Chemical Equilibrium – JEE Main ChemistryPractice Questions & PYQs

The decomposition reaction $\mathrm{N}_2\mathrm{O}_4(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}_2(\mathrm{g})$ has a positive enthalpy change of $\Delta H^{0} = +58 \text{ kJ}$, identifying it as an endothermic process that absorbs heat. When the temperature is decreased, Le Chatelier's principle dictates that the system will counteract the loss of thermal energy by shifting the equilibrium backward toward the reactants. Regarding the addition of $\mathrm{N}_2$, since it acts as an inert gas that does not participate in the chemical reaction, introducing it at a constant temperature produces no change in the equilibrium state.

The relationship between the equilibrium constant and temperature is governed by the van 't Hoff equation, which correlates the enthalpy change of a reaction to the equilibrium constants at two distinct temperatures:
$\ln \left( \frac{K_2}{K_1} \right) = \frac{\Delta H^{\circ}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$
By substituting the ratio value $\ln(10)$ and the temperatures $298 \text{ K}$ and $373 \text{ K}$ into this expression along with the gas constant $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$, the enthalpy change $\Delta H^{\circ}$ is determined to be $28.37 \text{ kJ mol}^{-1}$. The standard Gibbs free energy change is subsequently calculated using the expression $\Delta G^{\circ} = -RT \ln K$. Evaluating this thermodynamic relationship at $298 \text{ K}$ with $\ln(10)$ gives $\Delta G^{\circ} = -5.71 \text{ kJ mol}^{-1}$, whereas applying it at $373 \text{ K}$ with $\ln(100)$ results in a value of $\Delta G^{\circ} = -14.29 \text{ kJ mol}^{-1}$.

The equilibrium constant for a chemical reaction is defined by the ratio of product concentrations raised to their stoichiometric coefficients to reactant concentrations raised to their stoichiometric coefficients, evaluated at the point where the reaction reaches equilibrium. Given a vessel volume of 1 L, the initial molar concentrations correspond to the mole values provided: 1.0 mol/L for X, 1.5 mol/L for Y, and 0.5 mol/L for Z. As the concentration of Z increases from 0.5 mol/L to 1.0 mol/L at equilibrium, the total increase is 0.5 mol/L. According to the balanced equation $X + Y \rightleftharpoons 2Z$, the stoichiometry shows that producing 2 moles of Z requires consuming 1 mole of X and 1 mole of Y. Consequently, the production of 0.5 mol/L of Z consumes 0.25 mol/L of X and 0.25 mol/L of Y. This adjustment leads to equilibrium concentrations of 0.75 mol/L for X and 1.25 mol/L for Y. The equilibrium constant expression is $K_{\text{eq}} = \frac{[Z]^2}{[X][Y]}$. Substituting the equilibrium values results in $K_{\text{eq}} = \frac{1.0^2}{0.75 \times 1.25}$, which simplifies to $\frac{16}{15}$. Comparing this result to the requested form $\frac{x}{15}$, the value of x is 16.

The relationship between the equilibrium constant expressed in terms of partial pressure $K_P$ and that in terms of concentration $K_C$ depends entirely on the variation in gaseous moles between products and reactants, defined by the equation $K_P = K_C(RT)^{\Delta n_g}$. In the reaction provided, the gaseous product consists of $1$ mole of $\mathrm{NH}_3$, while the gaseous reactants consist of $\frac{3}{2}$ moles of $\mathrm{H}_2$, with solid species being excluded from the calculation of the change in gaseous moles.

Subtracting the reactant moles from the product moles gives $\Delta n_g = 1 - \frac{3}{2} = -\frac{1}{2}$. Applying this exponent to the relationship formula results in $K_P = K_C(RT)^{-\frac{1}{2}}$. Solving for $K_C$ by adjusting the terms yields $K_C = K_P(RT)^{\frac{1}{2}}$.

The addition of a non-volatile solute like glucose to water restricts the liquid surface by occupying space that would otherwise be available for solvent molecules to vaporize. This reduction in the effective surface area leads to a decrease in the rate at which water molecules can transition from the solution into the vapor phase, which directly explains why the overall vapor pressure of the solution is lowered.

Chemical equilibrium is defined by the condition where the rate of the forward reaction equals the rate of the reverse reaction, mathematically represented as $r_a = r_b$. For the reversible process $A \rightleftharpoons B$, the forward rate $r_a$ begins at a maximum and decreases over time as reactant $A$ is consumed. Simultaneously, the reverse rate $r_b$ starts from zero and increases as product $B$ accumulates, leading both rates to converge at the point of equilibrium.

The equilibrium constant of a chemical reaction is inherently linked to its stoichiometric coefficients and the direction in which the reaction proceeds. When a balanced equation is reversed, the new equilibrium constant becomes the reciprocal of the original value. Furthermore, if the stoichiometric coefficients of an equation are multiplied or divided by a specific factor, the equilibrium constant must be raised to the power of that factor.

For the initial reaction $\mathrm{N}_2(\mathrm{g}) + 3 \mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{g})$ with $K_C = 64$, reversing the process to form the reactants from the product yields a new constant of $\frac{1}{64}$. To reach the target reaction $\mathrm{NH}_3(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_2(\mathrm{g}) + \frac{3}{2} \mathrm{H}_2(\mathrm{g})$, the coefficients of this reversed equation are divided by two. Consequently, the equilibrium constant is calculated by taking the square root of the constant from the reversed reaction, which results in a final value of $\frac{1}{8}$.

When chemical reactions are added together to produce a net equation, the equilibrium constants of those individual reactions are multiplied to obtain the overall equilibrium constant. If a reaction must be reversed to obtain the target stoichiometry, its equilibrium constant is inverted, and multiplying stoichiometric coefficients by a value requires raising the equilibrium constant to the power of that value.

The target reaction $2\mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2\mathrm{SO}_3(g)$ is constructed by taking the second provided reaction and adding it to the reverse of the first reaction, which has been multiplied by 2. Specifically, the reaction $\mathrm{S}(s)+\mathrm{O}_2(g) \rightleftharpoons \mathrm{SO}_2(g)$ is reversed and scaled by 2 to produce $2\mathrm{SO}_2(g) \rightleftharpoons 2\mathrm{S}(s)+2\mathrm{O}_2(g)$, which transforms its equilibrium constant into $\frac{1}{K_1^2}$. Adding this to the reaction $2\mathrm{S}(s)+3\mathrm{O}_2(g) \rightleftharpoons 2\mathrm{SO}_3(g)$ cancels out the intermediate terms, leading to the following calculation:
$K_{\text{eq}} = \frac{K_2}{K_1^2} = \frac{10^{129}}{(10^{52})^2} = 10^{25}$

The equilibrium constant of a chemical reaction changes in a predictable manner when the equation is mathematically manipulated. Reversing a balanced equation transforms the original equilibrium constant into its reciprocal, while multiplying the coefficients of a reaction by a specific factor results in raising the equilibrium constant to the power of that factor. These fundamental rules allow for the derivation of a new equilibrium constant based on the modifications made to the stoichiometric ratios of the original reactants and products.

In the provided system, the first reaction is effectively reversed and then multiplied by 3 to obtain the coefficients present in the second reaction. Reversing the first equation changes the equilibrium constant from $K_1$ to $\frac{1}{K_1}$. Subsequent multiplication of this reversed equation by 3 to match the stoichiometry of the second reaction requires raising this reciprocal to the third power. This operation yields $(\frac{1}{K_1})^3$, which simplifies directly to $K_1^{-3}$, resulting in the relationship $K_2 = K_1^{-3}$.

The equilibrium constant for a heterogeneous reaction, such as the decomposition of solid ammonium hydrosulfide into gaseous ammonia and hydrogen sulfide, is determined by the product of the partial pressures of the gaseous products because the activity of a pure solid is constant. Start by calculating the initial moles of $NH_4SH$ by dividing the given 5.1 g mass by the molar mass of 51 g/mol, which equals 0.1 mol. Because 30% of this solid decomposes, the reaction generates 0.03 mol of $NH_3$ and 0.03 mol of $H_2S$ at equilibrium.

With the temperature at 327°C, which corresponds to 600 K, determine the partial pressure of each gas using the ideal gas equation $P = \frac{nRT}{V}$. Substituting the known values for either gas in the 3.0 L flask yields $P = \frac{0.03 \times 0.082 \times 600}{3} = 0.492$ atm. The equilibrium constant expression for this reaction is $K_p = P_{NH_3} \times P_{H_2S}$, so the final calculation is 0.492 multiplied by 0.492, which equals 0.242 atm².

The equilibrium constant for the reaction $A + 2B \rightleftharpoons 2C + D$ depends on the concentration of each species once the system reaches a stable state. If we represent the initial concentration of A as $a_0$, then the initial concentration of B is $1.5a_0$. By defining the reaction change in terms of a variable $x$, the remaining amounts at equilibrium are $a_0 - x$ for A and $1.5a_0 - 2x$ for B. Equating these two expressions according to the provided condition that the equilibrium concentrations of A and B are equal reveals that $x = 0.5a_0$.

Using this value for $x$, the specific equilibrium concentrations are $[A] = 0.5a_0$, $[B] = 0.5a_0$, $[C] = a_0$, and $[D] = 0.5a_0$. The equilibrium constant $K$ is determined by the law of mass action through the expression $K = \frac{[C]^2[D]}{[A][B]^2}$. Inserting these equilibrium concentrations into the formula results in $K = \frac{(a_0)^2(0.5a_0)}{(0.5a_0)(0.5a_0)^2}$. Simplifying this calculation, where the numerator $0.5a_0^3$ is divided by the denominator $0.125a_0^3$, yields a final value of 4.

The relationship between standard Gibbs free energy change and the equilibrium constant is governed by the equation $\Delta G^{\circ} = -RT\ln K$, where $R$ is the universal gas constant and $T$ is the absolute temperature. Because the term $-RT$ is consistently negative, the sign of $\Delta G^{\circ}$ depends entirely on the natural logarithm of $K$, which must be positive when $K > 1$, zero when $K = 1$, and negative when $K < 1$. Consequently, if $K < 1$, the value of $\ln K$ is negative, which forces $\Delta G^{\circ}$ to be positive, making the assertion that $\Delta G^{\circ} < 0$ when $K < 1$ mathematically incorrect.

The equilibrium constant for the dissociation of ammonia, $2\mathrm{NH}_3 \rightleftharpoons \mathrm{N}_2 + 3\mathrm{H}_2$, is the reciprocal of the given synthesis reaction constant, defined as $K_P^{-1} = \frac{P_{\mathrm{N}_2} P_{\mathrm{H}_2}^3}{P_{\mathrm{NH}_3}^2}$. Given the assumption $P_{\mathrm{NH}_3} \ll P$, where $P$ is the total equilibrium pressure, the dissociation is effectively complete, resulting in a stoichiometric ratio of $1:3$ for the nitrogen and hydrogen products. This implies that their partial pressures are $P_{\mathrm{N}_2} = \frac{1}{4}P$ and $P_{\mathrm{H}_2} = \frac{3}{4}P$.

Substituting these partial pressures into the dissociation equilibrium expression yields $K_P^{-1} = \frac{(\frac{1}{4}P)(\frac{3}{4}P)^3}{P_{\mathrm{NH}_3}^2} = \frac{(\frac{1}{4}P)(\frac{27}{64}P^3)}{P_{\mathrm{NH}_3}^2} = \frac{27 P^4}{256 P_{\mathrm{NH}_3}^2}$. Solving this equation for the partial pressure of ammonia results in $P_{\mathrm{NH}_3}^2 = \frac{27 P^4 K_P}{256}$. Taking the square root of both sides gives $P_{\mathrm{NH}_3} = \sqrt{\frac{27}{256}} P^2 K_P^{1/2} = \frac{3\sqrt{3} P^2 K_P^{1/2}}{16}$, which is equivalent to $\frac{3^{3/2} K_p^{1/2} P^2}{16}$.

The mathematical relationship between the equilibrium constants $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$, where $\Delta n_g$ represents the difference between the total number of moles of gaseous products and the total number of moles of gaseous reactants. Because the temperature and gas constant are non-zero, $K_p$ will only deviate from $K_c$ when $\Delta n_g$ is non-zero, a condition which requires the stoichiometry of gaseous species to be unequal on either side of the equilibrium. In the case involving solid carbon, one must exclude the solid phase when calculating the number of moles, which creates a disparity between the gaseous reactants and products that satisfies the condition for the equilibrium constants to be unequal.

The relationship between the equilibrium constant expressed in terms of partial pressure ($K_p$) and concentration ($K_c$) is derived from the ideal gas law and is governed by the equation $K_p = K_c(RT)^{\Delta n_g}$, which simplifies to the ratio $\frac{K_p}{K_c} = (RT)^{\Delta n_g}$. In this expression, $\Delta n_g$ represents the change in the number of moles of gaseous species, calculated by subtracting the sum of stoichiometric coefficients of gaseous reactants from those of the gaseous products.

For the reaction $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$, the value of $\Delta n_g$ is $2 - 2 = 0$, making the ratio $(RT)^0$, which equals 1. In the equilibrium $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$, the change in moles is $2 - 1 = 1$, resulting in a ratio of $(RT)^1$, or 24.62 dm^3 atm mol^-1. For the final reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, the difference in moles is $2 - 4 = -2$, so the ratio is $(RT)^{-2}$, or 1 divided by the square of 24.62, which yields approximately 1.65 × 10^-3 dm^-6 atm^-2 mol^2.

When multiple solids dissociate simultaneously, the gases generated must satisfy their respective equilibrium expressions while sharing the same space, which means the partial pressure of any common gas is determined by the contribution of both reactions. Defining the partial pressure of gas $B$ as $P_1$, the partial pressure of gas $E$ as $P_2$, and recognizing that gas $C$ is common to both reactions, its total partial pressure is $P_C = P_1 + P_2$. The equilibrium constants are established as $x = P_1(P_1 + P_2)$ and $y = P_2(P_1 + P_2)$. Summing these two expressions allows us to derive the relationship $x + y = (P_1 + P_2)(P_1 + P_2)$, which simplifies to $(P_1 + P_2)^2 = x + y$. The total pressure $P_T$ is the sum of all individual partial pressures, calculated as $P_T = P_B + P_C + P_E$, which reduces to $P_1 + (P_1 + P_2) + P_2$ or $2(P_1 + P_2)$. Substituting the square root of our summed equilibrium constants into this expression, the total pressure is $2\sqrt{x + y}$ atm.

The equilibrium constant $K_{c}$ indicates the extent to which a chemical reaction proceeds at equilibrium, but it provides no information regarding the reaction rate or the speed at which that equilibrium is achieved. A large value of $K_{c}$ confirms that the products are thermodynamically favored, yet the actual reaction rate is determined by kinetic factors such as activation energy, which a catalyst manages independently of the equilibrium position. Therefore, claiming that a large equilibrium constant removes the need for a catalyst is flawed, as it incorrectly conflates thermodynamic feasibility with the kinetic necessity of increasing the reaction speed.

According to Le Chatelier's principle, an increase in the volume of the reaction vessel decreases the system pressure, prompting the equilibrium to shift toward the side with a greater number of moles of gaseous species to counteract the pressure drop. In the reaction $2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)$, the product side contains 3 moles of gas while the reactant side contains 2 moles, resulting in a positive change in gaseous moles where $\Delta n_g = 3 - 2 = 1$. Because the number of gaseous moles is higher on the product side, an increase in volume shifts the equilibrium to favor the forward reaction.

Le Chatelier's principle dictates that a system at equilibrium shifts to counteract external changes. The decomposition reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ involves an increase in the total number of gaseous moles, making it favored by a reduction in effective system pressure. Adding an inert gas at constant pressure increases the total volume of the reaction vessel, which lowers the partial pressures of the reactants and products and shifts the equilibrium toward the side with more moles of gas to compensate for the decrease in pressure.

The equilibrium constant for the reaction $CO + Cl_2 \rightleftharpoons COCl_2$ is determined by the expression $K_c = \frac{[COCl_2]}{[CO][Cl_2]}$. With initial amounts of 2 moles of $CO$ and 3 moles of $Cl_2$ in 5 L, the reduction of $CO$ to 1 mole at equilibrium implies that 1 mole of $CO$ and 1 mole of $Cl_2$ are consumed to produce 1 mole of $COCl_2$. The resulting equilibrium amounts are 1 mole of $CO$, 2 moles of $Cl_2$, and 1 mole of $COCl_2$, which translate to equilibrium concentrations of 0.2 M, 0.4 M, and 0.2 M respectively. Substituting these concentrations into the equilibrium expression yields $K_c = \frac{0.2}{0.2 \times 0.4} = 2.5$.