Chemical Equilibrium – JEE Main ChemistryPractice Questions & PYQs

Physical equilibrium requires a closed system at a constant temperature to ensure that opposing processes proceed at equal rates, specifically $r_{f} = r_{b}$. While the macroscopic parameters of the system become constant at equilibrium, they do not necessarily equal each other, meaning the claim of parameter equality is incorrect. For the dissolution of solids in liquids, the solubility remains constant for a specific solute and solvent at a fixed temperature. Since the statements regarding the closed system, the equality of rates, and the constant nature of solubility are all factually correct, there are 3 correct statements in the provided list.

The dissociation of the weak electrolyte $A_2 B_3$ follows the chemical equation $A_2 B_3 \rightleftharpoons 2 A^{3+} + 3 B^{2-}$. For an initial concentration $c$ and a degree of dissociation $\alpha$, the equilibrium concentrations are $[A_2 B_3] = c(1-\alpha)$, $[A^{3+}] = 2c\alpha$, and $[B^{2-}] = 3c\alpha$. The equilibrium constant expression is K_{\text{eq}} = \frac{[A^{3+}]^2 \[B^{2-}]^3}{[A_2 B_3]}$, which becomes$K_{\text{eq}} = \frac{(2c\alpha)^2 (3c\alpha)^3}{c(1-\alpha)}$. Expanding the powers gives$K_{\text{eq}} = \frac{4c^2\alpha^2 × 27c^3\alpha^3}{c(1-\alpha)}$, simplifying to$\frac{108 c^5 \alpha^5}{c(1-\alpha)}$. Because the electrolyte is weak,$\alpha$is significantly smaller than$1$, permitting the approximation$1-\alpha \approx 1$. This simplifies the equation to$K_{\text{eq}} = 108 c^4 \alpha^5$, which, when solved for$\alpha$, yields the expression$\alpha = \left( \frac{K_{\text{eq}}}{108 c^4} \right)^{\frac{1}{5}}$.

Adding $\mathrm{Cl}_2$ to the mixture disturbs the existing equilibrium, causing the reaction to proceed in the forward direction to consume the excess reactant as per Le Chatelier's principle. Initially, the concentration of $\mathrm{Cl}_2$ increases to 0.3 mol L⁻¹ while the concentrations of $\mathrm{PCl}_3$ and $\mathrm{PCl}_5$ remain at 0.2 mol L⁻¹ and 0.4 mol L⁻¹, respectively. When a new equilibrium is established, let $x$ be the change in concentration; the expressions for the new concentrations become $[\mathrm{PCl}_3] = 0.2 - x$, $[\mathrm{Cl}_2] = 0.3 - x$, and $[\mathrm{PCl}_5] = 0.4 + x$. Setting the equilibrium constant expression equal to 20, we have the following equation:

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

Solving for $x$ yields $x \simeq 0.086$, which means the final concentration of $\mathrm{PCl}_5$ is $0.4 + 0.086 = 0.486$ mol L⁻¹, which is equivalent to $48.6 \times 10^{-2}$ mol L⁻¹.

The extent to which water decomposes at equilibrium is determined by the equilibrium constant and the total pressure of the system. For the reaction $\mathrm{H}_2\mathrm{O}(g) \rightleftharpoons \mathrm{H}_2(g) + \frac{1}{2} \mathrm{O}_2(g)$, we define $\alpha$ as the degree of dissociation. Given that the total pressure is 1 bar and the equilibrium constant $K_p$ is $2 \times 10^{-3}$, the partial pressures of the products at equilibrium are proportional to $\alpha$ and $\alpha/2$, while the reactant partial pressure remains approximately 1 bar due to the very small magnitude of the equilibrium constant.

This simplification allows us to express the equilibrium state as $K_p = \frac{\alpha \cdot (\alpha/2)^{1/2}}{1}$. By substituting the given value for $K_p$ into this expression, we arrive at $2 \times 10^{-3} = \frac{\alpha^{3/2}}{\sqrt{2}}$. Rearranging this equation gives $\alpha^{3/2} = 2^{3/2} \times 10^{-3}$, which simplifies to $\alpha = 0.02$. Converting this decimal value into a percentage indicates that the amount of water decomposing is 2 percent.

Chemical equilibrium is governed by the law of mass action, which establishes a constant ratio between product and reactant concentrations at a fixed temperature. Since the balanced chemical equation $\mathrm{CO}(g)+\mathrm{H}_2\mathrm{O}(g)\rightleftharpoons\mathrm{CO}_2(g)+\mathrm{H}_2(g)$ involves an equal number of moles of gas on both sides, the container volume does not affect the calculation and cancels out, allowing us to determine the equilibrium constant $K_e$ using mole amounts directly. Starting with 1 mole each of carbon monoxide and water, the consumption of 40% of the water reacts away 0.4 moles of both reactants, leaving 0.6 moles of each behind while simultaneously producing 0.4 moles of each product at equilibrium. The equilibrium constant expression is given by $K_e = \frac{[\mathrm{CO}_2][\mathrm{H}_2]}{[\mathrm{CO}][\mathrm{H}_2\mathrm{O}]}$, and substituting the equilibrium mole values yields $\frac{0.4 \times 0.4}{0.6 \times 0.6}$. This simplifies to the fraction $\frac{4}{9}$, or approximately 0.4444. Multiplying this result by 100 gives 44.44, which rounds to the nearest integer of 44.

Manipulating chemical equations and their equilibrium constants follows specific algebraic rules: reversing a reaction requires taking the reciprocal of its equilibrium constant, while multiplying a reaction by a stoichiometric coefficient requires raising the equilibrium constant to that same power. When multiple elementary reactions are combined to arrive at a target equation, the equilibrium constant for that final reaction is the product of the constants for the individual, manipulated reactions.

Manipulating the given equilibrium expressions reveals the path to the target equation $2\,\mathrm{NH}_3(\mathrm{g}) + \frac{5}{2} \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\,\mathrm{NO}(\mathrm{g}) + 3\,\mathrm{H}_2\mathrm{O}(\mathrm{g})$. Reversing the first equilibrium provides $2\,\mathrm{NH}_3(\mathrm{g})$ on the reactant side, necessitating an inversion of its equilibrium constant to $1/K_1$. Retaining the second equilibrium ensures $2\,\mathrm{NO}(\mathrm{g})$ appears correctly as a product with constant $K_2$. Multiplying the third equilibrium by a factor of three generates the required $3\,\mathrm{H}_2\mathrm{O}(\mathrm{g})$ and, crucially, provides sufficient $\mathrm{H}_2(\mathrm{g})$ to cancel the $\mathrm{H}_2(\mathrm{g})$ generated by the reversed first reaction, resulting in a constant of $K_3^3$.

The equilibrium constant for the overall process is calculated as $K_{\text{eq}} = \frac{K_2 \times K_3^3}{K_1}$. Substituting the numerical values yields $\frac{1.6 \times 10^{12} \times (1.0 \times 10^{-13})^3}{4 \times 10^5}$. Simplifying the exponents leads to $10^{12} \times 10^{-39} \times 10^{-5} = 10^{-32}$, and dividing the coefficients gives $\frac{1.6}{4} = 0.4$. Combining these, the final result is $0.4 \times 10^{-32}$, which is equivalent to $4 \times 10^{-33}$.

The equilibrium constant $K_c$ is defined by the ratio of product concentrations to reactant concentrations at equilibrium, where each species is raised to the power of its stoichiometric coefficient from the balanced equation $\mathrm{H}_2(\mathrm{g}) + \mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{HI}(\mathrm{g})$. Since the number of moles of product equals the total number of moles of reactants in this reaction, the volume of the vessel cancels out, allowing for the direct use of mole quantities in the equilibrium expression $\mathrm{K}_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}$.

The stoichiometry reveals that 3 moles of $\mathrm{HI}$ are produced by consuming 1.5 moles each of $\mathrm{H}_2$ and $\mathrm{I}_2$, leaving 3 moles of each reactant at equilibrium starting from an initial 4.5 moles. Substituting these equilibrium quantities into the expression $\frac{3^2}{3 \times 3}$ results in a final value of 1.

The conversion between the equilibrium constants $K_P$ and $K_c$ relies on the relationship $K_P = K_c(RT)^{\Delta n_g}$ where $\Delta n_g$ represents the change \in the number of gaseous moles. For the reaction $\mathrm{SO}_2(\mathrm{g})+\frac{1}{2}\mathrm{O}_2(\mathrm{g})\rightleftharpoons\mathrm{SO}_3(\mathrm{g})$, the value is $\Delta n_g = 1 - (1 + 0.5) = -0.5$. With the reaction at 300 K and a gas constant of 0.082, the product $RT$ equals 24.6.

Applying these values leads to the equation $2 \times 10^{12} = K_c(24.6)^{-0.5}$ Isolating $K_c$ results \in the calculation $K_c = 2 \times 10^{12} \times \sqrt{24.6}$ Since the square root of 24.6 is approximately 4.96, the result is $K_c = 9.92 × 10^{12}$, which is equivalent to $0.992 × 10^{13}$. When rounded to the nearest integer, the coefficient is 1.

The dissociation of $\mathrm{N}_2\mathrm{O}_4$ follows the equilibrium $\mathrm{N}_2\mathrm{O}_4 \rightleftharpoons 2\mathrm{NO}_2$. Starting with 1 mole of reactant, 50% dissociation yields 0.5 moles of $\mathrm{N}_2\mathrm{O}_4$ remaining and 1 mole of $\mathrm{NO}_2$ produced, resulting in a total of 1.5 moles at equilibrium. Under a total pressure of 1 atm, the partial pressures are calculated as $P_{\mathrm{N}_2\mathrm{O}_4} = \frac{0.5}{1.5}$ and $P_{\mathrm{NO}_2} = \frac{1}{1.5}$.

The equilibrium constant is defined by the equation $K_P = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2\mathrm{O}_4}}$, which simplifies to $\frac{(1/1.5)^2}{0.5/1.5} \approx 1.33$. Using the expression $\Delta G^{\circ} = -RT \ln K_P$ with temperature 300 K, gas constant $R = 8.31$, and the relationship $\ln K_P = \log K_P \times \ln 10$, the calculation is:
$\Delta G^{\circ} = -8.31 \times 300 \times (0.1239 \times 2.3)$
This result is -710.45, so the nearest integer value for $x$ is 710.

The equilibrium state in a container of fixed volume is determined by the total number of moles of all gaseous components, where inert gases contribute to the total pressure according to the ideal gas law $PV = nRT$. Starting with 5 moles of $\mathrm{PCl}_5$ and 4 moles of inert argon, the reaction $\mathrm{PCl}_5(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})$ reaches equilibrium with $x$ moles of reactant decomposed, resulting in a total of $(5-x) + x + x + 4 = 9+x$ moles. Solving the ideal gas equation $6 \times 100 = (9+x) \times 0.082 \times 610$ reveals that $9+x = 12$, confirming that $x = 3$.

With the number of moles at equilibrium known, the mole fractions for $\mathrm{PCl}_5$, $\mathrm{PCl}_3$, and $\mathrm{Cl}_2$ are $\frac{2}{12}$, $\frac{3}{12}$, and $\frac{3}{12}$ respectively. Multiplying these fractions by the total pressure of $6\,\text{atm}$ provides the partial pressures of $1\,\text{atm}$ for the reactant and $1.5\,\text{atm}$ for each product. The equilibrium constant $K_p$ is calculated as the ratio of product partial pressures to the reactant partial pressure: $K_p = \frac{1.5 \times 1.5}{1} = 2.25$.

Reversing a chemical equation results in an equilibrium constant equal to the reciprocal of the original, while halving the stoichiometric coefficients corresponds to taking the square root of that value. Consequently, the new equilibrium constant is calculated as $\frac{1}{\sqrt{2 \times 10^{15}}}$, which simplifies to $\frac{1}{\sqrt{20 \times 10^{14}}}$ or $\frac{1}{\sqrt{20}} \times 10^{-7}$. Expressing this coefficient in the required $x \times 10^{-8}$ format involves multiplying by $\frac{10}{10}$ to obtain $\frac{10}{\sqrt{20}} \times 10^{-8}$, which simplifies further to $\sqrt{5} \times 10^{-8}$. Given that $\sqrt{5}$ is approximately $2.236$, the value of $x$ rounded to the nearest integer is $2$.

The standard free energy change for a reversible reaction is defined by the relationship $\Delta G^{\circ} = -RT \ln K_p$, where $K_p$ is the equilibrium constant at a given temperature. For the decomposition reaction $2\mathrm{O}_3(\mathrm{g}) \rightleftharpoons 3\mathrm{O}_2(\mathrm{g})$, assuming an initial amount of 2 moles of $\mathrm{O}_3$ with 50 percent dissociation, the equilibrium mixture contains 1 mole of $\mathrm{O}_3$ and 1.5 moles of $\mathrm{O}_2$, resulting in 2.5 total moles of gas. Under a total pressure of 1 atm, the partial pressures are $P_{\mathrm{O}_3} = \frac{1}{2.5} = 0.4\ \mathrm{atm}$ and $P_{\mathrm{O}_2} = \frac{1.5}{2.5} = 0.6\ \mathrm{atm}$.

The equilibrium constant is calculated as $K_p = \frac{(P_{\mathrm{O}_2})^3}{(P_{\mathrm{O}_3})^2} = \frac{(0.6)^3}{(0.4)^2} = 1.35$. Substituting the values $R = 8.3\ \mathrm{J\ K^{-1}\ mol^{-1}}$, $T = 300\ \mathrm{K}$, and $\ln(1.35) = 0.3$ into the free energy formula, the calculation becomes $\Delta G^{\circ} = -(8.3)(300)(0.3) = -747\ \mathrm{J\ mol^{-1}}$. The missing value in the blank is 747.

The equilibrium constant for the reaction $A(g) \rightleftharpoons B(g) + \frac{1}{2} C(g)$ is determined by the partial pressures of the species at equilibrium, which depend on the degree of dissociation $\alpha$. Starting with 1 mole of $A$, the equilibrium mixture comprises $1-\alpha$ moles of $A$, $\alpha$ moles of $B$, and $\alpha/2$ moles of $C$, resulting in a total of $(2+\alpha)/2$ moles.

By expressing the partial pressure of each gas as its mole fraction multiplied by the total pressure $P$, the equilibrium expression $K = \frac{P_B (P_C)^{1/2}}{P_A}$ becomes $\frac{(\frac{2\alpha}{2+\alpha} P) (\frac{\alpha}{2+\alpha} P)^{1/2}}{\frac{2(1-\alpha)}{2+\alpha} P}$. Simplifying this fraction by cancelling common terms yields the final expression $K = \frac{\alpha^{3/2} P^{1/2}}{(1-\alpha)(2+\alpha)^{1/2}}$.

The standard Gibbs free energy change is related to the equilibrium constant by the equation $\Delta G^\ominus = -RT \ln K$, which serves as a bridge between thermodynamics and chemical kinetics. For the decomposition of hydrogen iodide, the balanced chemical equation is $\mathrm{HI} \rightleftharpoons \frac{1}{2} \mathrm{H}_2 + \frac{1}{2} \mathrm{I}_2$. When the degree of dissociation is 0.4, the concentrations at equilibrium are 0.6 for HI, and 0.2 for both H2 and I2. The equilibrium constant is defined by the expression $K = \frac{[\mathrm{H}_2]^{1/2} [\mathrm{I}_2]^{1/2}}{[\mathrm{HI}]}$, which simplifies to the equation $K = 1/3$. To calculate the final result, we use the formula $\Delta G^\ominus = RT \ln 3$. Substituting the values $R = 8.31$, $T = 300$, and $\ln 3 = 2.3 \times 0.477 = 1.0971$, the calculation $8.31 \times 300 \times 1.0971$ yields 2735 J/mol.

The equilibrium constant $K_p$ for a gas-phase reaction relates the partial pressures of products and reactants at equilibrium, raised to the power of their stoichiometric coefficients. With an initial mixture of 2 moles of NO and 1 mole of O2, the consumption of oxygen to a final 0.6 moles indicates that 0.4 moles were reacted. By reaction stoichiometry, this decrease leads to 1.2 moles of NO and 0.8 moles of NO2 at equilibrium, creating a system with a total of 2.6 moles.

Partial pressures are determined by multiplying the mole fraction of each gas by the total pressure of 1 atm, giving 1.2/2.6 for NO, 0.6/2.6 for O2, and 0.8/2.6 for NO2. Substituting these values into the equilibrium expression $K_p = \frac{(P_{\mathrm{NO}_2})^2}{(P_{\mathrm{NO}})^2 (P_{\mathrm{O}_2})}$ yields a calculation of 1.925, which rounds to 2.

Chemical equilibrium problems involving gas mixtures rely on determining the total moles present at equilibrium to calculate the partial pressure of each component. Because the total pressure of a system depends on the sum of all gaseous species present-including inert gases like nitrogen that do not participate in the reaction-one must account for all moles in the ideal gas equation before calculating the equilibrium constant.

To find the extent of dissociation, we denote the moles of $\mathrm{PCl}_5$ that react as $x$. The initial composition consists of 5 moles of $\mathrm{PCl}_5$ and 2 moles of $\mathrm{N}_2$. At equilibrium, the mixture contains $5-x$ moles of $\mathrm{PCl}_5$, $x$ moles of $\mathrm{PCl}_3$, $x$ moles of $\mathrm{Cl}_2$, and the original 2 moles of $\mathrm{N}_2$. Summing these values gives a total of $7+x$ moles at equilibrium. By applying the ideal gas law $PV = nRT$ with the given values $P = 2.46\ \mathrm{atm}$, $V = 200\ \mathrm{L}$, and $T = 600\ \mathrm{K}$, we arrive at the equation $2.46 \times 200 = (7+x) \times 0.082 \times 600$. Simplifying the right side where $0.082 \times 600$ equals $49.2$, we find $492 = (7+x) \times 49.2$, which leads to $7+x = 10$, or $x = 3$.

With the equilibrium mole count established, we determine the partial pressures using the relationship $P_i = \chi_i P_{total}$, where $\chi_i$ is the mole fraction. For the products $\mathrm{PCl}_3$ and $\mathrm{Cl}_2$, the mole fraction is $\frac{3}{10}$, and for the reactant $\mathrm{PCl}_5$, the mole fraction is $\frac{5-3}{10} = \frac{2}{10}$. The equilibrium constant expression is $K_p = \frac{P_{\mathrm{PCl}_3} P_{\mathrm{Cl}_2}}{P_{\mathrm{PCl}_5}}$. Substituting the partial pressures, we get $K_p = \frac{(\frac{3}{10} \times 2.46) \times (\frac{3}{10} \times 2.46)}{\frac{2}{10} \times 2.46}$. Canceling the common terms simplifies the expression to $\frac{3}{10} \times \frac{3}{2} \times 2.46$, which is $0.45 \times 2.46$, resulting in a value of $1.107$. This corresponds to $1107 \times 10^{-3}$.

The equilibrium constant of a chemical reaction changes in a specific way when the stoichiometric coefficients of the equation are multiplied by a constant factor $n$; specifically, the new equilibrium constant becomes the original constant raised to the power of $n$. Comparing the first reaction $2\mathrm{A}(g) \rightleftharpoons 2\mathrm{B}(g) + \mathrm{C}(g)$ associated with $K_1$ to the second reaction $\frac{3}{2}\mathrm{A}(g) \rightleftharpoons \frac{3}{2}\mathrm{B}(g) + \frac{3}{4}\mathrm{C}(g)$ associated with $K_2$, it is evident that every reactant and product coefficient in the first equation has been multiplied by a factor of $\frac{3}{4}$. Since the stoichiometric coefficients of the second equation are exactly $\frac{3}{4}$ times those of the first, we raise the original equilibrium constant $K_1$ to the power of $\frac{3}{4}$ to determine $K_2$, which yields $K_2=K_1^{3/4}$.

The transition of the entire liquid sample into a gaseous state allows us to treat the water vapour as an ideal gas that exerts an equilibrium pressure within the box. Before applying the ideal gas equation, ensure all units are consistent with the ideal gas constant by converting the vapour pressure to $P = \frac{32}{760} \, \text{atm}$, the temperature to $T = 27 + 273 = 300 \, \text{K}$, and calculating the quantity of gas in moles as $n = \frac{0.90}{18} = 0.05 \, \text{mol}$.

Using the ideal gas relationship
$V = \frac{nRT}{P}$
and substituting the known values yields
$V = \frac{0.05 \times 0.082 \times 300}{32/760}$
This expression simplifies to
$V = \frac{1.23 \times 760}{32}$
resulting in a volume of approximately $29.21 \, \text{L}$. Rounding to the nearest integer, the volume required is $29 \, \text{L}$.

The equilibrium constant $K_C$ is defined by the ratio of the product of the equilibrium concentrations of the products to that of the reactants, where each concentration is raised to the power of its stoichiometric coefficient. For the reaction $2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})$, an initial concentration of 2.0 M for $\mathrm{NOCl}$ in a 1 L vessel means that after 0.4 M of $\mathrm{NO}$ forms at equilibrium, an equal amount of $\mathrm{NOCl}$ has been consumed and half that amount, 0.2 M, of $\mathrm{Cl}_2$ has been generated.

The equilibrium concentrations are therefore 1.6 M for $\mathrm{NOCl}$, 0.4 M for $\mathrm{NO}$, and 0.2 M for $\mathrm{Cl}_2$. Substituting these into the equilibrium expression $K_C = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_2]}{[\mathrm{NOCl}]^2}$ allows us to compute the constant as $K_C = \frac{(0.4)^2 (0.2)}{(1.6)^2} = \frac{0.032}{2.56}$. This calculation results \in 0.0125, which is expressed as $125 × 10^{-4}$.

The equilibrium constant for a chemical reaction relates to the standard Gibbs free energy change through the equation $\Delta G^{\circ} = -RT \ln K$. By converting the standard Gibbs free energy change from kilojoules to joules, we obtain -9478 joules per mole. Substituting this value along with the temperature of 495 K and the gas constant of 8.314 J per mole per Kelvin into the equation yields $\ln K = 2.303$. Since $\ln 10 = 2.303$, it follows that the equilibrium constant $K = 10$.

For the reaction where A converts to B, we can determine the equilibrium composition using the starting amount of 22 millimoles. If $x$ millimoles of B are formed at equilibrium, then $22 - x$ millimoles of A remain. Expressing the equilibrium constant as the ratio of the amounts of products to reactants gives the equation $\frac{x}{22 - x} = 10$. This simplifies to $x = 220 - 10x$, which results in $11x = 220$, confirming that 20 millimoles of B are present at equilibrium.