Chemical Equilibrium – JEE Main ChemistryPractice Questions & PYQs

The relationship between the standard Gibbs free energy change and the equilibrium constant is defined by the equation $\Delta G^\circ = -RT \ln K_p$, which allows us to determine the thermodynamic feasibility of a reaction at a specific temperature. To find this value, one must first determine the equilibrium constant $K_p$ by assessing the system composition based on the stoichiometry of the reaction.

Assuming a starting amount of 1 mole of $\mathrm{N}_2\mathrm{O}_5$, the 50 percent dissociation results in 0.5 moles of $\mathrm{N}_2\mathrm{O}_5$ remaining, alongside the production of 0.5 moles of $\mathrm{N}_2\mathrm{O}_4$ and 0.25 moles of $\mathrm{O}_2$. The total number of moles at equilibrium is 1.25. Dividing the moles of each gas by the total mole count gives the respective mole fractions: 0.4 for $\mathrm{N}_2\mathrm{O}_5$, 0.4 for $\mathrm{N}_2\mathrm{O}_4$, and 0.2 for $\mathrm{O}_2$.

Using Dalton's Law, the partial pressure of each gas is the product of its mole fraction and the total pressure of 2 atm. Substituting these values into the equilibrium expression $K_p = \frac{p(\mathrm{N}_2\mathrm{O}_4)^2 p(\mathrm{O}_2)}{p(\mathrm{N}_2\mathrm{O}_5)^2}$ leads to $K_p = \frac{(0.4 \times 2)^2 \times (0.2 \times 2)}{(0.4 \times 2)^2}$, which simplifies to 0.4. Given the temperature 323 K, the standard free energy change is calculated as $\Delta G^\circ = -8.314 \times 323 \times \ln(0.4)$. Utilizing the approximation $\ln(0.4) \approx 2.303 \times (\log 4 - \log 10) = 2.303 \times (0.6 - 1) = -0.9212$, the final expression becomes $8.314 \times 323 \times 0.9212$, which yields approximately 2473.81 J mol^-1, rounding to 2474 as the nearest integer.

The dependence of the pressure equilibrium constant on temperature is governed by the van 't Hoff equation:
$\log_{10} K_{p} = -\frac{\Delta H^{\circ}}{2.303 R} \left( \frac{1}{T} \right) + \text{constant}$
This indicates that $\log_{10} K_{p}$ relates linearly to $1/T$ with a slope equal to $-\frac{\Delta H^{\circ}}{2.303 R}$. We determine the magnitude of $\Delta H^{\circ}/R$ by calculating the slope using two data points: $1/T = 0.05$ with $\log_{10} K_{p} = 3.5$ and $1/T = 0.06$ with $\log_{10} K_{p} = 2.5$.

By applying the formula $\log_{10} K_{P_{2}} - \log_{10} K_{P_{1}} = -\frac{\Delta H^{\circ}}{2.303 R} \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)$, we get $2.5 - 3.5 = -\frac{\Delta H^{\circ}}{2.303 R} (0.06 - 0.05)$, which simplifies to $-1.0 = -\frac{\Delta H^{\circ}}{2.303 R} (0.01)$. Rearranging for $\frac{\Delta H^{\circ}}{R}$ leads to $2.303 \times (1.0 / 0.01) = 230.3$, and rounding to the nearest integer provides 230.

The equilibrium constant $K_c$ for the reaction $P_2(g) + Q_2(g) \rightleftharpoons 2PQ(g)$ is defined by the ratio of product concentration to reactant concentrations, represented by the expression $K_c = \frac{n_{PQ}^2}{n_{P_2}n_{Q_2}}$, where the container volume $V$ cancels out because the change in stoichiometric coefficients is zero. Utilizing the initial equilibrium amounts of 2 moles for each component, $K_c = \frac{2^2}{2 \times 2} = 1$.

Upon adding one mole each of $P_2$ and $Q_2$, the initial amounts become 3 moles of $P_2$, 3 moles of $Q_2$, and 2 moles of $PQ$. Let $x$ represent the number of moles of $P_2$ (and $Q_2$) consumed to reach the new equilibrium, resulting in $n_{P_2} = 3-x$, $n_{Q_2} = 3-x$, and $n_{PQ} = 2+2x$. Substituting these values into the equilibrium constant expression gives:

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

Taking the square root of both sides yields $\frac{2+2x}{3-x} = 1$, which simplifies to $2+2x = 3-x$. Solving for $x$ gives $3x = 1$, or $x = 0.333$. Calculating the final amounts, the new number of moles for $P_2$, $Q_2$, and $PQ$ are $3 - 0.333 = 2.67$, $3 - 0.333 = 2.67$, and $2 + 2(0.333) = 2.67$, respectively.

For the chemical equilibrium
$\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g}) + \mathrm{C}(\mathrm{g})$
starting with $a$ moles of reactant A leads to an equilibrium mixture containing $(a-x)$ moles of A along with $x$ moles each of products B and C. The total number of moles present at equilibrium is the sum $(a-x) + x + x$, which simplifies to $(a+x)$. Because the partial pressure of each gas component is its mole fraction multiplied by the total pressure $P$, the partial pressures are calculated as $P_A = \frac{a-x}{a+x} P$, $P_B = \frac{x}{a+x} P$, and $P_C = \frac{x}{a+x} P$. Substituting these values into the equilibrium constant expression
$K_{p} = \frac{P_B \times P_C}{P_A} = \frac{\left(\frac{x}{a+x} P\right) \left(\frac{x}{a+x} P\right)}{\left(\frac{a-x}{a+x} P\right)}$
results in the final simplified expression $\frac{x^{2}}{a^{2}-x^{2}} \times P$.

The relationship between the standard Gibbs free energy change and the equilibrium constant is given by the equation $\Delta_{r}G^{\ominus} = -RT \ln K_{p}$. To determine the equilibrium constant, one must first identify the mole balance of the system at equilibrium. Starting with one mole of $\mathrm{X}_{2}\mathrm{Y}_{4}$ and a dissociation degree $\alpha = 0.75$, the reaction results in $1 - 0.75 = 0.25$ moles of reactant and $2 \times 0.75 = 1.5$ moles of product, yielding a total of $1.75$ moles in the flask.

Applying Dalton's Law of partial pressures with a total pressure of 1 atm, the partial pressure of $\mathrm{X}_{2}\mathrm{Y}_{4}$ is $0.25 / 1.75 = 1/7$ atm, and the partial pressure of $\mathrm{XY}_{2}$ is $1.5 / 1.75 = 6/7$ atm. The equilibrium constant is calculated as:
$K_{p} = \frac{(p_{\mathrm{XY}_{2}})^2}{p_{\mathrm{X}_{2}\mathrm{Y}_{4}}} = \frac{(6/7)^2}{1/7} = \frac{36}{7}$
Substituting these values into the Gibbs energy equation, $\Delta_{r}G^{\ominus} = -8.3 \times 600 \times \ln(36/7)$, results in a value of approximately $-8152\,\mathrm{J}\,\mathrm{mol}^{-1}$. Converting this to $\mathrm{kJ}\,\mathrm{mol}^{-1}$ and taking the magnitude yields 8 when rounded to the nearest integer.

The standard Gibbs energy change for the reaction $A_2(g) \rightleftharpoons 2A(g)$ is calculated as $\Delta G^{\circ} = 2 \Delta G_f^{\circ}(A) - \Delta G_f^{\circ}(A_2) = 2(-50.832) - (-100.00) = -1.664 \text{ kJ mol}^{-1}$, which equals $-1664 \text{ J mol}^{-1}$. Using the thermodynamic relation $\Delta G^{\circ} = -RT \ln K_p$, we find $-1664 = -(8)(300) \ln K_p$, which yields $\ln K_p = 1664 / 2400 \approx 0.6933$. Since $\ln 2 \approx 0.693$, the equilibrium constant is determined to be $K_p \approx 2$.

For the dissociation of 1 mole of $A_2$, the equilibrium mixture contains $1-\alpha$ moles of $A_2$ and $2\alpha$ moles of $A$, resulting in a total of $1+\alpha$ moles. The equilibrium constant in terms of the degree of dissociation $\alpha$ and total pressure $P_{total}$ is $K_p = \frac{(2\alpha / (1+\alpha))^2 P_{total}}{(1-\alpha) / (1+\alpha) P_{total}} = \frac{4\alpha^2}{1-\alpha^2} P_{total}$. Substituting $K_p = 2$ and $P_{total} = 1$ bar, the expression simplifies to $2 = \frac{4\alpha^2}{1-\alpha^2}$, which leads to $2 - 2\alpha^2 = 4\alpha^2$ or $6\alpha^2 = 2$. Thus, $\alpha^2 = 1/3 \approx 0.3333$. Comparing this with the given form $\alpha = (x \times 10^{-2})^{1/2}$, we have $\alpha^2 = x \times 10^{-2} = 0.3333$, implying $x = 33.33$. The nearest integer value for $x$ is 33.

The relationship between the equilibrium constants $K_p$ and $K_c$ for a gaseous reaction is defined by the equation $K_p = K_c(RT)^{\Delta n_g}$, where $\Delta n_g$ represents the change in the number of gaseous moles calculated as $y - x$, $R$ is the universal gas constant approximately $0.0821\text{ L atm K}^{-1}\text{ mol}^{-1}$, and $T$ is the temperature $400\text{ K}$. Calculating the product $RT$ yields $0.0821 \times 400 = 32.84$.

For case (i), substituting the given values $K_p = 85.87$ and $K_c = 2.586$ into the equation results in $85.87 = 2.586 \times (32.84)^{\Delta n_g}$, which simplifies to $(32.84)^{\Delta n_g} \approx 33.2$. This indicates that $\Delta n_g = 1$, as $32.84^1 \approx 33.2$. Given the options provided, assigning $x=1$ and $y=2$ satisfies the condition $\Delta n_g = 2 - 1 = 1$.

For case (ii), substituting $K_p = 0.862$ and $K_c = 28.62$ into the equation gives $0.862 = 28.62 \times (32.84)^{\Delta n_g}$, leading to $(32.84)^{\Delta n_g} \approx 0.0301$. Because $0.0301$ is approximately the reciprocal of $33.2$, $\Delta n_g$ must be $-1$. Assigning $x=2$ and $y=1$ satisfies the condition $\Delta n_g = 1 - 2 = -1$. Consequently, the values for $x$ and $y$ are $1, 2$ for the first case and $2, 1$ for the second case.

The equilibrium constant of a chemical reaction is inherently linked to its stoichiometric coefficients. If all coefficients in a balanced equation are multiplied by a scaling factor $n$, the new equilibrium constant becomes the original equilibrium constant raised to the power of $n$. In this case, the coefficients are multiplied by $\frac{1}{3}$, which means the new equilibrium constant is obtained by raising the original value, $2.7 \times 10^{-5}$, to the power of $\frac{1}{3}$.

Calculating the cube root of $2.7 \times 10^{-5}$ shows that $(3 \times 10^{-2})^3 = 27 \times 10^{-6} = 2.7 \times 10^{-5}$. Consequently, the new equilibrium constant is $3 \times 10^{-2}$.

The equilibrium constant for a net chemical reaction is calculated by combining the constants of intermediate reactions through algebraic operations corresponding to the manipulation of those equations. To isolate the target reaction, we invert $2xy \rightleftharpoons x_2 + y_2$ and multiply by $\frac{1}{2}$ to obtain $\frac{1}{2} x_2 + \frac{1}{2} y_2 \rightleftharpoons xy$, which transforms its constant to $\sqrt{\frac{1}{2.5 \times 10^5}} = \frac{1}{500}$. Adding this resulting equation to the second given reaction, $xy + \frac{1}{2} z_2 \rightleftharpoons xyz$, cancels the intermediate $xy$ species, and the overall equilibrium constant is determined by multiplying their respective constants: $K_3 = \frac{1}{500} \times 5 \times 10^{-3} = 1.0 \times 10^{-5}$.

The equilibrium constant $K_c$ is determined by the ratio of the square of the product concentration to the product of the reactant concentrations: $K_c = \frac{[Z]^2}{[X_2][Y_2]}$ Substituting the given equilibrium amounts for a 1 L flask, $K_c = \frac{9^2}{3 × 3} = 9$. Adding 10 mol of $Z(g)$ increases the total amount of $Z(g)$ to 19 mol, which shifts the equilibrium to the left. Let $2\alpha$ be the number of moles of $Z(g)$ that react as the system reaches the new equilibrium, resulting \in new amounts of $[X_2] = 3+\alpha$, $[Y_2] = 3+\alpha$, and $[Z] = 19-2\alpha$. Applying the equilibrium condition leads to the equation: $9 = \frac{(19-2\alpha)^2}{(3+\alpha)^2}$ Taking the square root of both sides gives $3 = \frac{19-2\alpha}{3+\alpha}$, which simplifies to $9+3\alpha = 19-2\alpha$. Solving this equation, $5\alpha = 10$, so $\alpha = 2$. The final amount of $Z(g)$ is $19 - 2(2) = 15$ mol.

Multiple simultaneous equilibria share gaseous species, meaning the partial pressure of a shared gas must satisfy every equilibrium expression in the system concurrently. For the decomposition of calcium carbonate, $\mathrm{CaCO}_3 \rightleftharpoons \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})$, the solid components are excluded from the equilibrium constant, so $K_{P_1} = P_{\mathrm{CO}_2}$, which directly provides a value of $0.08 \text{ atm}$ for the partial pressure of carbon dioxide.

With the carbon dioxide pressure fixed at $0.08 \text{ atm}$, the second reaction $\mathrm{C}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})$ determines the equilibrium partial pressure of carbon monoxide. The equilibrium constant expression for this second process is $K_{P_2} = \frac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO}_2}}$, and substituting the given constant of $2$ results in the equation $2 = \frac{(P_{\mathrm{CO}})^2}{0.08}$. Multiplying these values yields $(P_{\mathrm{CO}})^2 = 0.16$, or $16 \times 10^{-2}$, and calculating the square root reveals the final partial pressure to be $4 \times 10^{-1} \text{ atm}$.

The pressure equilibrium constant $K_p$ for the dissociation reaction $\mathrm{NH}_3(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_2(g) + \frac{3}{2} \mathrm{H}_2(g)$ is defined in terms of the degree of dissociation $\alpha$ and total pressure $P$ as:
$K_p = \frac{3\sqrt{3} P \alpha^2}{4(1-\alpha^2)}$
Substituting the given values $K_p = 9 \text{ atm}$ and $P = \sqrt{3} \text{ atm}$ into the expression yields:
$9 = \frac{3\sqrt{3} (\sqrt{3}) \alpha^2}{4(1-\alpha^2)} = \frac{9 \alpha^2}{4(1-\alpha^2)}$
Simplifying this equation results in $1 = \frac{\alpha^2}{4(1-\alpha^2)}$, which leads to $4(1-\alpha^2) = \alpha^2$, or $5\alpha^2 = 4$, yielding $\alpha^2 = 0.8$. Given the expression $\alpha = (x \times 10^{-2})^{-\frac{1}{2}}$, squaring both sides gives $\alpha^2 = (x \times 10^{-2})^{-1} = \frac{100}{x}$. Equating $\frac{100}{x} = 0.8$ results in $x = 125$.

The equilibrium constant $K_c$ for the reaction $A(g) \rightleftharpoons B(g)$ is determined by the expression $K_c = [B]/[A]$. Substituting the initial values gives $0.375 / 0.5 = 0.75$. Upon adding 0.1 moles of $A$ to the 1 L flask, the concentration of $A$ increases to 0.6 M, while the concentration of $B$ remains 0.375 M. Letting $x$ represent the change in concentration required to re-establish equilibrium, the new concentrations become $[A] = 0.6 - x$ and $[B] = 0.375 + x$.

These values must satisfy the equilibrium expression, leading to the following equation:
$\frac{0.375 + x}{0.6 - x} = 0.75$
Solving for $x$ yields $0.375 + x = 0.45 - 0.75x$, which simplifies to $1.75x = 0.075$, resulting in $x = 0.04286$ M. The new equilibrium concentrations are $[A] = 0.6 - 0.04286 = 0.557$ M and $[B] = 0.375 + 0.04286 = 0.418$ M.

The ideal gas law $PV = nRT$ allows for the direct calculation of partial pressures for non-reacting gases. Because Helium is inert, its partial pressure in the 10 L flask at 400 K is determined using its initial 1 mole and the given constant 0.082 L atm K^-1 mol^-1. Calculating this result by evaluating $P = \frac{nRT}{V}$ leads to a pressure of 3.28 atm.

For the reversible reaction between gases A and B, the equilibrium constant $K_c = \frac{[B]}{[A]}$ relates the molar concentrations at equilibrium. Given that the volume cancels out in the expression, it simplifies to $4 = \frac{x}{1-x}$, where $x$ represents the moles of B produced from the initial 1 mole of A. Solving this equation yields $x = 0.8$ moles of B. Using the ideal gas law once more to determine the pressure of B at these conditions, we find the partial pressure is 2.624 atm.

According to Le Chatelier's principle, when the total pressure on a gaseous equilibrium system is increased, the system shifts toward the side with fewer moles of gas to minimize the stress. In the reaction $A(\mathrm{g}) \rightleftharpoons B(\mathrm{g}) + C(\mathrm{g})$, the forward direction results in an increase in the number of gaseous moles, meaning that increasing the total pressure $p$ drives the equilibrium in the backward direction. Consequently, this shift toward the reactant side reduces the extent of the reaction, causing the degree of dissociation $\alpha$ to decrease.

The equilibrium constant $K_p^\theta$ is determined by the partial pressures of all gaseous species at equilibrium, which are derived using the ideal gas law $PV = nRT$. By calculating the total moles of gas at equilibrium using $n_{\text{total}} = \frac{PV}{RT}$, we find 0.25 moles, and given 0.04 moles of $\mathrm{CH}_3\mathrm{OH}$ are produced from an initial 0.1 moles of $\mathrm{CO}$, it follows that 0.06 moles of $\mathrm{CO}$ and 0.15 moles of $\mathrm{H}_2$ remain. Applying the partial pressure formula $P_i = \frac{n_i RT}{V}$ with these molar amounts results in equilibrium pressures of 0.8 bar for $\mathrm{CH}_3\mathrm{OH}$, 1.2 bar for $\mathrm{CO}$, and 3 bar for $\mathrm{H}_2$. Substituting these into the equilibrium expression $K_p = \frac{P_{\mathrm{CH}_3\mathrm{OH}}}{P_{\mathrm{CO}} (P_{\mathrm{H}_2})^2}$ yields a final value of 0.07407, which corresponds to $74 \times 10^{-3}$ as the nearest integer.

As the reaction $\mathrm{H}_2(g) + \mathrm{I}_2(g) \rightleftharpoons 2\,\mathrm{HI}(g)$ proceeds, the molar concentrations of the reactants $\mathrm{H}_2$ and $\mathrm{I}_2$ decrease over time, while the concentration of the product $\mathrm{HI}$ simultaneously increases. Eventually, the system achieves a state of equilibrium where the concentrations of all species involved stop changing and remain constant. Therefore, a correct plot must show the reactant concentrations descending and the product concentration ascending until they all reach a flat, unchanging level.

The equilibrium constant $K_c$ is a thermodynamic property determined solely by the temperature of the system, meaning a catalyst cannot modify its value regardless of the reaction pathway provided. Because a catalyst increases the rate of both forward and backward reactions by an identical factor, it allows the system to reach equilibrium more efficiently without shifting the equilibrium position or changing the final composition of the mixture. Consequently, since a catalyst has no impact on either the equilibrium position or the constant, the first statement is true and the second statement is false.

When an inert gas like xenon is added to a gaseous equilibrium system at constant temperature and pressure, the container volume must expand to maintain the total pressure, which reduces the partial pressure of each gaseous species. According to Le Chatelier's principle, the system will shift in the direction that produces more moles of gas to compensate for this overall decrease in partial pressure. In the reaction $\mathrm{PCl}_5(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})$, the product side contains two moles of gas while the reactant side contains only one, causing the equilibrium to shift forward and thereby increasing the amounts of $\mathrm{PCl}_3$ and $\mathrm{Cl}_2$ while decreasing the amount of $\mathrm{PCl}_5$.

The equilibrium constant $K_p$ for a reaction defines the ratio of product partial pressures to reactant partial pressures at a constant temperature. For the decomposition of water, $\mathrm{H}_2\mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{g}) + \frac{1}{2}\mathrm{O}_2(\mathrm{g})$, we start with 1 mole of reactant and define the degree of dissociation as $\alpha$. At equilibrium, the molar amounts of $\mathrm{H}_2\mathrm{O}$, $\mathrm{H}_2$, and $\mathrm{O}_2$ are $(1-\alpha)$, $\alpha$, and $\frac{\alpha}{2}$, respectively. Because $\alpha$ is negligible with respect to 1, the total number of moles at equilibrium remains essentially 1, allowing us to approximate the partial pressure of each gas as simply its molar amount when the total pressure is 1 bar.

Substituting these values into the equilibrium expression yields the relationship between the constant and the degree of dissociation:

$K_p = \frac{P_{\mathrm{H}_2} \cdot (P_{\mathrm{O}_2})^{1/2}}{P_{\mathrm{H}_2\mathrm{O}}} \approx \frac{\alpha \cdot (\frac{\alpha}{2})^{1/2}}{1} = \frac{\alpha^{3/2}}{\sqrt{2}}$

Equating this expression to the provided value of $8.0 \times 10^{-3}$ gives $\frac{\alpha^{3/2}}{\sqrt{2}} = 8.0 \times 10^{-3}$, which simplifies to $\alpha^{3/2} = 8 \sqrt{2} \times 10^{-3}$. Raising both sides to the power of $\frac{2}{3}$ allows us to isolate $\alpha$:

$\alpha = (8 \sqrt{2} \times 10^{-3})^{2/3}$

Since $8 \sqrt{2}$ is equivalent to $2^3 \cdot 2^{0.5} = 2^{3.5}$ or $2^{7/2}$, the expression becomes $\alpha = (2^{7/2} \times 10^{-3})^{2/3} = 2^{7/3} \times 10^{-2}$. Calculating the numerical value of $2^{7/3}$ yields approximately 5.04, which results in a degree of dissociation of $5 \times 10^{-2}$ when rounded to the nearest integer.