Chemical Equilibrium – JEE Main ChemistryPractice Questions & PYQs

Le Chatelier's principle dictates that altering the concentration of a species in a chemical equilibrium shifts the reaction to counteract the change, but this principle only applies to species in gaseous or aqueous phases whose molar concentrations can vary. In the reaction $Fe_2O_3(s)+3CO(g) \rightleftharpoons 2Fe(l)+3CO_2(g)$, the reactant $Fe_2O_3$ is a solid and the product $Fe$ is a liquid, both of which have constant activities that remain unchanged regardless of the amount of substance present. Therefore, the addition of solid $Fe_2O_3$ does not change its effective concentration and will not disturb the equilibrium, whereas modifying the concentrations of gaseous $CO$ or $CO_2$ would directly alter the reaction quotient and induce a shift.

The position of equilibrium for a chemical reaction is defined by the equilibrium constant, $K_{eq}$, which represents the ratio of the product of concentrations of products to that of reactants at a given temperature. In the reaction $A + B \rightleftharpoons C + D$, since every species begins with a concentration of 1 M, introducing a variable $x$ accounts for the change in concentration as the system shifts toward equilibrium.

At equilibrium, the concentrations are $[A] = 1-x$, $[B] = 1-x$, $[C] = 1+x$, and $[D] = 1+x$. Substituting these values into the equilibrium expression $\frac{[C][D]}{[A][B]} = 100$ results in the equation $\frac{(1+x)^2}{(1-x)^2} = 100$. Taking the square root of both sides gives $\frac{1+x}{1-x} = 10$. Expanding this leads to $1+x = 10-10x$, which simplifies to $11x = 9$, or $x = \frac{9}{11}$. Consequently, the final equilibrium concentration of $D$ is $1+x$, which is $1 + \frac{9}{11}$, totaling $1.818$ mol $L^{-1}$.

The equilibrium constant $K_p$ for a heterogeneous reaction is determined by the partial pressures of the gaseous species involved, while solids are excluded from the expression. Given the reaction $\mathrm{XY}(s) \rightleftharpoons \mathrm{X}(g) + \mathrm{Y}(g)$, the solid reactant does not contribute to the pressure, meaning the total pressure is generated entirely by the gases $\mathrm{X}$ and $\mathrm{Y}$. Because these gases are produced in a stoichiometric ratio of 1:1, they must exert equal partial pressures, with each contributing half of the total 10 bar, resulting in a value of 5 bar for each gas. The equilibrium constant is found by taking the product of these partial pressures, which results in $K_p = (P_X)(P_Y) = 5 \times 5 = 25$.

The direction in which a chemical reaction proceeds is determined by comparing the reaction quotient, $Q_c$, to the equilibrium constant, $K_c$. If the value of $Q_c$ exceeds $K_c$, the system currently has a higher concentration of products relative to reactants than is maintained at equilibrium, which forces the reaction to shift toward the reactants to re-establish balance.

To determine the current state, first calculate $K_c$ using the standard Gibbs energy equation $\Delta G^{\circ} = -RT \ln K$. Substituting the given values, $2494.2 = -8.314 \times 300 \ln K$, leads to $\ln K = -1$, which results in $K_c = e^{-1} \approx 0.3679$. Next, find the reaction quotient using the given concentrations $[A] = \frac{1}{2}$, $[B] = 2$, and $[C] = \frac{1}{2}$ by calculating $Q_c = \frac{[B][C]}{[A]^2} = \frac{2 \times \frac{1}{2}}{(\frac{1}{2})^2} = 4$. Comparing these results, $Q_c = 4$ is greater than $K_c \approx 0.3679$, confirming that the reaction proceeds in the reverse direction.

The relationship between the equilibrium constant expressed in partial pressures $K_P$ and molar concentrations $K_C$ is governed by the expression $K_P = K_C(RT)^{\Delta n_g}$ where $\Delta n_g$ represents the difference between the \sum of the stoichiometric coefficients of the gaseous products and the gaseous reactants. For the reaction $\mathrm{SO}_{2(g)} + \frac{1}{2}\mathrm{O}_{2(g)} \rightleftharpoons \mathrm{SO}_{3(g)}$, we determine $\Delta n_g$ by subtracting the \sum of the reactant coefficients from the product coefficient. Given there is 1 mole of product and the \sum of the gaseous reactants is 1 plus 0.5 moles, this calculation becomes $1 - (1 + \frac{1}{2}) = -\frac{1}{2}$ which gives the change \in gaseous moles. By comparing this result to the given equation $K_P = K_C(RT)^x$, it follows that $x = -\frac{1}{2}$.

The equilibrium constant $K_C$ for a chemical reaction changes in a specific, predictable manner when the equation itself is modified. If a reaction is reversed, the new equilibrium constant becomes the reciprocal of the original, and if the stoichiometric coefficients are multiplied by a factor, the constant is raised to the power of that factor.

The target reaction is the reverse of the original equation, and all stoichiometric coefficients have been divided by two. Consequently, the new equilibrium constant is calculated as the reciprocal of the square root of the original constant. Substituting the given value into the expression $1 / \sqrt{4 \times 10^{-4}}$ results in 50.

The chemical equilibrium between carbon dioxide and solid graphite is governed by the law of mass action, which relates the partial pressures of gaseous species at equilibrium. Since solid carbon has a constant activity, it is excluded from the equilibrium constant expression, meaning $K_p$ depends solely on the partial pressures of the carbon dioxide and carbon monoxide gases.

Initially, the vessel contains carbon dioxide at a pressure of 0.5 atm. As the reaction proceeds to equilibrium, let $x$ represent the change in pressure. According to the stoichiometry of the reaction, the partial pressure of carbon dioxide decreases by $x$, while the partial pressure of carbon monoxide increases by $2x$. The total pressure at equilibrium is the sum of these partial pressures, which simplifies to 0.5 + $x$. By equating this sum to the provided equilibrium pressure of 0.8 atm, we find that $x = 0.3$.

With this value of $x$, the partial pressure of carbon dioxide at equilibrium is 0.2 atm, and the partial pressure of carbon monoxide is 0.6 atm. The equilibrium constant is defined by the formula $K_p = \frac{(P_{\text{CO}})^2}{P_{\text{CO}_2}}$ Substituting the determined partial pressures into this expression, we calculate $K_p$ as $\frac{(0.6)^2}{0.2}$, which yields a final value of 1.8 atm.

Carbonic acid undergoes a two-stage ionization process where the first step, $\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$, is governed by the ionization constant $K_1 = 4.2 \times 10^{-7}$. Because this dissociation follows a $1:1$ stoichiometry and acts as the primary source of ions, the concentration of $\text{H}^+$ is approximately equal to that of $\text{HCO}_3^-$ in a $0.034\text{ M}$ solution. The second ionization step, $\text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-}$, occurs with a much smaller constant of $K_2 = 4.8 \times 10^{-11}$, which is negligible in comparison to the first step. Consequently, this secondary process does not significantly alter the concentrations established by the initial equilibrium, confirming that the concentration of $\text{H}^+$ and $\text{HCO}_3^-$ remain essentially identical.

When chemical equations are summed, their respective equilibrium constants are multiplied to find the equilibrium constant for the overall reaction. Adding reaction (a), $\mathrm{CO}(\mathrm{g}) + \mathrm{H_2O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{g}) + \mathrm{H_2}(\mathrm{g})$, and reaction (b), $\mathrm{CH}_4(\mathrm{g}) + \mathrm{H_2O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g}) + 3\mathrm{H_2}(\mathrm{g})$, causes the intermediate $\mathrm{CO}(\mathrm{g})$ to cancel from both sides. The resulting net reaction, $\mathrm{CH}_4(\mathrm{g}) + 2\mathrm{H_2O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{g}) + 4\mathrm{H_2}(\mathrm{g})$, matches reaction (c), which leads to the expression $K_3 = K_1 \cdot K_2$.

The equilibrium constant $K_P$ relates the partial pressures of chemical species in a gaseous mixture, which depend on both the total pressure of the system and the mole fractions of the individual components. For the reaction $X \rightleftharpoons 2Y$, assuming one mole of $X$ initially and a degree of dissociation $\alpha$, the equilibrium mixture consists of $1-\alpha$ moles of $X$ and $2\alpha$ moles of $Y$. These sum to $1+\alpha$ total moles, allowing the equilibrium constant to be expressed as:
$K_{P1} = \frac{4\alpha^2 P_{T1}}{1-\alpha^2}$

Applying the same approach to the reaction $Z \rightleftharpoons P+Q$ with an identical degree of dissociation $\alpha$ produces $1-\alpha$ moles of $Z$ and $\alpha$ moles each of $P$ and $Q$. This reaction mixture also contains a total of $1+\alpha$ moles, resulting in an equilibrium constant of:
$K_{P2} = \frac{\alpha^2 P_{T2}}{1-\alpha^2}$

Since the ratio of the equilibrium constants is 1:9, dividing the expression for $K_{P1}$ by $K_{P2}$ yields the relation $4 \frac{P_{T1}}{P_{T2}} = \frac{1}{9}$. Rearranging this equation confirms that the ratio of total pressure at these equilibria is 1:36.

Calculating partial pressure relies on the principle that the partial pressure of a gas is its mole fraction multiplied by the total system pressure. By considering an initial amount of one mole of $\mathrm{PCl}_5$, the degree of dissociation $x$ leads to an equilibrium mixture containing $1-x$ moles of $\mathrm{PCl}_5$, $x$ moles of $\mathrm{PCl}_3$, and $x$ moles of $\mathrm{Cl}_2$. The total number of moles at equilibrium is the sum of these, resulting in $1+x$ moles. Because the mole fraction of $\mathrm{PCl}_3$ is defined as its equilibrium moles divided by the total moles, its partial pressure is $\left( \frac{x}{1+x} \right) \mathrm{P}$.

When chemical equations are modified, the equilibrium constant $K_c$ changes according to specific mathematical rules: reversing a reaction takes the reciprocal of the equilibrium constant, while multiplying the stoichiometric coefficients by a factor raises the constant to that power. The target reaction $2 \mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{g})$ is obtained by reversing the original reaction $\mathrm{SO}_3(\mathrm{g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g})$ and multiplying the coefficients by two. Thus, the new equilibrium constant is the reciprocal of the original constant squared. This is calculated as:

$K_c' = \frac{1}{(4.9 \times 10^{-2})^2}$

Squaring the value $4.9 \times 10^{-2}$ gives $24.01 \times 10^{-4}$, and finding its reciprocal results in 416.

The relationship between the equilibrium constants $K_p$ and $K_c$ is defined by the equation $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ represents the change in the number of gaseous moles between products and reactants. For the decomposition reaction $2 \mathrm{NO}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) + \mathrm{O}_2(\mathrm{g})$, there are three moles of gaseous products and two moles of gaseous reactants, resulting in $\Delta n = 1$. Converting the temperature to Kelvin gives $T = 273 + 184 = 457 \text{ K}$. Applying the given values $K_c = 1.8 \times 10^{-6}$ and $R = 0.0831 \text{ kJ}/(\text{mol} \cdot \mathrm{K})$, the equation yields $K_p = 1.8 \times 10^{-6} \times 0.0831 \times 457 = 6.836 \times 10^{-6}$, which confirms that $K_p$ is greater than $K_c$.

The equilibrium constant for a heterogeneous decomposition reaction is determined by the product of the partial pressures of the gaseous products, as the solid phase is excluded from the expression. For the reaction $\mathrm{NH}_4\mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_3(\mathrm{g}) + \mathrm{H}_2\mathrm{S}(\mathrm{g})$, this is defined by the expression $K_p = P_{\mathrm{NH}_3} \cdot P_{\mathrm{H}_2\mathrm{S}}$.

Initially, the flask contains 0.50 atm of ammonia gas and 0 atm of hydrogen sulfide. As the reaction proceeds to reach equilibrium, an increase in pressure, denoted by $x$, occurs for both gaseous products due to the 1:1 stoichiometry of the decomposition. The equilibrium partial pressure of ammonia becomes $(0.50 + x)$ atm, while the partial pressure of hydrogen sulfide is $x$ atm. Given that the total pressure in the flask is 0.84 atm at equilibrium, we sum these individual pressures: $(0.50 + x) + x = 0.84$. Solving this equation gives $2x = 0.34$, which means $x = 0.17$. With these values established, the partial pressure of $\mathrm{NH}_3$ is 0.67 atm and the partial pressure of $\mathrm{H}_2\mathrm{S}$ is 0.17 atm. The final equilibrium constant is calculated as $0.67 \times 0.17 = 0.1139$, which rounds to 0.11.

Le Chatelier's principle dictates that a system at equilibrium will shift to mitigate any disturbance, such as the addition of a reactant in the exothermic process $\mathrm{Cl}_2(\mathrm{g}) + 3 \mathrm{F}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{ClF}_3(\mathrm{g})$ with $\Delta H = -329 \mathrm{kJ}$. By increasing the concentration of $\mathrm{F}_2$, the system acts to counteract this change by consuming the added reactant, which drives the reaction forward toward the production of more $\mathrm{ClF}_3$. Because the system seeks to re-establish balance, this shift effectively increases the quantity of the product in the mixture.

The relationship between the equilibrium constants $K_p$ and $K_c$ is defined by the equation $K_p = K_c (RT)^{\Delta n}$, where $\Delta n$ represents the total change in the number of moles of gas as the reaction proceeds. For the reaction $\mathrm{CO}(g) + \mathrm{Cl}_2(g) \rightleftharpoons \mathrm{COCl}_2(g)$, this change is calculated by subtracting the two moles of gaseous reactants from the single mole of gaseous product, yielding $\Delta n = 1 - 2 = -1$. Substituting this value into the expression gives $K_p = K_c (RT)^{-1}$, and rearranging this equation directly provides the ratio $\frac{K_p}{K_c} = \frac{1}{RT}$.

The equilibrium constant expression is determined by the ratio of product concentrations to reactant concentrations, with each concentration raised to the power of its stoichiometric coefficient in the balanced chemical equation. A crucial principle in this calculation is that pure solids and pure liquids are assigned a concentration of unity and are therefore excluded from the expression. Because $\mathrm{P}_4$ and $\mathrm{P_4O_{10}}$ are solids in this reaction, they are omitted, meaning only the gaseous oxygen reactant affects the value of the equilibrium constant. Following these rules for the provided reaction leads to the final expression $K_c = \frac{1}{[\mathrm{O}_2]^5}$.

Equilibrium constants follow specific mathematical rules when reaction coefficients are adjusted or when a reaction is reversed. Reversing a chemical equation results in the reciprocal of the original constant, while multiplying the stoichiometry by a fraction corresponds to raising that constant to the power of that fraction. Since the target reaction is the reverse of the primary reaction multiplied by a factor of $\frac{1}{2}$, its equilibrium constant is determined by taking the reciprocal of the square root of the initial constant. Calculating this gives $1 / \sqrt{4 \times 10^{-4}}$, which simplifies to $1 / (2 \times 10^{-2})$ and results in a final value of 50.

Le Chatelier's principle states that a system at equilibrium adjusts to counteract any imposed disturbance. Since the reaction is exothermic with $\Delta H^{\circ} = -198 \text{ kJ}$, reducing the temperature shifts the equilibrium to favor product formation. Additionally, because the reaction involves a reduction in the total number of gaseous moles from 3 to 2, increasing the pressure pushes the system toward the forward reaction to occupy less volume.

The equilibrium constant $K_c$ represents the ratio of product concentrations raised to their stoichiometric coefficients over reactant concentrations raised to theirs. For the reaction $\mathrm{N}_2\mathrm{O}_4(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}_2(\mathrm{g})$, the equilibrium expression is:

$K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]}$

Substituting the provided values requires squaring the concentration of $\mathrm{NO}_2$, which is 1.2 x 10^-2, to obtain 1.44 x 10^-4. Dividing this result by the concentration of $\mathrm{N}_2\mathrm{O}_4$, which is 4.8 x 10^-2, gives a final value of 3 x 10^-3 mol/L.