Chemical Equilibrium | JEE Main Chemistry Chapter Wise Question Papers

Q 1 :

For the given reaction, choose the correct expression of $K_{C}$ from the following:

${Fe}_{(a q)}^{3 +} + {SCN}_{(a q)}^{-} ⇌ {(FeSCN)}_{(a q)}^{2 +}$ [2024]

$K_{C} = \frac{{[{FeSCN}^{2 +}]}^{2}}{[{Fe}^{3 +}] [{SCN}^{-}]}$
$K_{C} = \frac{[{FeSCN}^{2 +}]}{{[{Fe}^{3 +}]}^{2} {[{SCN}^{-}]}^{2}}$
$K_{C} = \frac{[{Fe}^{3 +}] [{SCN}^{-}]}{[{FeSCN}^{2 +}]}$
$K_{C} = \frac{[{FeSCN}^{2 +}]}{[{Fe}^{3 +}] [{SCN}^{-}]}$

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(4)

${Fe}^{3 +} (a q) + {SCN}^{-} (a q) ⇌ {(FeSCN)}^{2 +} (a q)$

$K_{c} = \frac{[{FeSCN}^{2 +}]}{[{Fe}^{3 +}] [{SCN}^{-}]}$

Q 2 :

The equilibrium constant for the reaction ${SO}_{3} (g) ⇌ {SO}_{2} (g) + \frac{1}{2} O_{2} (g)$ is $K_{C} = 4.9 \times 10^{- 2}$ . The value of $K_{C}$ for the reaction given below is

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ is : [2024]

4.9
49
41.6
416

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(4)

${SO}_{3} (g) ⇌ {SO}_{2} (g) + \frac{1}{2} O_{2} (g), K_{c} = 4.9 \times 10^{- 2}$ - I

Reverse the above equation. On reversing a chemical equation, equilibrium constant gets inversed.

${SO}_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {SO}_{3} (g), K_{c}^{1} = \frac{1}{4.9 \times 10^{- 2}}$ - II

Multiply II by 2. On multiplying an equation by $n$ , equilibrium constant gets raised by a power $n$ .

$2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g), K_{c}^{11} = {(\frac{1}{4.9 \times 10^{- 2}})}^{2} = 416.49$

Q 3 :

At –20°C and 1 atm pressure, a cylinder is filled with equal number of $H_{2}$ , $I_{2}$ and HI molecules for the reaction $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ , the $K_{p}$ for the process is $x \times 10^{- 1}$ , $x =$ _______ . [Given : R = 0.082 L atm $K^{- 1}$ ${mol}^{- 1}$ ] [2024]

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10
0.01

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(3)

$n_{HI} = n_{H_{2}} = n_{I_{2}} = x$

$H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$

$K_{p} = K_{c} = \frac{{[HI]}^{2}}{[H_{2}] [I_{2}]} = \frac{{(n_{HI} / V)}^{2}}{(n_{H_{2}} / V) (n_{I_{2}} / V)} = \frac{{(n_{HI})}^{2}}{n_{H_{2}} \times n_{I_{2}}} = \frac{{(x)}^{2}}{x \times x} = 10 \times 10^{- 1}$

Q 4 :

The ratio $\frac{K_{P}}{K_{C}}$ for the reaction:

${CO}_{(g)} + \frac{1}{2} O_{2 (g)} ⇌ {CO}_{2 (g)}$ is : [2024]

$R T$
${(R T)}^{1 / 2}$
$\frac{1}{\sqrt{R T}}$
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(3)

$CO (g) + \frac{1}{2} O_{2} (g) ⇌ {CO}_{2} (g)$

$Δ n_{g} = Gaseous mol of product - gaseous mol of reactant$

$= 1 - (1 + 0.5) = - 0.5$

$K_{p} = K_{c} {(R T)}^{Δ n_{g}}$

$\frac{K_{p}}{K_{c}} = {(R T)}^{- 0.5} = \frac{1}{\sqrt{R T}}$

Q 5 :

For the given hypothetical reactions, the equilibrium constants are as follows:

$X ⇌ Y; K_{1} = 1.0$

$Y ⇌ Z; K_{2} = 2.0$

$Z ⇌ W; K_{3} = 4.0$

The equilibrium constant for the reaction $X ⇌ W$ is [2024]

12.0
6.0
8.0
7.0

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(3)

On adding reactions, equilibrium constants are multiplied

$X ⇌ Y, K_{1} = \frac{[Y]}{[X]}$

$Y ⇌ Z, K_{2} = \frac{[Z]}{[Y]}$

$Z ⇌ W, K_{3} = \frac{[W]}{[Z]}$

___________________

Adding all three:

$X ⇌ W, K = \frac{[W]}{[X]} = \frac{[W] [Z] [Y]}{[Z] [Y] [X]} = K_{1} K_{2} K_{3} = 1 \times 2 \times 4 = 8$

Q 6 :

For the reaction $N_{2} O_{4 (g)} ⇄ 2 {NO}_{2 (g)},$ $K_{p} = 0.492$ atm at 300 K. $K_{c}$ for the reaction at same temperature is _______ $\times 10^{- 2}$ . (Given: R = 0.082 L atm ${mol}^{- 1}$ $K^{- 1}$ ) [2024]

(2)

$N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$

$Δ n_{g} =$ gaseous moles of products - gaseous moles of reactants

$= 2 - 1 = 1$

$K_{p} = K_{c} {(R T)}^{Δ n_{g}}$

$0.492 atm = K_{c} \times {(0.082 atm L {mol}^{- 1} K^{- 1} \times 300 K)}^{1}$

$K_{c} = \frac{0.492}{0.082 \times 300} mol L^{- 1}$

$= 0.02 mol L^{- 1} = 2 \times 10^{- 2} mol L^{- 1}$

Q 7 :

The following concentrations were observed at 500 K for the formation of ${NH}_{3}$ from $N_{2}$ and $H_{2}$ . At equilibrium; $[N_{2}] = 2 \times 10^{- 2} M, [H_{2}] = 3 \times 10^{- 2} M and [{NH}_{3}] = 1.5 \times 10^{- 2} M .$ Equilibrium constant for the reaction is _____________ . [2024]

(417)

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$

$K_{c} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} = \frac{{(1.5 \times 10^{- 2})}^{2}}{[2 \times 10^{- 2}] {[3 \times 10^{- 2}]}^{3}} = 417$

Q 8 :

$A_{(g)} ⇌ B_{(g)} + \frac{C}{2} (g)$ The correct relationship between $K_{p},$ $α$ and equilibrium pressure P is [2024]

$K_{P} = \frac{α^{1 / 2} P^{1 / 2}}{{(2 + α)}^{3 / 2}}$
$K_{P} = \frac{α^{1 / 2} P^{3 / 2}}{{(2 + α)}^{3 / 2}}$
$K_{P} = \frac{α^{3 / 2} P^{1 / 2}}{{(2 + α)}^{1 / 2} (1 - α)}$
$K_{P} = \frac{α^{1 / 2} P^{1 / 2}}{{(2 + α)}^{1 / 2}}$

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(3)

$K_{p} = \frac{P_{C}^{1 / 2} \times P_{B}}{P_{A}}$

$= \frac{{(\frac{α P}{2 + α})}^{1 / 2} \times (\frac{2 α}{2 + α}) P}{(\frac{2 - 2 α}{2 + α}) P} = \frac{α^{3 / 2} P^{1 / 2}}{{(2 + α)}^{1 / 2} (1 - α)}$

Q 9 :

A vessel at 1000 K contains ${CO}_{2}$ with a pressure of 0.5 atm. Some of ${CO}_{2}$ is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then $K_{p}$ is: [2025]

3 atm
1.8 atm
0.18 atm
0.3 atm

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(2)

$\begin{matrix} {CO}_{2} (g) & + & C (s) & ⟶ & 2 CO (g) \\ t = 0 & 0.5 atm & 0 & 0 \\ t = \infty & (0.5 - p) atm & 0 & 2 p atm \end{matrix}$

$Total pressure at equilibrium = P_{{CO}_{2}} + P_{CO} = 0.5 - p + 2 p = 0.5 + p = 0.8,$ $p = 0.3 atm$

Hence, $P_{{CO}_{2}} = 0.5 - p = (0.5 - 0.3) = 0.2 atm$

$P_{CO} = 2 p = 2 \times 0.3 = 0.6 atm$

$K_{p} = \frac{P_{CO}^{2}}{P_{{CO}_{2}}} = \frac{{(0.6)}^{2}}{0.2} = 1.8$

Q 10 :

For a reaction $N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$ in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]

7/4 times of initial pressure
5/2 times of initial pressure
5 times of initial pressure
7/2 times of initial pressure

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(1)

$N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$

$\begin{matrix} t = 0 & p & 0 & 0 \\ t = \infty & p - 0.5 p & 2 \times 0.5 p & \frac{1}{2} \times 0.5 p \\ = 0.5 p & = p & = 0.25 p \end{matrix} \begin{matrix} Total = 1.75 p \\ = \frac{7}{4} p \end{matrix}$

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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(3)

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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(1)

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

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

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

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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(1)

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

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

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

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

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

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]

(74)

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

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]

(962)

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]

(5)

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

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

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

$α^{3} = 128 \times 10^{- 6}$

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

Q 18 :

For a concentrated solution of a weak electrolyte ( $K_{eq} =$ equilibrium constant) $A_{2} B_{3}$ , of concentration $' c'$ , the degree of dissociation $' α'$ is [2023]

${(\frac{K_{eq}}{6 c^{5}})}^{1 / 5}$
${(\frac{K_{eq}}{25 c^{2}})}^{1 / 5}$
${(\frac{K_{eq}}{5 c^{4}})}^{1 / 5}$
${(\frac{K_{eq}}{108 c^{4}})}^{1 / 5}$

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$\begin{matrix} A_{2} B_{3} & ⇌ & 2 A & + & 3 B \\ Initial conc. & C & 0 & 0 \\ C - C α & 2 C α & 3 C α \end{matrix}$

$K = \frac{{[A]}^{2} {[B]}^{3}}{[A_{2} B_{3}]}$

$K = \frac{108 C^{5} α^{5}}{C} = 108 C^{4} α^{5}$

$α = {[\frac{K}{108 C^{4}}]}^{1 / 5}$

Q 19 :

$(i) X (g) ⇌ Y (g) + Z (g), K p_{1} = 3$

$(ii) A (g) ⇌ 2 B (g), K p_{2} = 1$

If the degree of dissociation and initial concentration of both the reactants X(g) and A(g) are equal, then the ratio of the total pressure at equilibrium $(\frac{P_{1}}{P_{2}})$ is equal to $x$ : 1. The value of $x$ is ________ (Nearest integer). [2023]

(12)

$X (g) ⇌ Y (g) + Z (g)$ $K_{P_{1}} = \frac{α^{2}}{1 - α^{2}} \times P_{1} = 3$

$A (g) ⇌ 2 B (g)$ $K_{P_{2}} = \frac{4 α^{2}}{1 - α^{2}} \times P_{2} = 1$

$\frac{K_{P_{1}}}{K_{P_{2}}} = \frac{P_{1}}{4 P_{2}} = \frac{3}{1}$

So, $(\frac{P_{1}}{P_{2}}) = 12$

Q 20 :

The equilibrium composition for the reaction

${PCl}_{3} + {Cl}_{2} ⇌ {PCl}_{5} at 298 K is given below:$

${[{PCl}_{3}]}_{eq} = {0.2 mol L}^{- 1}, {[{Cl}_{2}]}_{eq} = {0.1 mol L}^{- 1}, {[{PCl}_{5}]}_{eq} = {0.40 mol L}^{- 1}$

If 0.2 mol of ${Cl}_{2}$ is added at the same temperature, the equilibrium concentration of ${PCl}_{5}$ is $_____\times 10^{- 2} {mol L}^{- 1}$

Given: $K_{c}$ for the reaction at 298 K is 20. [2023]

(49)

$\begin{matrix} {PCl}_{3} (g) & + & {Cl}_{2} (g) & ⇌ & {PCl}_{5} (g) \\ At equilibrium & 0.2 \frac{mol}{Lit} & 0.1 \frac{mol}{Lit} & 0.4 \frac{mol}{Lit} \\ New equilibrium & (0.1 - x) & (0.3 - x) & (0.4 + x) \end{matrix}$

$K_{c} = (\frac{0.4 + x}{(0.2 - x) (0.3 - x)}) \Rightarrow 20 = \frac{0.4 + x}{(0.2 - x) (0.3 - x)}$

$(0.4 + x) = 20 (0.3 - x) (0.2 - x)$

$∴$ $0.4 + x = 20 (0.06 + x^{2} - 0.5 x)$

$20 x^{2} - 11 x + 0.8 = 0$

$x = \frac{11 \pm \sqrt{121 - 64}}{40}$

$x \approx 0.08625$

$∴$ $({PCl}_{5}) = 0.48625 \approx 49 \times 10^{- 2}$

Q 21 :

The number of correct statement/s involving equilibria in physical processes from the following is ________ [2023]

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

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

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

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

(3)

Statements given in A, B and D are correct.

(C), when equilibrium is attained at a given temperature, the value of all its parameters becomes constant.

Q 22 :

$A (g) ⇌ 2 B (g) + C (g)$

For the given reaction, if the initial pressure is 450 mm Hg and the pressure at time t is 720 mm Hg at a constant temperature T and constant volume V. The fraction of A(g) decomposed under these conditions is $x \times 10^{- 1}$ . The value of $x$ is ______ (Nearest Integer) [2023]

(3)

$\begin{matrix} A(g) & ⇌ & 2 B(g) & + & C(g) \\ t = 0 & 450 & 0 & 0 \\ t = t & 450 - y & 2 y & y \end{matrix}$

$720 = 450 - y + 2 y + y$

$∴$ $y = 135$

$Now, fraction of$ $A_{(g)}$ $decomposed = \frac{135}{450} = 0.3 = 3 \times 10^{- 1} = x \times 10^{- 1}$

$∴$ $x = 3$

Q 23 :

A mixture of 1 mole of $H_{2} O$ and 1 mole of CO is taken in a 10 litre container and heated to 725 K. At equilibrium 40% of water by mass reacts with carbon monoxide according to the equation:

$CO (g) + H_{2} O (g) ⇌ {CO}_{2} (g) + H_{2} (g)$

The equilibrium constant $K_{c} \times 10^{2}$ for the reaction is ________. (Nearest integer) [2023]

(44)

$\begin{matrix} CO & + & H_{2} O & ⇌ & {CO}_{2} & + & H_{2} \\ t = 0 & 1 & 1 & 0 & 0 \\ eq. & 1 - 0.4 & 1 - 0.4 & 0.4 & 0.4 \end{matrix}$

$V = 10 L$ $K_{C} = \frac{0.4 \times 0.4}{0.6 \times 0.6} = \frac{4}{9} = 0.44$

$K_{C} \times 100 = 44$

Q 24 :

4.5 moles each of hydrogen and iodine is heated in a sealed ten litre vessel. At equilibrium, 3 moles of HI were found. The equilibrium constant for $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ is ________. [2023]

(1)

$\begin{matrix} H_{2} (g) & + & I_{2} (g) & ⇌ & 2 HI (g) \\ Initial moles & 4.5 & 4.5 & 0 \\ Moles at eq. & 4.5 - 1.5 = 3 & 4.5 - 1.5 = 3 & 3 \end{matrix}$

$K_{equilibrium} = \frac{{[HI]}^{2}}{{[H_{2}]}^{1} {[I_{2}]}^{1}} = \frac{{[3]}^{2}}{3 \times 3} = 1$

Q 25 :

Consider the general reaction given below at 400 K

$x A (g) ⇌ y B (g)$

The values of $K_{p}$ and $K_{c}$ are studied under the same condition of temperature but variation in $x$ and $y$ .

(i) $K_{p} = 85.87$ and $K_{c} = 2.586$ (appropriate units)
(ii) $K_{p} = 0.862$ and $K_{c} = 28.62$ (appropriate units)

The values of $x$ and $y$ in (i) and (ii) respectively are: [2026]

(i) 1, 2 (ii) 2, 1
(i) 4, 1 (ii) 4, 1
(i) 3, 1 (ii) 3, 1
(i) 1, 3 (ii) 2, 1

1

2

3

4

(1)

For reaction (i) : $K_{P} > K_{C}$

$Δ n_{g} > 0$

$y - x > 0$

$85.87 = 2.586 {(0.0821 \times 400)}^{y - x}$

Solving $y - x = 1$

For reaction (ii) : $K_{P} < K_{C}$

$y - x < 0$

Q 26 :

For the following gas phase equilibrium reaction at constant temperature,

${NH}_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$

if the total pressure is $\sqrt{3}$ atm and the pressure equilibrium constant $(K_{p})$ is 9 atm, then the degree of dissociation is given as ${(x \times 10^{- 2})}^{- 1 / 2} .$ The value of $x$ is ______.(nearest integer) [2026]

(125)

$\begin{matrix} {NH}_{3 (g)} & ⇌ & \frac{1}{2} N_{2 (g)} & + & \frac{3}{2} H_{2 (g)} \\ t = 0 & 1 mole & - & - \\ t = t_{e q} & 1 - α & α / 2 & 3 α / 2 \end{matrix}$

$K_{p} = \frac{{(\frac{α}{2})}^{1 / 2} {(\frac{3 α}{2})}^{3 / 2}}{(1 - α)} {[\frac{P_{T}}{1 + α}]}^{1}$ $[∵ P_{T} = \sqrt{3} atm]$

$9 = \frac{{(\frac{α}{2})}^{1 / 2} {(\frac{3 α}{2})}^{3 / 2}}{(1 - α)} \times \frac{{(3)}^{1 / 2}}{1 + α}$

$9 = \frac{9 {(\frac{α}{2})}^{2}}{1 - α^{2}}$

$1 - α^{2} = \frac{α^{2}}{4}$

$\frac{5 α^{2}}{4} = 1$

$α^{2} = 0.8$

$α = {(0.8)}^{1 / 2}$

$α = {[\frac{1}{0.8}]}^{- 1 / 2}$

$α = {[125 \times 10^{- 2}]}^{- 1 / 2}$

$x = 125$

Q 27 :

Consider the following gaseous equilibrium in a closed container of volume 'V' at T(K).

$P_{2} (g) + Q_{2} (g) ⇌ 2 PQ (g)$

2 moles each of $P_{2} (g)$ , $Q_{2} (g)$ and $PQ (g)$ are present at equilibrium. Now one mole each of $' P_{2}'$ and $' Q_{2}'$ are added to the equilibrium keeping the temperature at T(K). The number of moles of $P_{2}$ , $Q_{2}$ and $PQ$ at the new equilibrium, respectively, are [2026]

2.67, 2.67, 2.67
1.21, 2.24, 1.56
1.66, 1.66, 1.66
2.56, 1.62, 2.24

1

2

3

4

(1)

$\begin{matrix} P_{2} (g) & + & Q_{2} (g) & ⇌ & 2 PQ (g) \\ t = t_{e q} & 2 mole & 2 mole & 2 mole \end{matrix}$

$K_{e q} = \frac{2^{2}}{2.2} = 1$

$Now 1 mole of each P_{2} and Q_{2} is added$

$So reaction will move in forward direction$

$\begin{matrix} P_{2} (g) & + & Q_{2} (g) & ⇌ & 2 PQ (g) \\ t = t'_{e q} & 3 - x & 3 - x & 2 + 2 x \end{matrix}$

$K_{c} = 1 = \frac{{(2 + 2 x)}^{2}}{(3 - x) (3 - x)}$

$\frac{2 + 2 x}{3 - x} = 1$

$2 + 2 x = 3 - x$

$x = \frac{1}{3}$

$At new equilibrium:$

$Moles of P_{2} = \frac{8}{3} = 2.67$

$Moles of Q_{2} = \frac{8}{3} = 2.67$

$Moles of P Q = \frac{8}{3} = 2.67$

Topic Question Set

Q 1 :

For the given reaction, choose the correct expression of $K_{C}$ from the following:

${Fe}_{(a q)}^{3 +} + {SCN}_{(a q)}^{-} ⇌ {(FeSCN)}_{(a q)}^{2 +}$ [2024]

Q 2 :

The equilibrium constant for the reaction ${SO}_{3} (g) ⇌ {SO}_{2} (g) + \frac{1}{2} O_{2} (g)$ is $K_{C} = 4.9 \times 10^{- 2}$ . The value of $K_{C}$ for the reaction given below is

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ is : [2024]

Q 3 :

At –20°C and 1 atm pressure, a cylinder is filled with equal number of $H_{2}$ , $I_{2}$ and HI molecules for the reaction $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ , the $K_{p}$ for the process is $x \times 10^{- 1}$ , $x =$ _______ . [Given : R = 0.082 L atm $K^{- 1}$ ${mol}^{- 1}$ ] [2024]

Q 4 :

The ratio $\frac{K_{P}}{K_{C}}$ for the reaction:

${CO}_{(g)} + \frac{1}{2} O_{2 (g)} ⇌ {CO}_{2 (g)}$ is : [2024]

Q 5 :

For the given hypothetical reactions, the equilibrium constants are as follows:

$X ⇌ Y; K_{1} = 1.0$

$Y ⇌ Z; K_{2} = 2.0$

$Z ⇌ W; K_{3} = 4.0$

The equilibrium constant for the reaction $X ⇌ W$ is [2024]

Q 6 :

For the reaction $N_{2} O_{4 (g)} ⇄ 2 {NO}_{2 (g)},$ $K_{p} = 0.492$ atm at 300 K. $K_{c}$ for the reaction at same temperature is _______ $\times 10^{- 2}$ . (Given: R = 0.082 L atm ${mol}^{- 1}$ $K^{- 1}$ ) [2024]

Q 8 :

$A_{(g)} ⇌ B_{(g)} + \frac{C}{2} (g)$ The correct relationship between $K_{p},$ $α$ and equilibrium pressure P is [2024]

Q 9 :

A vessel at 1000 K contains ${CO}_{2}$ with a pressure of 0.5 atm. Some of ${CO}_{2}$ is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then $K_{p}$ is: [2025]

Q 10 :

For a reaction $N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$ in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]

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]

Q 18 :

For a concentrated solution of a weak electrolyte ( $K_{eq} =$ equilibrium constant) $A_{2} B_{3}$ , of concentration $' c'$ , the degree of dissociation $' α'$ is [2023]

Q 24 :

4.5 moles each of hydrogen and iodine is heated in a sealed ten litre vessel. At equilibrium, 3 moles of HI were found. The equilibrium constant for $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ is ________. [2023]

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Topic Question Set

Q 1 : For the given reaction, choose the correct expression of KC from the following: Fe(aq)3++SCN(aq)-⇌(FeSCN)(aq)2+ [2024]

Q 2 : The equilibrium constant for the reaction SO3(g)⇌SO2(g)+12O2(g) is KC=4.9×10-2. The value of KC for the reaction given below is 2SO2(g)+O2(g)⇌2SO3(g) is : [2024]

Q 3 : At –20°C and 1 atm pressure, a cylinder is filled with equal number of H2, I2 and HI molecules for the reaction H2(g)+I2(g)⇌2HI(g), the Kp for the process is x×10-1, x= _______ . [Given : R = 0.082 L atm K-1 mol-1] [2024]

Q 4 : The ratio KPKC for the reaction: CO(g)+12O2(g)⇌CO2(g) is : [2024]

Q 5 : For the given hypothetical reactions, the equilibrium constants are as follows: X⇌Y; K1=1.0 Y⇌Z; K2=2.0 Z⇌W; K3=4.0 The equilibrium constant for the reaction X⇌W is [2024]

Q 6 : For the reaction N2O4(g)⇄2NO2(g), Kp=0.492 atm at 300 K. Kc for the reaction at same temperature is _______ ×10-2. (Given: R = 0.082 L atm mol-1 K-1) [2024]

Q 7 : The following concentrations were observed at 500 K for the formation of NH3 from N2 and H2. At equilibrium; [N2]=2×10-2 M, [H2]=3×10-2 M and [NH3]=1.5×10-2 M. Equilibrium constant for the reaction is _____________ . [2024]

Q 8 : A(g)⇌B(g)+C2(g) The correct relationship between Kp, α and equilibrium pressure P is [2024]

Q 9 : A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of CO2 is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then Kp is: [2025]

Q 10 : For a reaction N2O5(g) ⟶ 2NO2(g)+12O2(g) in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]

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]

Q 18 : For a concentrated solution of a weak electrolyte (Keq= equilibrium constant) A2B3, of concentration 'c', the degree of dissociation 'α' is [2023]

Q 19 : (i) X(g)⇌Y(g)+Z(g), Kp1=3 (ii) A(g)⇌2B(g), Kp2=1 If the degree of dissociation and initial concentration of both the reactants X(g) and A(g) are equal, then the ratio of the total pressure at equilibrium (P1P2) is equal to x : 1. The value of x is ________ (Nearest integer). [2023]

Q 22 : A(g)⇌2B(g)+C(g) For the given reaction, if the initial pressure is 450 mm Hg and the pressure at time t is 720 mm Hg at a constant temperature T and constant volume V. The fraction of A(g) decomposed under these conditions is x×10-1. The value of x is ______ (Nearest Integer) [2023]

Q 24 : 4.5 moles each of hydrogen and iodine is heated in a sealed ten litre vessel. At equilibrium, 3 moles of HI were found. The equilibrium constant for H2(g)+I2(g)⇌2HI(g) is ________. [2023]

Q 26 : For the following gas phase equilibrium reaction at constant temperature, NH3(g)⇌12N2(g)+32H2(g) if the total pressure is 3 atm and the pressure equilibrium constant (Kp) is 9 atm, then the degree of dissociation is given as (x×10-2)-1/2. The value of x is ______.(nearest integer) [2026]

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Q 1 :

For the given reaction, choose the correct expression of $K_{C}$ from the following:

${Fe}_{(a q)}^{3 +} + {SCN}_{(a q)}^{-} ⇌ {(FeSCN)}_{(a q)}^{2 +}$ [2024]

Q 2 :

The equilibrium constant for the reaction ${SO}_{3} (g) ⇌ {SO}_{2} (g) + \frac{1}{2} O_{2} (g)$ is $K_{C} = 4.9 \times 10^{- 2}$ . The value of $K_{C}$ for the reaction given below is

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ is : [2024]

Q 3 :

At –20°C and 1 atm pressure, a cylinder is filled with equal number of $H_{2}$ , $I_{2}$ and HI molecules for the reaction $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ , the $K_{p}$ for the process is $x \times 10^{- 1}$ , $x =$ _______ . [Given : R = 0.082 L atm $K^{- 1}$ ${mol}^{- 1}$ ] [2024]

Q 4 :

The ratio $\frac{K_{P}}{K_{C}}$ for the reaction:

${CO}_{(g)} + \frac{1}{2} O_{2 (g)} ⇌ {CO}_{2 (g)}$ is : [2024]

Q 5 :

For the given hypothetical reactions, the equilibrium constants are as follows:

$X ⇌ Y; K_{1} = 1.0$

$Y ⇌ Z; K_{2} = 2.0$

$Z ⇌ W; K_{3} = 4.0$

The equilibrium constant for the reaction $X ⇌ W$ is [2024]

Q 6 :

For the reaction $N_{2} O_{4 (g)} ⇄ 2 {NO}_{2 (g)},$ $K_{p} = 0.492$ atm at 300 K. $K_{c}$ for the reaction at same temperature is _______ $\times 10^{- 2}$ . (Given: R = 0.082 L atm ${mol}^{- 1}$ $K^{- 1}$ ) [2024]

Q 8 :

$A_{(g)} ⇌ B_{(g)} + \frac{C}{2} (g)$ The correct relationship between $K_{p},$ $α$ and equilibrium pressure P is [2024]

Q 9 :

A vessel at 1000 K contains ${CO}_{2}$ with a pressure of 0.5 atm. Some of ${CO}_{2}$ is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then $K_{p}$ is: [2025]

Q 10 :

For a reaction $N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$ in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]

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]

Q 18 :

For a concentrated solution of a weak electrolyte ( $K_{eq} =$ equilibrium constant) $A_{2} B_{3}$ , of concentration $' c'$ , the degree of dissociation $' α'$ is [2023]

Q 24 :

4.5 moles each of hydrogen and iodine is heated in a sealed ten litre vessel. At equilibrium, 3 moles of HI were found. The equilibrium constant for $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ is ________. [2023]