Chemical Equilibrium

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.