Chemical Equilibrium

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 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.

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.

Total number of moles $=0.5+1=1.5 \mathrm{~mol}$

$\begin{aligned} & p_{\mathrm{N}_2 \mathrm{O}_4}=\frac{0.5}{1.5} \times 1 \mathrm{~atm}=\frac{1}{3} \mathrm{~atm} \\ & p_{\mathrm{NO}_2}=\frac{1}{1.5} \times 1 \mathrm{~atm}=\frac{2}{3} \mathrm{~atm} \end{aligned}$

According to law of chemical equilibrium

$K_p=\frac{p^2 \mathrm{NO}_2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{\left(\frac{2}{3}\right)^2}{\left(\frac{1}{3}\right)}=1.33 \mathrm{~atm}$

We know,

$\Delta G \Upsilon=-2.303 R T \log K_p$ ...... (i)

Substituting value in Eq. (i), we get

$\begin{aligned} \Delta G \Upsilon & =-2.303 \times 0.082 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 333 \mathrm{~K} \\ & =-7.79 \mathrm{~atm} \text { litre } \quad \end{aligned}$

$\begin{aligned} & 1 \mathrm{~atm} \text { litre }=101.32500 \text { joules } \\ & \begin{aligned} -7.79 \mathrm{~atm} \text { litre } & =-7.79 \times 101.32 \text { joules } \\ & =-789.2 \mathrm{~J} \mathrm{~mol}^{-1} \approx-790 \mathrm{~J} \mathrm{~mol}^{-1} \end{aligned} \end{aligned}$.

Le-Chatelier principle is not applicable to pure solids and liquids because of the reason that they experience negligible change in concentration during chemical equilibrium which means that adding or removing a solid form system at equilibrium has no effect on equilibrium.

$\mathrm{Fe}(s)+\mathrm{S}(s) \rightleftharpoons \mathrm{FeS}(s)$ is in solid equilibrium phase, hence Le-Chatelier principle is not applicable.

$\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{NO}_2(g)$

Given, $\Delta_f H^{\circ}[\mathrm{NO}(g)]=90.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\begin{aligned} & \Delta_f H^{\circ}\left[\mathrm{NO}_2(g)\right]=32.48 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta H^{\circ}=\Sigma \Delta H^{\circ} \text { product }-\Sigma \Delta_f H^{\circ}(\mathrm{NO}) \\ & =3248-90.4 \\ & =-57.92 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \end{aligned}$

We know that,

Standard Gibb's free energy, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$

$\begin{gathered} =-57.92-298 \times(-70.8) \\ =-57.92+21.098=-36.82 \mathrm{~kJ} / \mathrm{mol} \\ =-38.82 \times 10^3 \mathrm{~J} / \mathrm{mol} \end{gathered}$

We also know, $\Delta G^{\circ}=-2.303 R T \log K_{\text {eq }}$

Here, $K_{\mathrm{eq}}=$ Equilibrium constant

$\begin{aligned} -36.82 \times 10^3 & =-2303 \times 8.314 \times 298 \times \log K_{\mathrm{eq}} \\ \log K_{\mathrm{eq}} & =\frac{36.82}{5.70584} \\ \log K_{\mathrm{eq}} & =0.501 \\ K_{\mathrm{eq}} & =\text { antilog } 0.501 \\ & =3.162 \times 10^6 \end{aligned}$

$\text { } \begin{aligned} & \frac{1}{2} X_2+\frac{3}{2} Y_2 \rightleftharpoons X Y_3 ; \Delta H=-30 \mathrm{~kJ} \\ & \Delta S_{\text {reaction }}=\sum_\limits{i=1} \Delta S_{\text {product }}-\sum \Delta S_{\text {reactant }} \\ & X_2+3 Y_2 \rightleftharpoons 2 X Y_3 \\ & \Delta S_{\text {reaction }}=2 \times 50-3 \times 40-1 \times 60 \\ &=-80 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \end{aligned}$

Using Gibb's free energy, $\Delta G=\Delta H-T \Delta S$

$\begin{aligned} \Delta G & =0 \\ 0 & =\Delta H-T \Delta S \Rightarrow \Delta H=T \Delta S \\ 1000 \times(-60) & =T \times(-80) \Rightarrow T=750 \mathrm{~K} .\end{aligned}$

$\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{SO}_3(g)$

The above equilibrium reaction is exothermic reaction. When pressure is lowered, the forward reaction is favoured.

On going from $p_1$ to $p_3$ yield of $\mathrm{SO}_3$ is increasing because lower pressure favours forward reaction.

The relationship between K$_p$ and K$_c$ is

${K_p} = {K_c}{(RT)^{\Delta n}}$ ..... (i)

Here, $\Delta n$ = (number of moles of product of gas) $-$ (number of moles of reactant of gas)

K$_p$ = equilibrium constant at constant pressure

K$_c$ = equilibrium constant at constant concentration

R = rate constant

$2 A(g) \rightleftharpoons B(g)+C(g)$

The value of $\Delta n$ is $=2-2=0$

Put value of $\Delta n$ in Eq. (i) we get,

$K_p=K_C(R T)^0 \Rightarrow\left[K_p=K_C\right] $