Equilibrium – NEET UG ChemistryPractice Questions & PYQs

The equilibrium constant of a chemical reaction is strictly determined by its stoichiometric coefficients, such that multiplying a balanced equation by a factor $n$ results in the original equilibrium constant being raised to the power of $n$. In this scenario, the second reaction is obtained by multiplying the coefficients of the first reaction by a factor of $\frac{1}{2}$. Applying this rule, the new equilibrium constant is calculated by raising the initial constant $K$ to the power of $\frac{1}{2}$, resulting in $K^{\frac{1}{2}}$.

The relationship between the standard Gibbs free energy change and the solubility product constant is defined by the thermodynamic equation $\Delta G^\circ = -2.303RT \log K_{sp}$. Because the Gibbs free energy for the reaction is 63.3 kJ, it must be converted to 63300 J to ensure consistency with the gas constant $R$ of 8.314 J / (K mol) and the temperature $T$ of 298 K.

Substituting these values into the formula produces the equation:
$63300 = -2.303 \times 8.314 \times 298 \times \log K_{sp}$
Simplifying the product of the constants on the right side results in:
$63300 = -5708 \times \log K_{sp}$
Dividing 63300 by -5708 gives a logarithm of approximately -11.09, and calculating the antilogarithm of this value yields $K_{sp} = 8.0 \times 10^{-12}$.

The pH of a salt solution in water is determined by the strength of the parent acid and base from which the salt is derived. Salts such as $\mathrm{KCl}$ and $\mathrm{NaCl}$ are formed from a strong acid and a strong base, resulting in a neutral solution with a pH of 7, while $\mathrm{CuSO}_4$, derived from a strong acid and a weak base, creates an acidic solution with a pH less than 7. In contrast, $\mathrm{Na}_2\mathrm{CO}_3$ is a salt of a strong base ($\mathrm{NaOH}$) and a weak acid ($\mathrm{H}_2\mathrm{CO}_3$), which undergoes hydrolysis to release $\mathrm{OH}^-$ ions in water and thereby produces a basic solution with a pH greater than 7.

According to Le Chatelier's Principle, a system at equilibrium adjusts to counteract any imposed external changes. Because the forward reaction $N_2(g)+3H_2(g)\rightleftharpoons2\mathrm{NH}_3(g)$ involves a reduction in the total number of gas moles from four to two, increasing the pressure favors the product side to relieve the stress. Additionally, as this process is exothermic and releases heat, decreasing the temperature encourages the system to shift forward to produce more heat and regain stability.

According to Le Chatelier's principle, treating heat as a product in an exothermic reaction implies that increasing the temperature drives the system toward the reactants, thereby reducing the equilibrium constant. Because the reaction is exothermic, the equilibrium constant at the higher temperature $T_2$, represented as $K_P'$, must be smaller than the value $K_P$ measured at the lower temperature $T_1$. This inverse relationship between temperature and the equilibrium constant ensures that $K_P > K_P'$.

The acidity or basicity of a solution is fundamentally defined by the auto-ionization of water, represented by the equilibrium constant $K_w =$[H^+]$[OH^-]$. Because pure water is always neutral, the concentration of hydrogen ions must equal the concentration of hydroxide ions, which implies$[H^+] = \sqrt{K_w}$. As the temperature increases, the extent of water dissociation rises, shifting$K_w$ from its standard value at 25 degrees Celsius to a higher value.

Given that $K_w$ at 100 degrees Celsius is 55 times its value at 25 degrees Celsius, where $K_w$ is $1 \times 10^{-14}$, the new equilibrium constant becomes $55 \times 10^{-14}$. The concentration of hydrogen ions is obtained by taking the square root, resulting in $[H^+] = \sqrt{55 \times 10^{-14}}$. Applying the definition $\text{pH} = -\log [H^+]$ leads to the expression $\text{pH} = -\frac{1}{2} \log (55 \times 10^{-14})$. Expanding this logarithm into $\text{pH} = -\frac{1}{2} (\log 55 + \log 10^{-14})$ and substituting 1.74 for $\log 55$ produces the calculation $-\frac{1}{2} (1.74 - 14)$, which results in a final pH value of 6.13.

The dissociation constant for a weak acid defines the equilibrium established when the acid partially ionizes in solution. For a weak electrolyte like lactic acid, we apply Ostwald's dilution law, $K_a = C \alpha^2$, to link the molar concentration $C$ and the degree of dissociation $\alpha$. First, express the $3.7 \%$ dissociation as the decimal $\alpha = 0.037$. Substituting this and the $0.10 \ \mathrm{M}$ concentration into the equation gives $K_a = 0.10 \times (0.037)^2$, which calculates to approximately $1.4 \times 10^{-4}$.

According to the principles of chemical equilibrium, removing a product from a reaction system drives the equilibrium in the forward direction. To drive the completion of $3MnO_4^{2-} + 2H_2O \rightleftharpoons 2MnO_4^{-} + MnO_2 + 4OH^{-}$, we must neutralize the $OH^-$ ions produced. While $HCl$ and $SO_2$ are acidic, they are also strong reducing agents that would degrade the $MnO_4^-$ product, making $CO_2$ the better choice as it provides the necessary acidic medium to neutralize $OH^-$ without acting as a reducing agent.

A Lewis base acts as an electron pair donor, while a Lewis acid functions as an electron pair acceptor because it is electron-deficient. In the case of $BF_3$, the central boron atom possesses an incomplete octet with only six valence electrons, which forces it to accept electron pairs rather than donate them. Consequently, unlike $PF_3$, $CO$, and $F^-$, which all possess lone pairs available for donation, $BF_3$ functions as a Lewis acid and is therefore the least likely to act as a Lewis base.

When two sparingly soluble salts sharing a common ion exist together in a saturated aqueous solution, the total concentration of that common ion is determined by the combined solubility equilibria of both substances. Let the total concentration of the calcium ion be represented by $S$. Since both calcium carbonate and calcium oxalate dissociate into one calcium ion per formula unit, the solubility product expressions for each salt effectively involve the total calcium ion concentration.

Specifically, the equilibrium for calcium carbonate satisfies the relation $K_{sp(\mathrm{CaCO}_{3})} =$[\mathrm{Ca}^{2+}]$[\mathrm{CO}{3}^{2-}]$, while the equilibrium for calcium oxalate satisfies$K{sp(\mathrm{CaC}{2} \mathrm{O}{4})} = [\mathrm{Ca}^{2+}]\[\mathrm{C}{2} \mathrm{O}{4}^{2-}]$. Because the calcium ions are shared, we can sum the two solubility product expressions. By factoring out the total calcium ion concentration, we arrive at the relationship$S^2 = K_{sp(\mathrm{CaCO}{3})} + K{sp(\mathrm{CaC}{2} \mathrm{O}{4})}$.

Substituting the given values into this derived equation yields $S^2 = (4.7 \times 10^{-9}) + (1.3 \times 10^{-9})$, which simplifies to $S^2 = 6.0 \times 10^{-9}$. Calculating the square root of $6.0 \times 10^{-9}$, which is equivalent to $\sqrt{60 \times 10^{-10}}$, results in a total calcium ion concentration of $7.746 \times 10^{-5} \mathrm{M}$.

The Henderson-Hasselbalch equation defines the relationship between the pH of a buffer solution, the dissociation constant of the weak acid, and the concentrations of the salt and acid components. This relationship is expressed as $\text{pH} = pK_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$ where pKa is the negative logarithm of the acid dissociation constant, Ka. Since Ka equals $1 × 10^{-4}$, the value of pKa is calculated as 4. Substituting the given pH of 5 and the calculated pKa into the equation yields $5 = 4 + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$ Rearranging this expression to isolate the logarithm term leads to $1 = \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$ Calculating the antilog of both sides provides the ratio of salt to acid, which is 10, resulting in a final ratio of 10:1.

For a reversible chemical reaction at equilibrium, the equilibrium constant is derived from the law of mass action, which establishes a quantitative relationship between the concentrations of reactants and products. This constant is defined as the product of the molar concentrations of the substances on the product side, each raised to the power of its stoichiometric coefficient, divided by the product of the molar concentrations of the reactants, also raised to their respective stoichiometric coefficients.

Applying this to the reaction $\mathrm{A}_2(\mathrm{g}) + \mathrm{B}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{AB}(\mathrm{g})$, the expression for the equilibrium constant is $K_C = \frac{[\mathrm{AB}]^2}{[\mathrm{A}_2][\mathrm{B}_2]}$. Substituting the provided values into this equation gives $K_C = \frac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3}) \times (4.2 \times 10^{-3})}$. Because the powers of ten in the numerator and denominator cancel out, the expression simplifies to calculating $\frac{2.8 \times 2.8}{3.0 \times 4.2}$, resulting in a final value of 0.62.

When an equilibrium constant is given for a reaction, modifying the equation fundamentally alters that value according to the specific operations performed on the stoichiometric coefficients. Reversing the direction of the reaction inverts the equilibrium constant, while dividing all coefficients by a constant factor requires raising the equilibrium constant to the power of that factor. The initial reaction is $\mathrm{2SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\rightleftharpoons\mathrm{2SO}_{3}(\mathrm{g})$ with an equilibrium constant of 278, which becomes $\frac{1}{278}$ after reversing the reaction to $\mathrm{2SO}_{3}(\mathrm{g})\rightleftharpoons\mathrm{2SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$. Dividing the coefficients by 2 to reach $\mathrm{SO}_{3}(\mathrm{g})\rightleftharpoons\mathrm{SO}_{2}(\mathrm{g})+\frac{1}{2}\mathrm{O}_{2}(\mathrm{g})$ requires taking the square root of this value, calculated as $\sqrt{\frac{1}{278}}$, which equals approximately 0.0599 or $6.0 \times 10^{-2}$.

Buffer solutions consist of a weak acid and its conjugate base, or a weak base and its conjugate acid, which react with any added strong acid or base to form unionized, weakly dissociated species. By converting added $H^+$ or $OH^-$ ions into these poorly dissociated forms, the buffer prevents significant changes in the concentration of free ions in the solution. This mechanism ensures that the pH remains essentially constant despite the addition of external acidic or basic reagents.

The pH of an equimolar salt solution is primarily determined by the strength of the parent base involved in its reaction with $HCl$. These salts are derived from the hydroxides $Ba(OH)_2$, $Al(OH)_3$, $LiOH$, and $Be(OH)_2$, where the resulting pH varies according to the basicity of the parent metal hydroxide. Because $Ba(OH)_2$ is the strongest base among the given options, the $BaCl_2$ solution remains the least acidic, thus recording the highest pH value.

The Henderson-Hasselbalch equation defines the relationship between the alkalinity of a basic buffer solution and the concentrations of its component species. Calculating $pK_b$ by finding $-\log(1.8 \times 10^{-5})$ provides the constant value of $4.74$. Integrating the concentrations $[\mathrm{NH}_4^+] = 0.20 \, \mathrm{M}$ and $[\mathrm{NH}_3] = 0.30 \, \mathrm{M}$ into the expression pOH = pK_b + \log \left( \frac{[\mathrm{NH}_4^+]}}{[\mathrm{NH}_3]} \right)$results \in$pOH = 4.74 + \log \left( \frac{0.20}{0.30} \right)$, which simplifies to$4.56$. Final acidity is obtained by subtracting this$pOH$from$14$, resulting \in a$pH$of$9.43$.

According to Le Chatelier's Principle, the direction in which a chemical system shifts to counteract an external stress depends on the heat of reaction and the change in gaseous moles. Since the reaction is exothermic with $\Delta H < 0$, lowering the temperature encourages the system to move forward to generate heat. Because the formation of $\mathrm{XY}_{4}$ results in a net decrease in gaseous volume-from 5 moles of reactants to 2 moles of products-increasing the pressure compels the reaction to shift toward the side with fewer gas molecules, where $\Delta n_g < 0$, to effectively alleviate the imposed stress.

When chemical reactions are combined, their equilibrium constants can be determined using algebraic rules: adding reactions causes their constants to multiply, reversing a reaction results in the reciprocal of its constant, and applying a stoichiometric coefficient requires raising the constant to that power. To determine the constant for the target reaction, we first add the given processes $\mathrm{N}_2 + \mathrm{O}_2 \rightleftharpoons 2\mathrm{NO}$ with constant $K_1$ and $2\mathrm{NO} + \mathrm{O}_2 \rightleftharpoons 2\mathrm{NO}_2$ with constant $K_2$. This addition cancels the intermediate $\mathrm{NO}$ species, resulting in the reaction $\mathrm{N}_2 + 2\mathrm{O}_2 \rightleftharpoons 2\mathrm{NO}_2$ with a combined constant of $K_1K_2$.

The target reaction, $\mathrm{NO}_2 \rightleftharpoons \frac{1}{2}\mathrm{N}_2 + \mathrm{O}_2$, is the reverse of this combined equation divided by two. First, we invert the combined constant to get $\frac{1}{K_1K_2}$ for the reaction $2\mathrm{NO}_2 \rightleftharpoons \mathrm{N}_2 + 2\mathrm{O}_2$. Finally, reducing the stoichiometric coefficients by half necessitates taking the square root of this value, which yields $\left[\frac{1}{K_1K_2}\right]^{1/2}$.

A Lewis base is characterized by its ability to donate a lone pair of electrons, whereas a Lewis acid functions as an electron-pair acceptor. Molecules such as $H_2O$, $NH_3$, and $OH^-$ possess lone pairs of electrons, which allow them to effectively donate these pairs and serve as Lewis bases. Because $BF_3$ is an electron-deficient species with an incomplete octet on the central boron atom, it functions as an electron acceptor rather than a donor, making it the least likely substance in the group to behave as a Lewis base.

The separation of metal ions in group I relies on the principle of solubility product equilibrium, where the product of the ion concentrations in a saturated solution remains constant at a given temperature. When hydrochloric acid is introduced, the common ion effect from the chloride ions shifts the equilibrium toward the formation of solid precipitates, thereby significantly reducing the concentration of dissolved metal ions. The equilibrium for silver ions is governed by the expression $K_{\text{sp}} =$[\text{Ag}^+]$[\text{Cl}^-]$, while the equilibrium for lead ions follows$K_{\text{sp}} = [\text{Pb}^{2+}]\$$[\text{Cl}^-]^2.

With the chloride ion concentration established at $0.10 \text{ M}$, the remaining equilibrium concentration for silver ions is determined to be $1.8 \times 10^{-11} \text{ M}$. Following the same equilibrium principles, the concentration of lead ions in the presence of the established chloride concentration is $1.7 \times 10^{-4} \text{ M}$. These values represent the point at which the dissolved metal ion concentrations are at equilibrium with their respective solid chloride salts under the specified conditions.