Chemical Equilibrium | JEE Main Chemistry Chapter Wise Question Papers

Q 11 :
For the reaction,

$H_{2 (g)} + I_{2 (g)} ⇌ 2 HI (g)$

Attainment of equilibrium is predicted correctly by: [2025]

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During the time interval from start of the reaction till equilibrium, concentration of reactants ( $H_{2}$ and $I_{2}$ ) decrease and concentration of products (HI) increase. Once equilibrium is reached concentration of each species becomes constant.

Q 12 :
At temperature T, compound $A B_{2} (g)$ dissociates as $A B_{2} (g) ⇌ A B (g) + \frac{1}{2} B_{2 (g)}$ having degree of dissociation $x$ (small compared to unity). The correct expression for $x$ in terms of $K_{p}$ and $p$ is [2025]

$\sqrt[3]{\frac{2 K_{p}^{2}}{p}}$
$\sqrt{K_{p}}$
$\sqrt[3]{\frac{2 K_{p}}{p}}$
$\sqrt[4]{\frac{2 K_{p}}{p}}$

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$\begin{matrix} {AB}_{2} (g) & ⇌ & AB (g) & + & \frac{1}{2} B_{2} (g) \\ t=0 (moles) & a & 0 & 0 \\ t =∞ (moles) & a (1 - x) & a x & a x / 2 \\ t =∞ (mole fraction) & \frac{a (1 - x)}{a (1 + \frac{x}{2})} & \frac{a x}{a (1 + \frac{x}{2})} & \frac{a x}{a (1 + \frac{x}{2})} \\ t =∞ (partial pressure) & \frac{a (1 - x)}{a (1 + \frac{x}{2})} p & \frac{a x}{a (1 + \frac{x}{2})} p & \frac{a x / 2}{a (1 + \frac{x}{2})} p \\ \approx p & \approx x p & \approx \frac{x p}{2} \end{matrix} \begin{matrix} Total = a (1 + x / 2) \end{matrix}$
$K_{p} = \frac{P_{AB} \times {(P_{B_{2}})}^{1 / 2}}{P_{{AB}_{2}}} = \frac{x p \times {(\frac{x p}{2})}^{1 / 2}}{p}$

${(K_{p})}^{2} = x^{3} p / 2$

$x = \sqrt[3]{\frac{2 {(K_{p})}^{2}}{p}}$

Q 13 :
Consider the fraction

$X_{2} Y (g) = X_{2} (g) + \frac{1}{2} Y_{2} (g)$

The equation representing correct relationship between the degree of dissociation $(x)$ of $X_{2} Y$ (g) with its equilibrium constant $Kp$ is _______ .

Assume $x$ to be very small. [2025]

$x = \sqrt[3]{\frac{2 K p^{2}}{p}}$
$x = \sqrt[3]{\frac{K p}{2 p}}$
$x = \sqrt[3]{\frac{K p}{p}}$
$x = \sqrt[3]{\frac{2 K p}{p}}$

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$\begin{matrix} X_{2} Y (g) & ⇌ & X_{2} (g) & + & \frac{1}{2} Y_{2} (g) \\ t=0 (moles) & a & 0 & 0 \\ t =∞ (moles) & a (1 - x) & a x & \frac{a x}{2} \\ t =∞ (mole fraction) & \frac{a (1 - x)}{a (1 + \frac{x}{2})} & \frac{a x}{a (1 + \frac{x}{2})} & \frac{a x}{2 a (1 + \frac{x}{2})} \\ = (\frac{1 - x}{1 + \frac{x}{2}}) & = (\frac{x}{1 + \frac{x}{2}}) & \frac{x}{2 (1 + \frac{x}{2})} \\ \approx 1 & \approx x & \approx \frac{x}{2} \\ Partial pressures & 1 \times P & x \times P & \frac{x \times P}{2} \end{matrix} \begin{matrix} Total = a (1 + \frac{x}{2}) \end{matrix}$

$K_{p} = \frac{P_{X_{2}} \times {(P_{Y_{2}})}^{1 / 2}}{P_{X_{2} Y}} = \frac{x P \times {(\frac{x P}{2})}^{1 / 2}}{P}$

$K_{p} = \frac{x^{3 / 2} P^{1 / 2}}{2^{1 / 2}}$

$K_{p}^{2} = \frac{x^{3} P}{2}$

$x = \sqrt[3]{\frac{2 K_{p}^{2}}{P}}$

Q 14 :
Consider the following chemical equilibrium of the gas phase reaction at a constant temperature:

$A (g) ⇌ B (g) + C (g)$

If $p$ being the total pressure, $K_{p}$ is the pressure equilibrium constant and $α$ is the degree of dissociation, then which of the following is true at equilibrium? [2025]

If $p$ value is extremely high compared to $K_{p}, α \approx 1$
When $p$ increases $α$ decreases
If $K_{p}$ value is extremely high compared to $p, α$ becomes much less than unity
When $p$ increases $α$ increases

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$\begin{matrix} A (g) & ⇌ & B (g) & + & C (g) \\ t=0 (moles) & a & 0 & 0 \\ t =∞ (moles) & a (1 - α) & a α & a α \\ t =∞ (mole fraction) & \frac{1 - α}{1 + α} & \frac{α}{1 + α} & \frac{α}{1 + α} \\ t =∞ (partial pressure) & (\frac{1 - α}{1 + α}) P & (\frac{α}{1 + α}) P & (\frac{α}{1 + α}) P \end{matrix} \begin{matrix} Total = a (1 + α) \end{matrix}$

$K_{p} = \frac{P_{B} \times P_{C}}{P_{A}} = \frac{(\frac{α}{1 + α}) P \times (\frac{α}{1 + α}) P}{(\frac{1 - α}{1 + α}) P} = \frac{α^{2} P}{1 - α^{2}}$

As $K_{p}$ is constant for a constant temperature, as P increases, $α$ decreases.

Q 15 :
Consider the following equilibrium,

$CO(g) + 2 H_{2} (g) \to {CH}_{3} OH(g)$

0.1 mol of CO along with a catalyst is present in a 2 ${d m}^{3}$ flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of $C H_{3} O H$ is formed. The $K_{p}^{0}$ is ______ $\times 10^{- 3}$ (nearest integer).

Given : R = 0.08 ${d m}^{3}$ bar $K^{- 1}$ ${mol}^{- 1}$

Assume only methanol is formed as the product and the system follows ideal gas behaviour. [2025]

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$\begin{matrix} CO(g) & + & 2 H_{2} (g) & ⇌ & {CH}_{3} OH(g) \\ t=0 (moles) & 0.1 & x & 0 \\ t=∞ (moles) & 0.1 - y & x - 2 y & y \end{matrix}$

As moles of $C H_{3} O H = y = 0.04$ mol

$\begin{matrix} CO(g) & + & 2 H_{2} (g) & ⇌ & {CH}_{3} OH(g) \\ t =∞ (moles) & 0.1 - 0.04 & x - 0.08 & 0.04 \\ = 0.06 \end{matrix}$

Total moles = $0.06 + x - 0.08 + 0.04 = (0.02 + x)$ mol

By ideal gas equation at equilibrium:

$P V = n R T$

$5 \times 2 = (0.02 + x) \times 0.08 \times 500$

$x = 0.23 mol$

Thus:

$\begin{matrix} CO(g) & + & 2 H_{2} (g) & ⇌ & {CH}_{3} OH(g) \\ t =∞ (moles) & 0.06 & 0.23 - 0.08 = 0.15 & 0.04 \\ t =∞ (mole fraction) & \frac{6}{25} & \frac{15}{25} & \frac{4}{25} \\ t =∞ (partial pressure) & \frac{6}{25} \times 5 = \frac{6}{5} & \frac{15}{25} \times 5 = 3 & \frac{4}{25} \times 5 = \frac{4}{5} \end{matrix} \begin{matrix} Total=0.25 \end{matrix}$

$K_{p} = \frac{P_{{CH}_{3} OH}}{P_{CO} \times {(P_{H_{2}})}^{2}}$ $= \frac{4 / 5}{(\frac{6}{5}) \times {(3)}^{2}} = 74 \times 10^{- 3}$

Q 16 :
37.8 g of $N_{2} O_{5}$ was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K

$2 N_{2} O_{5} (g) ⇌ 2 N_{2} O_{4} (g) + O_{2} (g)$

The total pressure at equilibrium was found to be 18.65 bar.

Then, $K_{p}$ = ______ $\times 10^{- 2}$ [nearest integer]

Assume $N_{2} O_{5}$ to behave ideally under these conditions.

Given: R = 0.082 bar L ${mol}^{- 1}$ $K^{- 1}$ [2025]

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Initial moles of $N_{2} O_{5} (n_{N_{2} O_{5}})$ $= \frac{Given mass of N_{2} O_{5}}{Molar mass of N_{2} O_{5}} = \frac{37.8}{108} mol = 0.35 mol$

Initial pressure of $N_{2} O_{5} = \frac{n_{N_{2} O_{5}} R T}{V}$

$= \frac{0.35 \times 0.082 \times 500}{1} bar = 14.35 bar$

$\begin{matrix} 2 N_{2} O_{5} (g) & ⇌ & 2 N_{2} O_{4} (g) & + & O_{2} (g) \\ t = 0 & 14.35 bar & 0 & 0 \\ t = \infty & 14.35 - 2 p & 2 p & p \end{matrix} \begin{matrix} total = 14.35 + p \end{matrix}$

Total pressure at equilibrium = 14.35 + p = 18.65, p = 4.3 bar

$K_{p} = \frac{P_{O_{2}} \times P_{N_{2} O_{4}}^{2}}{P_{N_{2} O_{5}}^{2}} = \frac{p \times {(2 p)}^{2}}{{(14.35 - 2 p)}^{2}}$

$= \frac{4.3 \times {(2 \times 4.3)}^{2}}{{(14.35 - 2 \times 4.3)}^{2}} = \frac{4.3 \times {(2 \times 4.3)}^{2}}{{(5.75)}^{2}} ≃ 9.62 = 962 \times 10^{- 2}$

Q 17 :
The equilibrium constant for decomposition of $H_{2} O (g)$
$H_{2} O(g) ⇌ H_{2} (g) + \frac{1}{2} O_{2} (g) (Δ G ° = 92.34 {kJ mol}^{- 1})$

is $8.0 \times 10^{- 3}$ at 2300 K and total pressure at equilibrium is 1 bar.
Under this condition, the degree of dissociation $(α)$ of water is $_____\times 10^{- 2}$ (nearest integer value).

[Assume $α$ is negligible with respect to 1] [2025]

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$\begin{matrix} H_{2} O (g) & ⇌ & H_{2} (g) & + & \frac{1}{2} O_{2} (g) \\ t =0 (mol) & a & 0 & 0 \\ t =∞ (mol) & a (1 - α) & a α & \frac{a α}{2} & Total= a (1 + \frac{α}{2}) \\ \approx a (as α ≪ 1) \\ \approx a (as α ≪ 1) \\ t =∞ (mol fraction) & \frac{a}{a} = 1 & \frac{a α}{a} = α & \frac{a α}{2 a} = \frac{α}{2} \\ t =∞ & 1 \times P & α \times P & \frac{α}{2} \times P \\ (partial & = 1 \times 1 & = α \times 1 & = \frac{α}{2} \times 1 \\ pressure) & = 1 & = α & = \frac{α}{2} \end{matrix}$

$K_{p} = \frac{{(P_{O_{2}})}^{1 / 2} \times P_{H_{2}}}{P_{H_{2} O}} = \frac{{(\frac{α}{2})}^{1 / 2} \times α}{1} = \frac{α^{3 / 2}}{\sqrt{2}}$

$8 \times 10^{- 3} = \frac{α^{3 / 2}}{\sqrt{2}}$

$α^{3} = 128 \times 10^{- 6}$

$α = {(128)}^{1 / 3} \times 10^{- 2} = 5.03 \times 10^{- 2}$

Topic Question Set

Q 11 :
For the reaction,

$H_{2 (g)} + I_{2 (g)} ⇌ 2 HI (g)$

Attainment of equilibrium is predicted correctly by: [2025]

Q 12 :
At temperature T, compound $A B_{2} (g)$ dissociates as $A B_{2} (g) ⇌ A B (g) + \frac{1}{2} B_{2 (g)}$ having degree of dissociation $x$ (small compared to unity). The correct expression for $x$ in terms of $K_{p}$ and $p$ is [2025]

Q 13 :
Consider the fraction

$X_{2} Y (g) = X_{2} (g) + \frac{1}{2} Y_{2} (g)$

The equation representing correct relationship between the degree of dissociation $(x)$ of $X_{2} Y$ (g) with its equilibrium constant $Kp$ is _______ .

Assume $x$ to be very small. [2025]

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Topic Question Set

Q 11 : For the reaction, H2(g)+I2(g) ⇌ 2HI(g) Attainment of equilibrium is predicted correctly by: [2025]

Q 12 : At temperature T, compound AB2(g) dissociates as AB2(g)⇌AB(g)+12B2(g) having degree of dissociation x (small compared to unity). The correct expression for x in terms of Kp and p is [2025]

Q 13 : Consider the fraction X2Y(g) = X2(g)+12Y2(g) The equation representing correct relationship between the degree of dissociation (x) of X2Y(g) with its equilibrium constant Kp is _______ . Assume x to be very small. [2025]

Q 14 : Consider the following chemical equilibrium of the gas phase reaction at a constant temperature: A(g)⇌B(g)+C(g) If p being the total pressure, Kp is the pressure equilibrium constant and α is the degree of dissociation, then which of the following is true at equilibrium? [2025]

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Q 11 :
For the reaction,

$H_{2 (g)} + I_{2 (g)} ⇌ 2 HI (g)$

Attainment of equilibrium is predicted correctly by: [2025]

Q 12 :
At temperature T, compound $A B_{2} (g)$ dissociates as $A B_{2} (g) ⇌ A B (g) + \frac{1}{2} B_{2 (g)}$ having degree of dissociation $x$ (small compared to unity). The correct expression for $x$ in terms of $K_{p}$ and $p$ is [2025]

Q 13 :
Consider the fraction

$X_{2} Y (g) = X_{2} (g) + \frac{1}{2} Y_{2} (g)$

The equation representing correct relationship between the degree of dissociation $(x)$ of $X_{2} Y$ (g) with its equilibrium constant $Kp$ is _______ .

Assume $x$ to be very small. [2025]