= 9 + x

At equilibrium, total pressure (p_T) = 6 atm

Volume (v) = 100 L

Temperature = 610 K

$\therefore$ Using ideal gas equation,

${P_T}V = {n_T}RT$

$ \Rightarrow 6 \times 100 = (9 + x) \times 0.082 \times 610$

$ \Rightarrow x = 3$

Now,

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

$ = {{\left[ {{3 \over {9 + 3}} \times 6} \right] \times \left[ {{3 \over {9 + 3}} \times 6} \right]} \over {\left[ {{{5 - 3} \over {9 + 3}} \times 6} \right]}}$

$ = {{27} \over {12}}$

$ = {9 \over 4}$

= 2.25 atm

Note :

Inert gas always contribute to total mole and pressure calculation.

Now,

${K_P}$ or $K = {{{P_B} \times {{\left( {{P_C}} \right)}^{{1 \over 2}}}} \over {{P_A}}}$

$ = {{\left( {{\alpha \over {1 + {\alpha \over 2}}}} \right)P \times {{\left[ {\left( {{{{\alpha \over 2}} \over {1 + {\alpha \over 2}}}} \right)P} \right]}^{{1 \over 2}}}} \over {\left( {{{1 - \alpha } \over {1 + {\alpha \over 2}}}} \right)P}}$

$ = {{\left( {{{2\alpha } \over {2 + \alpha }}} \right)P \times {{\left[ {\left( {{\alpha \over {2 + \alpha }}} \right)P} \right]}^{{1 \over 2}}}} \over {\left( {{{2(1 - \alpha )} \over {2 + \alpha }}} \right)P}}$

$= {\alpha \over {1 - \alpha }} \times {\left( {{{\alpha P} \over {2 + \alpha }}} \right)^{{1 \over 2}}}$

$ = {{{\alpha ^{{3 \over 2}}}\,.\,{P^{{1 \over 2}}}} \over {(1 - \alpha ){{(2 + \alpha )}^{{1 \over 2}}}}}$

$ \begin{aligned} & \text { } 4 \mathrm{NH}_3+5 \mathrm{O}_2(g) \rightleftharpoons 4 \mathrm{NO}(g) \begin{aligned} & +6 \mathrm{H}_2 \mathrm{O}(g) \end{aligned} \\ & \begin{aligned} K_p & =\frac{\left(p_{\mathrm{NO}}\right)^4\left(p_{\mathrm{H}_2 \mathrm{O}}\right)^6}{\left(p_{\mathrm{O}_2}\right)^5\left(p_{\mathrm{NH}_3}\right)^4} \Rightarrow \frac{(0.5)^4(0.5)^6}{(0.5)^5(0.5)^4} \\ & =0.5 \mathrm{~atm} \end{aligned} \end{aligned} $

Formation of ammonia is given by

$ \text {} \begin{aligned} Now, \,\, {\left[\mathrm{N}_2\right] } & =\frac{n_{\mathrm{N}_2}}{V}=\frac{(1-x)}{V} \\ {\left[\mathrm{H}_2\right] } & =\frac{n_{\mathrm{H}_2}}{V}=\frac{(3-3 x)}{V} \\ {\left[\mathrm{NH}_3\right] } & =\frac{n_{\mathrm{NH}_3}}{V}=\frac{2 x}{V} \end{aligned} $

$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} K_C & =\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}=\frac{\left(\frac{2 x}{V}\right)^2}{\left(\frac{1-x}{V}\right)\left(\frac{3-3 x}{V}\right)^3} \\ & =\frac{4 x^2}{V^2} \times \frac{V^4}{(1-x)(3-3 x)^3}=\frac{4 x^2 V^2}{27(1-x)^4} . \end{aligned} \end{aligned} $

The given reaction is

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

$k_p$ can be calculated as

$ k_p=\frac{\left[p_{\mathrm{SO}_3}\right]^2}{\left[p_{\mathrm{SO}_2}\right]^2\left[p_{\mathrm{O}_2}\right]} $

[ $p_{\mathrm{SO}_3}=$ partial pressure of $\mathrm{SO}_3$,

$p_{\mathrm{SO}_2}=$ partial pressure of $\mathrm{SO}_2$,

$p_{\mathrm{O}_2}=$ partial pressure of $\mathrm{O}_2$ ]

As, $\quad\left[p_{\mathrm{SO}_3}\right]=\left[p_{\mathrm{SO}_2}\right]$

$ \begin{aligned}So,\,\,\,\, 5 & =\frac{1}{\left[p_{\mathrm{O}_2}\right]} \\ {\left[p_{\mathrm{O}_2}\right] } & =\frac{1}{5}=0.2 \end{aligned} $

So, partial pressure of oxygen will be 0.2 atm .

On increasing pressure, equilibrium shifts towards lesser number of gaseous moles. It means gaseous moles on product are less than gaseous moles on reactant side, i.e. $\Delta n_g<0$.

(i) $X_2(g)+3 Y_2(g) \rightleftharpoons 2 X Y_3(g) \Delta n_g=2-4=-2$;

(ii) $X_2(g)+Y_2(g) \rightleftharpoons 2 X Y(g)\Delta n_g=2-2=0$;

(iii) $X_2(g)+Z_2(g) \rightleftharpoons 2 X Z(g)$; \Delta n_g=2-2=0

(iv) $X_2(g)+Y_4(g) \rightleftharpoons 2 X Y_2(g)\Delta n_g=2-2=0 $;

$\Delta n_g<0$ only for reaction (i) hence, position of equilibrium shifts towards the products if total pressure is increased.

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

$ K_p=K_C(R T)^{\Delta n_g} $

$\Delta n_g=$ moles of gaseous products moles of gaseous reactants.

$ \begin{aligned} \Delta n_g & =n_p(g)-n_r(g)=2-(1+3)=-2 \\ K_p & =K_C(R T)^{-2} \Rightarrow \frac{K_p}{K_C}=\frac{1}{(R T)^2} \end{aligned} $

$ \begin{aligned} &\text {For the reaction at equilibrium, }\\ &\begin{aligned} & A(s) \rightleftharpoons B(s)+C(g) \\ & K=\frac{[B][C]}{[A]} \end{aligned} \end{aligned} $

The concentrations of pure solids, pure liquids, and solvents are omitted from equilibrium constant expressions because they do not change significantly during reactions when enough amount is present to reach equilibrium. Therefore, the expression is written as, $K_C=[C] K_p=p C(g)$, where $p C(g)$ is the partial pressure of $C$.

$\Delta H$ is dependent on temperature, if the reaction is exothermic, and the temperature is increased, the reaction enthalpy becomes more negative (more heat energy is released). If the reaction is endothermic, and the temperature is increased, the reaction enthalpy becomes more positive (more heat energy is absorbed).

$\Delta H$ is independent of catalyst addition. The catalyst only decreases the activation energy of a reaction, thereby increases

the reaction rate but does not affect the enthalpy of reaction.

So, statements (A), (B) and (D) are only correct.

At 500 K, for the reaction

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

the equilibrium constant $K_p$ is $0.036 \mathrm{~atm}^{-2}$. To find the equilibrium constant $K_C$ in $\mathrm{L}^2 \mathrm{~mol}^{-1}$, given that the gas constant $R$ is $0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, we utilize the relationship between $K_p$ and $K_C$. The formula is:

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

For this reaction:

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

The change in moles of gas, $\Delta n$, is calculated as follows:

$\Delta n = ( \text{moles of products} ) - ( \text{moles of reactants} ) = 2 - (3 + 1) = 2 - 4 = -2$

Given that:

$K_p = 0.036 \text{ (provided)}$

We can now use the relationship to find $K_C$:

$0.036 = K_C (0.082 \times 500)^{-2}$

Solving for $K_C$ gives us:

$K_C = \frac{0.036}{(0.082 \times 500)^{-2}} = 60.516 \mathrm{~L}^2 \mathrm{~mol}^{-2}$

$3 \mathrm{H}_2(g)+\mathrm{N}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) ; \Delta H=+\mathrm{ve}$,

As the formation of ammonia from its constituent elements is an exothermic reaction. Thus, energy is released during the reaction. So, on increasing the temperature, the reaction is favoured in backward direction. As per Le-Chatelier's principle, change made in one direction/site is opposed by movement of reaction in opposite direction.

NH₄OH $\rightleftharpoons$ NH$_{4}^+$ + OH^$-$

K_b = $\frac{[NH_{4}^+][OH^-]}{[NH_4OH]}$

1.8 $\times$ 10^$-$5 = $\frac{[NH_{4}^+][0.01]}{[0.02]}$

$\therefore$ [NH$_{4}^+$] = 3.6 $\times$ 10^$-$5 M

van't Hoff equation giving variation of equilibrium constant, K with temperature T is

log_e K = $-$ $\frac{\Delta H^\circ}{RT}+\frac{\Delta S^\circ}{R}$

Given, log_e K = $-$ $\frac{2000}{T}+4.0$

$\therefore$ $\Delta$S$^\circ$/R = 4 $\Rightarrow$ $\Delta$S$^\circ$ = 4R