Chemical Equilibrium – JEE Main ChemistryPractice Questions & PYQs

The equilibrium constant $K_p$ for the dissociation reaction $\mathrm{X_2Y(g)} \rightarrow \mathrm{X_2(g)} + \frac{1}{2}\mathrm{Y_2(g)}$ is determined by the partial pressures of the reactant and products at equilibrium. Starting with 1 mole of $\mathrm{X_2Y}$, the equilibrium composition involves $1-x$ moles of reactant, $x$ moles of $\mathrm{X_2}$, and $\frac{x}{2}$ moles of $\mathrm{Y_2}$, making the total number of moles $1+\frac{x}{2}$. Since $x$ is very small, we approximate the total moles as $1$, which simplifies the reactant partial pressure to $p$, while the product partial pressures become $xp$ and $\frac{x}{2}p$ respectively. The equilibrium constant expression is given by:

$K_p = \frac{(xp)(\frac{x}{2}p)^{1/2}}{p}$

Simplifying this relationship leads to $K_p = \frac{x^{3/2}p^{1/2}}{\sqrt{2}}$. Squaring both sides yields $K_p^2 = \frac{x^3 p}{2}$, which allows us to solve for the degree of dissociation as:

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

The equilibrium constant for the reaction $\mathrm{CO}_2(g) + \mathrm{C}(s) \rightleftharpoons 2\mathrm{CO}(g)$ is derived from the partial pressures of the gases present, as solid carbon does not contribute to the total pressure. If the $\mathrm{CO}_2$ pressure decreases by x, the $\mathrm{CO}$ pressure increases by 2x, leading to equilibrium partial pressures of $(0.5 - x)$ for $\mathrm{CO}_2$ and $(2x)$ for $\mathrm{CO}$. Given the total equilibrium pressure of 0.8 atm, the equation $(0.5 - x) + 2x = 0.8$ reveals that $x = 0.3$, which corresponds to equilibrium partial pressures of 0.2 atm for $\mathrm{CO}_2$ and 0.6 atm for $\mathrm{CO}$. Applying the formula $K_p = \frac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO}_2}}$ with these values gives $K_p = \frac{(0.6)^2}{0.2}$, which equals 1.8 atm.

When external pressure is applied to a gaseous equilibrium system at constant temperature, Le Chatelier's principle dictates that the reaction shifts to minimize the stress by proceeding toward the side with fewer moles of gas. For the reaction $\mathrm{CO}(g) + 3\mathrm{H}_2(g) \rightleftharpoons \mathrm{CH}_4(g) + \mathrm{H}_2\mathrm{O}(g)$, there are two moles of product for every four moles of reactant, represented by $\Delta n_{g} = -2$, which drives the system to shift in the forward direction. Simultaneously, the decrease in volume caused by the increased pressure results in an increase in the molar concentrations of both the reactants and the products. Since the equilibrium constant depends exclusively on temperature, it remains unchanged despite these shifts in position and concentration.

The equilibrium constant $K_p$ for a gas-phase reaction relates the partial pressures of reactants and products once a system reaches a steady state, which can be determined by tracking pressure changes from the initial conditions. First, calculate the initial number of moles of $\mathrm{N}_2\mathrm{O}_5$ using its molar mass of $108 \mathrm{g/mol}$, yielding $\frac{37.8}{108} = 0.35 \mathrm{mol}$. Using the ideal gas law $P = \frac{nRT}{V}$ with the provided values, the initial pressure of $\mathrm{N}_2\mathrm{O}_5$ in the $1 \mathrm{L}$ vessel is $\frac{0.35 \times 0.082 \times 500}{1} = 14.35 \mathrm{bar}$.

With the reaction $2\mathrm{N}_2\mathrm{O}_5(\mathrm{g}) \rightleftharpoons 2\mathrm{N}_2\mathrm{O}_4(\mathrm{g}) + \mathrm{O}_2(\mathrm{g})$, the stoichiometric coefficients imply that for every $2x$ decrease in the partial pressure of the reactant, the products increase by $2x$ and $x$ respectively. At equilibrium, the partial pressures are $14.35 - 2x$ for $\mathrm{N}_2\mathrm{O}_5$, $2x$ for $\mathrm{N}_2\mathrm{O}_4$, and $x$ for $\mathrm{O}_2$. Since the total pressure at equilibrium is the sum of these partial pressures, we have $14.35 - 2x + 2x + x = 14.35 + x = 18.65 \mathrm{bar}$, which gives $x = 4.3 \mathrm{bar}$.

Substituting $x = 4.3$ back into the partial pressure expressions, the equilibrium partial pressures are $P_{\mathrm{N}_2\mathrm{O}_5} = 14.35 - 8.6 = 5.75 \mathrm{bar}$, $P_{\mathrm{N}_2\mathrm{O}_4} = 8.6 \mathrm{bar}$, and $P_{\mathrm{O}_2} = 4.3 \mathrm{bar}$. The expression for the equilibrium constant is $K_p = \frac{(P_{\mathrm{N}_2\mathrm{O}_4})^2 \times P_{\mathrm{O}_2}}{(P_{\mathrm{N}_2\mathrm{O}_5})^2}$, which equals $\frac{(8.6)^2 \times 4.3}{(5.75)^2}$. Calculating this yields $9.62$, which corresponds to $962 \times 10^{-2}$.

For the dissociation of $\mathrm{A B}_2(\mathrm{g})$ into $\mathrm{A B}(\mathrm{g})$ and $\frac{1}{2} \mathrm{B}_2(\mathrm{g})$, we begin by establishing the partial pressures of each species at equilibrium. If the initial pressure of the reactant is $p_0$, then at equilibrium, the partial pressures are $p_0(1-x)$ for $\mathrm{A B}_2$, $p_0 x$ for $\mathrm{A B}$, and $\frac{1}{2} p_0 x$ for $\mathrm{B}_2$. The total pressure $p$ is the sum of these partial pressures, giving $p = p_0(1 - x + x + \frac{x}{2}) = p_0(1 + \frac{x}{2})$.

Since the degree of dissociation $x$ is very small compared to unity, we can apply the approximation $1 + \frac{x}{2} \approx 1$, which implies that the initial pressure is approximately equal to the total pressure, or $p_0 \approx p$. Similarly, in the expression for the equilibrium constant $K_p = \frac{p_{\mathrm{AB}} (p_{\mathrm{B}_2})^{1/2}}{p_{\mathrm{AB}_2}}$, the term $1-x$ in the denominator is approximately equal to 1. Substituting the partial pressures into the expression, we get $K_p \approx \frac{(p_0 x) (\frac{1}{2} p_0 x)^{1/2}}{p_0}$.

Replacing $p_0$ with $p$ in this simplified expression leads to $K_p \approx \frac{p x \cdot p^{1/2} x^{1/2}}{p \sqrt{2}} = \frac{p^{1/2} x^{3/2}}{\sqrt{2}}$. To solve for $x$, we square both sides of the equation to obtain $K_p^2 = \frac{p x^3}{2}$. Isolating $x^3$ yields $x^3 = \frac{2 K_p^2}{p}$, which confirms that $x = \sqrt[3]{ \frac{2 K_p^2}{p} }$.

The equilibrium constant $K_C$ for a reaction is determined by the law of mass action, which expresses the ratio of product concentrations raised to their stoichiometric coefficients to reactant concentrations raised to theirs. For the formation of ammonia, $\mathrm{N}_2 + 3\mathrm{H}_2 \rightleftharpoons 2\mathrm{NH}_3$, this translates to the mathematical relationship $K_C = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3}$. Substituting the specified molar concentrations gives $K_C = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2}) \times (3 \times 10^{-2})^3}$. Evaluating the powers in the expression leads to $K_C = \frac{2.25 \times 10^{-4}}{5.4 \times 10^{-7}}$, which simplifies to 417.

The connection between the pressure-based equilibrium constant $K_p$ and the concentration-based equilibrium constant $K_c$ is established by the equation:

$K_p = K_c(RT)^{\Delta n_g}$

In this expression, $\Delta n_g$ represents the difference between the sum of the stoichiometric coefficients of the gaseous products and the gaseous reactants. For the decomposition reaction $\mathrm{N}_2\mathrm{O}_4(\mathrm{g}) \rightleftharpoons 2\,\mathrm{NO}_2(\mathrm{g})$, there are two moles of gaseous products for every one mole of reactant, which makes $\Delta n_g = 2 - 1$, or 1.

Given that the temperature is 300 K, the value of the gas constant $R$ is 0.082, and the equilibrium constant $K_p$ is 0.492, we determine $K_c$ by rearranging the formula to $K_c = \frac{K_p}{(RT)^1}$. Substituting these values gives $K_c = \frac{0.492}{0.082 \times 300}$. Since the product in the denominator is 24.6, the equation simplifies to $K_c = \frac{0.492}{24.6}$. Calculating this division results in 0.02, which is equivalent to $2 \times 10^{-2}$. Thus, the required value for the blank is 2.

The magnitude of an equilibrium constant depends on the direction and stoichiometry of the chemical equation. Reversing a reaction transforms the equilibrium constant into its reciprocal, and multiplying all coefficients by a constant factor raises the original equilibrium constant to that factor as a power.

To obtain the target equation $2 \mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{g})$ from $\mathrm{SO}_3(\mathrm{g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g})$, we must invert the reaction and multiply the coefficients by two. This results in a new equilibrium constant calculated as $( \frac{1}{K_c} )^2$, where $K_c = 4.9 \times 10^{-2}$. Performing this calculation yields a value of approximately 416.

According to Le Chatelier's principle, the equilibrium position of a reaction is sensitive only to changes in the concentration or partial pressure of substances that appear in the equilibrium constant expression, which excludes pure solids and liquids. Because $\mathrm{Fe_2O_{3(s)}}$ exists in a solid state, adding more of it to the reaction $\mathrm{Fe_2O_{3(s)}} + 3\mathrm{CO_{(g)}} \rightleftharpoons \mathrm{Fe_{(l)}} + 3\mathrm{CO_{2(g)}}$ does not alter its effective concentration or chemical potential. Consequently, the addition of this solid reactant results in no change to the system, leaving the equilibrium undisturbed.

The law of mass action determines the equilibrium constant, $K_{C}$, for a reversible reaction by calculating the ratio of the molar concentrations of the products to the molar concentrations of the reactants, with each concentration raised to the power of its stoichiometric coefficient. For the reaction between iron(III) and thiocyanate ions, the only product is $(\mathrm{FeSCN})^{2+}$, while the reactants are $\mathrm{Fe}^{3+}$ and $\mathrm{SCN}^{-}$. Because each species in this balanced chemical equation has a stoichiometric coefficient of one, the resulting expression is simply K_{C}=\frac{[\mathrm{FeSCN}^{2+}]\}{[\mathrm{Fe}^{3+}]$$[\mathrm{SCN}^{-}]$}$.

The equilibrium constant $K_P$ for a gas-phase reaction relates the partial pressures of the products and reactants, and its behavior is dictated by the change in the number of gaseous moles, denoted as $\Delta n_g$. For the reaction $\mathrm{H}_2(\mathrm{g}) + \mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{HI}(\mathrm{g})$, we calculate $\Delta n_g = 2 - (1 + 1) = 0$. Because the change in gaseous moles is zero, the contribution of the total pressure to the equilibrium expression is nullified, making the equilibrium constant independent of the system's pressure.

Since the problem specifies that there are an equal number of molecules for each species, let $n$ represent the amount of $\mathrm{H}_2$, $\mathrm{I}_2$, and $\mathrm{HI}$. Substituting these into the equilibrium expression gives $K_P = \frac{n^2}{n \cdot n} = 1$. By equating this calculated value to the given expression $K_P = x \times 10^{-1}$, we find $1 = x \times 10^{-1}$, which implies $x = 10$.

The fundamental relationship connecting the equilibrium constants expressed in terms of partial pressure and concentration is given by the equation $K_P = K_C(RT)^{\Delta n_g}$, where $\Delta n_g$ signifies the change in the number of moles of gaseous species during the reaction. For the reaction $\mathrm{CO}_{(\mathrm{g})} + \frac{1}{2} \mathrm{O}_{2(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{g})}$, the number of moles of gaseous products is 1, and the total number of moles of gaseous reactants is $1 + 0.5 = 1.5$. By subtracting the reactant moles from the product moles, the value of $\Delta n_g$ is calculated as $1 - 1.5 = -0.5$, or $-1/2$. Rearranging the relationship to isolate the ratio $\frac{K_P}{K_C}$ yields $(RT)^{-1/2}$, which simplifies to $\frac{1}{\sqrt{RT}}$.

The equilibrium constant $K_P$ for a gaseous reaction is defined by the ratio of the partial pressures of the products to the reactants, each raised to the power of its respective stoichiometric coefficient. For the dissociation $A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$, we begin with one mole of reactant and allow a degree of dissociation $\alpha$. This results in equilibrium molar amounts of $(1-\alpha)$ for $A$, $\alpha$ for $B$, and $\alpha/2$ for $C$. Summing these quantities, the total number of moles at equilibrium is $(2+\alpha)/2$.

Since the partial pressure of a gas is the product of its mole fraction and the total pressure $P$, the partial pressures for the species are $P_A = \frac{1-\alpha}{1+\alpha/2} P$, $P_B = \frac{\alpha}{1+\alpha/2} P$, and $P_C = \frac{\alpha/2}{1+\alpha/2} P$. Applying these expressions to the equilibrium constant formula $K_P = \frac{P_B \cdot P_C^{1/2}}{P_A}$ and simplifying the resulting algebraic terms leads to $K_P = \frac{\alpha^{3/2} P^{1/2}}{(1-\alpha)(2+\alpha)^{1/2}}$.

When sequential chemical reactions are summed to obtain a net reaction, the equilibrium constant for the final equation is determined by multiplying the equilibrium constants of the individual steps. By combining the reactions $X \rightleftharpoons Y$, $Y \rightleftharpoons Z$, and $Z \rightleftharpoons W$, the intermediate species $Y$ and $Z$ appear on both sides of the equations and cancel out to yield the overall reaction $X \rightleftharpoons W$. Mathematically, this combination requires multiplying the individual constants, represented as $K_{net} = K_1 \times K_2 \times K_3$. Substituting the given values results in the calculation $1.0 \times 2.0 \times 4.0 = 8.0$.

The precipitation of salts in a saturated solution upon the addition of $\mathrm{HCl}$ gas is governed by the common ion effect and the law of mass action. When $\mathrm{HCl}$ gas is passed through a saturated solution of $\mathrm{BaCl}_2$, the substantial increase in chloride ion concentration forces the equilibrium to shift to the left, leading to the precipitation of $\mathrm{BaCl}_2$ in the form of white turbidity. However, the assertion that $\mathrm{HCl}$ gas precipitates $\mathrm{NaCl}$ from a saturated solution under these conditions is inaccurate, as such precipitation is not observed at room temperature despite the presence of the common ion.

The equilibrium constant $K_p$ for a gas-phase reaction relates to the total pressure $p$ and the degree of dissociation $\alpha$ through the mole fractions of the reactants and products. For the dissociation of $\mathrm{X}(g)$ into $\mathrm{Y}(g)$ and $\mathrm{Z}(g)$, starting with initial moles $n$, the equilibrium composition consists of $n(1-\alpha)$ moles of $\mathrm{X}$, $n\alpha$ moles of $\mathrm{Y}$, and $n\alpha$ moles of $\mathrm{Z}$, resulting in a total of $n(1+\alpha)$ moles. Using these values to express the partial pressures, the equilibrium constant is $K_{p1} = \frac{\left( \frac{\alpha}{1+\alpha} p_1 \right)^2}{\frac{1-\alpha}{1+\alpha} p_1} = \frac{\alpha^2 p_1}{1-\alpha^2} = 3.$ Following the same logic for the dissociation of $\mathrm{A}(g)$ into $2\mathrm{B}(g)$, the equilibrium moles are $n(1-\alpha)$ for $\mathrm{A}$ and $2n\alpha$ for $\mathrm{B}$, with a total of $n(1+\alpha)$ moles. The equilibrium constant for this second reaction is $K_{p2} = \frac{\left( \frac{2\alpha}{1+\alpha} p_2 \right)^2}{\frac{1-\alpha}{1+\alpha} p_2} = \frac{4\alpha^2 p_2}{1-\alpha^2} = 1.$ Because the degree of dissociation $\alpha$ is given as equal for both reactions, we can find the ratio of the total pressures by dividing the first equation by the second, which allows the common terms containing $\alpha$ to cancel out. This simplifies to $\frac{3}{1} = \frac{p_1}{4 p_2},$ indicating that the ratio of the total pressures $\frac{p_1}{p_2}$ is 12. Therefore, the value of $x$ is 12.

According to Le Chatelier's Principle, a system at equilibrium shifts to counteract external disturbances, which determines how one can optimize the yield of $\mathrm{SO}_3$ for the exothermic reaction $2\,\mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\,\mathrm{SO}_3(\mathrm{g})$ characterized by $\Delta H = -190\,\mathrm{kJ}$. Increasing the temperature shifts the equilibrium toward the reactants, while adding a catalyst only accelerates the rate at which equilibrium is reached without changing the final amounts of products, so neither of these factors increases the yield. However, increasing the total pressure shifts the position toward the side with fewer gaseous moles, and adding more of either reactant, $\mathrm{SO}_2$ or $\mathrm{O}_2$, drives the reaction forward to restore equilibrium, meaning these three specific actions effectively increase the production of $\mathrm{SO}_3$.

When an inert gas like helium is added to the system $\mathrm{PCl}_5(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})$ at constant volume, the total pressure within the container increases, yet the partial pressures and concentrations of the individual reacting species remain unchanged. Because the active mass of each reactant and product stays constant, the reaction quotient does not shift relative to the equilibrium constant. Consequently, the equilibrium state is not affected by the presence of the inert gas.

Determining the standard Gibbs energy change begins with finding the equilibrium constant, which is the ratio of the forward rate constant to the reverse rate constant. Dividing $10^3$ by $10^2$ gives an equilibrium constant of $10$. This constant relates to the standard Gibbs energy change through the thermodynamic equation $\Delta_{r} G^{\circ} = -RT \ln K_{\text{eq}}$.

First, convert the temperature from $27^{\circ}\text{C}$ to the absolute scale to get $300\ \text{K}$. Substituting the values $R = 8.3\ \text{J K}^{-1}\text{mol}^{-1}$, $T = 300\ \text{K}$, and $\ln 10 = 2.3$ into the equation produces $\Delta_{r} G^{\circ} = -(8.3 \times 300 \times 2.3)\ \text{J mol}^{-1}$. Calculating this results in $-5727\ \text{J mol}^{-1}$, which is $-5.727\ \text{kJ mol}^{-1}$. Since the expression already includes a negative sign, the nearest integer magnitude required to fill the blank is $6$.

In a reaction involving gases at constant volume and temperature, the total pressure is directly proportional to the total number of gaseous moles, allowing us to track the reaction progress using partial pressures. Consider the reactant $\mathrm{A}$ beginning with an initial pressure of 450 mmHg. As the reaction proceeds, let $P$ represent the decrease in the partial pressure of $\mathrm{A}$. According to the stoichiometry of $\mathrm{A(g)} \rightleftharpoons 2\mathrm{B(g)} + \mathrm{C(g)}$, the partial pressures of the products $\mathrm{B}$ and $\mathrm{C}$ at time $t$ will be $2P$ and $P$, respectively, while the remaining pressure of $\mathrm{A}$ will be $450 - P$.

The total pressure at any time $t$ is the sum of these partial pressures:
$450 - P + 2P + P = 720$
Simplifying this expression leads to $450 + 2P = 720$, which reveals that $2P = 270$, so $P = 135$. The fraction of $\mathrm{A}$ decomposed is the ratio of the amount reacted, $P$, to the initial amount, 450. Calculating this gives $135 / 450 = 0.3$, which corresponds to $3 \times 10^{-1}$. Comparing this to $x \times 10^{-1}$, we find that the value of $x$ is 3.