Equilibrium – NEET UG ChemistryPractice Questions & PYQs

Equilibrium constants are determined by the specific stoichiometry of a reaction, meaning they change in predictable ways when reactions are reversed or their coefficients are multiplied. For the first equilibrium, the constant is defined as $K_1 = \frac{p_{\mathrm{NO}_2}}{p_{\mathrm{NO}} \cdot (p_{\mathrm{O}_2})^{1/2}}$. By examining the expression for $K_2 = \frac{(p_{\mathrm{NO}})^2 \cdot p_{\mathrm{O}_2}}{(p_{\mathrm{NO}_2})^2}$, we can identify that this equation represents the inverse of the first reaction after its coefficients have been doubled. Calculating the square root of $K_2$ simplifies the expression to $\sqrt{K_2} = \frac{p_{\mathrm{NO}} \cdot (p_{\mathrm{O}_2})^{1/2}}{p_{\mathrm{NO}_2}}$, which is equivalent to $\frac{1}{K_1}$, demonstrating that the relationship is $K_2 = \frac{1}{K_1^2}$.

For a weak base, the concentration of hydroxyl ions at equilibrium is determined by the relationship $[\mathrm{OH}^{-}] = \sqrt{K_b C}$, where $K_b$ is the dissociation constant and $C$ is the initial molar concentration. Substituting the values $K_b = 1.0 \times 10^{-12}$ and $C = 0.01$ M into the equation provides $[\mathrm{OH}^{-}] = \sqrt{1.0 \times 10^{-12} \times 0.01}$. Evaluating the expression inside the radical leads to $[\mathrm{OH}^{-}] = \sqrt{1.0 \times 10^{-14}}$, which results in a concentration of $1.0 \times 10^{-7}$ mol L$^{-1}$.

This behavior is governed by the common ion effect, which suppresses the dissociation of the weak electrolyte hydrogen sulphide in the presence of strong acid. When dilute HCl is added, the resulting high concentration of $H^+$ ions shifts the equilibrium $H_2S \rightleftharpoons 2H^+ + S^{2-}$ to the left, drastically reducing the availability of free $S^{2-}$ ions in the solution. Since group II cations form sulphides with very low solubility products, they precipitate at this low sulphide concentration, while group IV cations, which have higher solubility products, remain unaffected and stay in the solution.

The equilibrium of a sparingly soluble salt like $AX_2$ depends on the ratio of its constituent ions in a saturated solution. Upon dissolution, the salt releases one cation $A^{2+}$ and two anions $X^-$ for every unit of the compound, which establishes a stoichiometric relationship where the concentration of the cation is $s$ and the concentration of the anion is $2s$, with $s$ representing the molar solubility.

The solubility product constant $K_{sp}$ is defined by the expression $K_{sp} =$[A^{2+}]$$[X^-]^2$. Substituting the equilibrium concentrations into this formula gives $K_{sp} = (s)(2s)^2$, which simplifies to $4s^3$. By setting $4s^3$ equal to $3.2 × 10^{-11}$, the equation $s^3 = 0.8 × 10^{-11}$ or $8 × 10^{-12}$ is obtained. Taking the cube root of this result yields $s = 2 × 10^{-4}$ moles/L.

An acid-base indicator, symbolized as $HIn$, acts as a weak acid that dissociates in aqueous solution according to the equilibrium $HIn \rightleftharpoons H^+ + In^-$. The equilibrium constant expression for this dissociation is defined as
K_{In} = \frac{[H^+]}$\$[In^-]$}{[HIn]}
which relates the concentration of hydrogen ions to the ratio of the indicator's conjugate base and acid forms. By rearranging this expression to solve for $[H^+]$ and taking the negative logarithm of both sides, we derive the relationship pH = pK_{In} + \log \frac{[In^-]}}{[HIn]}. Rearranging this final equation to isolate the logarithmic term results in the expression $$\log \frac{[In^-]}}{[HIn]} = pH - pK_{In}$$
which dictates how the indicator's ratio of forms shifts with the acidity of the solution.

The reaction between dihydrogen and dinitrogen is defined by the stoichiometric equation $3\text{H}_2 + \text{N}_2 \rightarrow 2\text{NH}_3$. According to this equation, every 3 volumes of $\text{H}_2$ require 1 volume of $\text{N}_2$ to produce 2 volumes of $\text{NH}_3$. Given 30 litres of both reactants, hydrogen acts as the limiting reagent because it would be completely consumed first if the reaction went to completion. Theoretically, 30 litres of hydrogen should yield 20 litres of $\text{NH}_3$, but since the efficiency is only 50%, the actual production is reduced to 10 litres of $\text{NH}_3$.

To produce 10 litres of $\text{NH}_3$, the stoichiometry dictates that half of the theoretical requirement is consumed. This means 15 litres of $\text{H}_2$ and 5 litres of $\text{N}_2$ are utilized in the process. Subtracting these consumed amounts from the initial volumes leaves 15 litres of $\text{H}_2$ and 25 litres of $\text{N}_2$ remaining in the container, alongside the 10 litres of $\text{NH}_3$ formed.

The strength of hydrohalic acids depends on the ease with which the hydrogen-halogen bond can dissociate to release a proton ($H^+$) in solution. Because of the exceptionally high bond energy of the hydrogen-fluorine bond, the proton is held tightly and is not released easily, which makes HF the weakest acid rather than the strongest. Since bond dissociation energy decreases as you move down the group, the hydrogen-chlorine bond in HCl is significantly easier to break, making HCl a stronger acid than HF.

To find the equilibrium constant for the target reaction, we treat the given equations as algebraic components where reversing a reaction requires taking the reciprocal of its equilibrium constant, and multiplying a reaction by a coefficient requires raising the equilibrium constant to that power. We begin by reversing the first reaction to place $2NH_3$ on the reactant side, providing a modified equilibrium constant of $\frac{1}{K_1}$, and we retain the second reaction as is to ensure $2NO$ appears on the product side with its constant $K_2$. To obtain $3H_2O$ on the product side, we multiply the third reaction by a factor of $3$, which results in a modified equilibrium constant of $K_3^3$. When these three equations are summed, the intermediate species $N_2$ and $3H_2$ cancel out on both sides, yielding the target stoichiometry and an overall equilibrium constant of $\frac{K_2 K_3^3}{K_1}$.

The direction of a reversible reaction such as $\mathrm{N}_{2(\mathrm{g})}+3 \mathrm{H}_{2(\mathrm{g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{g})}$ is determined by comparing the reaction quotient $Q$ with the equilibrium constant $K_{\mathrm{c}}$. A state where $Q > K_{\mathrm{c}}$ implies that the current concentration of products is higher than the ratio required for equilibrium conditions. To restore balance, the reaction shifts in the reverse direction, moving from right to left to convert excess products back into reactants until the ratio drops to equal $K_{\mathrm{c}}$.

This problem illustrates the common ion effect, where the presence of a common ion decreases the solubility of a sparingly soluble salt because the equilibrium shifts toward the solid phase. For the dissolution of silver iodide, the equilibrium expression $AgI(s) \rightleftharpoons Ag^+(aq) + I^-(aq)$ is governed by the solubility product constant K_{sp} = \[Ag^+]\[I^-]$= 1.0 × 10^{-16}$. In a $10^{-4} \text{ mol L}^{-1}$ solution of potassium iodide, the concentration of iodide ions is $10^{-4} \text{ mol L}^{-1}$, so the expression becomes $1.0 × 10^{-16} = [Ag^+](10^{-4})$. Solving this gives $[Ag^+] = 1.0 × 10^{-12} \text{ mol L}^{-1}$, which represents the solubility of silver iodide in this environment.

For the reaction $\text{BaO}_{2(s)} \rightleftharpoons \text{BaO}_{(s)} + \text{O}_{2(g)}$, the equilibrium constant expression involves only the gaseous component because the activity of pure solids remains constant throughout the process. Consequently, the partial pressure of $\text{O}_2$ at equilibrium is independent of the masses of the solid reactants and products. Since the reaction is endothermic, applying Le Chatelier's Principle dictates that increasing the temperature shifts the equilibrium toward the product side to absorb the additional heat, thereby increasing the dissociation of $\text{BaO}_2$ and raising the partial pressure of $\text{O}_2$.

The solubility product of an electrolyte represents the equilibrium established between its solid form and its dissolved ions in a saturated solution. For an $MX_{2}$ type electrolyte, the dissociation process follows the equation $MX_{2} \rightleftharpoons M^{2+} + 2X^{-}$, where the concentration of the cation $M^{2+}$ is equal to the solubility $s$, and the concentration of the anion $X^{-}$ is twice the solubility, $2s$. The solubility product constant is determined by the expression $K_{sp} =$[M^{2+}]$$[X^{-}]^{2}$, which simplifies to $K_{sp} = (s)(2s)^{2}$, or $K_{sp} = 4s^{3}$. By substituting the solubility $s = 0.5 × 10^{-4}$ mole/lit into this expression, the calculation becomes $K_{sp} = 4 × (0.5 × 10^{-4})^{3}$, resulting \in $K_{sp} = 5 × 10^{-13}$.

The alkalinity of a salt solution is determined by the hydrolysis of its constituent ions, where salts derived from a strong base and a weak acid yield the highest pH levels. Among the choices, $\text{NH}_4\text{Cl}$ creates an acidic solution, $\text{NaNO}_3$ results in a neutral solution with a pH of 7, and $\text{CH}_3\text{COOK}$ forms a weakly alkaline solution with a pH of approximately 8.8. In contrast, $\text{Na}_2\text{CO}_3$ undergoes hydrolysis to yield a strongly alkaline environment with a pH greater than 10, as represented by the chemical equation $\text{Na}_2\text{CO}_3 + 2\text{H}_2\text{O} \rightarrow 2(\text{Na}^+ + \text{OH}^-) + \text{H}_2\text{CO}_3$

The mixture of ammonium hydroxide and ammonium chloride functions as a basic buffer. For such systems, the relationship between the acidity of the solution and the dissociation constant of the base is governed by the Henderson-Hasselbalch equation expressed as:
$\mathrm{pH} = 14 - \mathrm{p}K_b - \log \frac{[\text{salt}]}{[\text{base}]}$
Because the normality of the salt and the base are identical, the ratio of their concentrations is unity, making the logarithmic term equal to zero. With the $\mathrm{pH}$ provided as 9.25, the calculation simplifies to $\mathrm{p}K_b = 14 - 9.25$, leading directly to a value of 4.75.

The solubility product constant, $K_{sp}$, quantifies the equilibrium of a sparingly soluble salt by the product of its dissociated ion concentrations raised to the power of their stoichiometric coefficients. For the specific dissociation $M_{2}S \rightleftharpoons 2M^{+} + S^{2-}$, the ion concentrations are proportional to the solubility $s$, resulting in $[M^{+}] = 2s$ and $[S^{2-}] = s$.

Substituting these values into the equilibrium expression gives $K_{sp} =$[M^{+}]^{2} $[S^{2-}]$= (2s)^{2} (s) = 4s^{3}$. Applying the known solubility of $3.5 × 10^{-6}$ into this equation results \in $4 × (3.5 × 10^{-6})^{3}$, which simplifies to $1.7 × 10^{-16}$.

The ionization constant for a weak acid provides a mathematical relationship between the concentrations of dissociated ions and the remaining undissociated acid molecules at chemical equilibrium. For the dissociation reaction $\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-} + \mathrm{H}^{+}$, this constant is defined by the expression:

K_{\text{ion}} = \frac{[\mathrm{CH}_{3} \mathrm{COO}^{-}]}$ \$[\mathrm{H}^{+}]$}{[\mathrm{CH}_{3} \mathrm{COOH}]}

Because acetic acid dissociates in a one-to-one ratio, the concentration of acetate ions is equivalent to the hydrogen ion concentration, which is $[\mathrm{H}^{+}] = 3.4 \times 10^{-4}$. Substituting these values into the equilibrium equation allows us to isolate the initial concentration of the acid:

$[\mathrm{CH}_{3} \mathrm{COOH}] = \frac{(3.4 \times 10^{-4}) \times (3.4 \times 10^{-4})}{1.7 \times 10^{-5}}$

Solving this calculation leads to a final concentration of $[\mathrm{CH}_{3} \mathrm{COOH}] = 6.8 \times 10^{-3}$.

Basicity, or the proton-accepting tendency, is dictated by the ability to donate a lone pair to a proton, a capacity that intensifies in amines like $\mathrm{R}-\mathrm{NH}_{2}$ due to the electron-releasing inductive effect of the alkyl group compared to the un-substituted $\mathrm{NH}_{3}$. For anionic species, basicity diminishes as ionic size increases because the larger ions form weaker associations with the proton, ranking the small nucleophile $\mathrm{HS}^{-}$ as more basic than the very large and weakly interacting $\mathrm{I}^{-}$. This hierarchy leads to an overall proton-accepting tendency of $\mathrm{R}-\mathrm{NH}_{2} > \mathrm{NH}_{3} > \mathrm{HS}^{-} > \mathrm{I}^{-}$.

The equilibrium constant for a chemical reaction is defined as the ratio of the product concentrations to the reactant concentrations raised to their stoichiometric powers at equilibrium. When a reaction is written in the reverse direction, the former products become the reactants and the former reactants become the products, which effectively inverts the equilibrium constant expression. By the principles of chemical equilibrium, the constant for this reverse process is simply the reciprocal of the original constant, resulting in an equilibrium constant of $1/K$.

The equilibrium constant, denoted as $K_{eq}$, characterizes a specific chemical reaction and depends solely on temperature rather than the initial amounts or concentrations of the participating substances. For a reversible reaction represented by $A+B \rightleftharpoons C+D$, the constant is defined by the ratio $K_{eq} = \frac{[C][D]}{[A][B]}$. While increasing the concentration of reactants shifts the equilibrium position to favor the formation of more products, the system adjusts in such a way that the ratio of product concentrations to reactant concentrations remains unchanged, leaving $K_{eq}$ unaffected.

In chemical equilibrium, the expression for the equilibrium constant $K_p$ is derived from the partial pressures of gaseous species, while pure solids are assigned an activity of unity because they do not contribute to the pressure of the system. For the reaction $\mathrm{MgCO}_{3(s)} \rightleftharpoons \mathrm{MgO}_{(s)} + \mathrm{CO}_{2(g)}$, the standard equilibrium expression would involve the ratio of the products to reactants, but the solids are excluded from the calculation. Since $\mathrm{MgCO}_{3}$ and $\mathrm{MgO}$ are solids, their contribution is treated as 1, leaving only the partial pressure of carbon dioxide in the final expression, which simplifies to $K_p = \mathrm{P}_{\mathrm{CO}_{2}}$.