Equilibrium – NEET UG ChemistryPractice Questions & PYQs

The equilibrium constant for a reversible reaction is fundamentally defined as the ratio of the rate constant for the forward reaction to the rate constant for the backward reaction. Given that the backward reaction rate constant is 2500 times greater than the forward rate constant, the concentration-based equilibrium constant is calculated as $K_C = \frac{k_f}{k_b} = \frac{1}{2500}$.

To find the pressure-based equilibrium constant $K_P$, we relate it to $K_C$ using the expression $K_P = K_C(RT)^{\Delta n_g}$, where $\Delta n_g$ represents the difference between the moles of gaseous products and gaseous reactants. For the reaction $\mathrm{A}(g) \rightleftharpoons 2 \mathrm{B}(g)$, the change in the number of gaseous moles is $\Delta n_g = 2 - 1 = 1$. Substituting the given values $K_C = \frac{1}{2500}$, $R = 0.0831 \, \text{L atm} \, \text{mol}^{-1} \, \text{K}^{-1}$, and $T = 1000 \, \text{K}$ into this relationship yields $K_P = \frac{1}{2500} \times (0.0831 \times 1000)^1 = \frac{83.1}{2500}$, which simplifies to 0.033.

Calculating the degree of dissociation for a weak acid requires comparing its observed molar conductivity at a specific concentration to its theoretical molar conductivity at infinite dilution. Kohlrausch's law provides the foundation for this by stating that the molar conductivity at infinite dilution is the sum of the contributions of the individual ions, expressed as $\Lambda_m^{\circ} = \Lambda_{+}^{\circ} + \Lambda_{-}^{\circ}$.

Adding the ionic contributions of 349.6 and 50.4 results in an infinite dilution value of 400 S cm² mol⁻¹. The degree of dissociation, denoted as $\alpha$, follows the relationship $\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$. Substituting the measured molar conductivity of 90 S cm² mol⁻¹ into this expression means dividing 90 by 400, which results in 0.225.

The equilibrium constants $K_p$ and $K_c$ are related by the expression
$K_p = K_c(RT)^{\Delta n_g}$
where $\Delta n_g$ is the change in the number of gaseous moles, defined as the difference between the total moles of gaseous products and gaseous reactants. When $\Delta n_g$ is zero, the term $(RT)^{\Delta n_g}$ becomes one, making $K_p$ equal to $K_c$. For the decomposition reaction $\mathrm{PCl}_{5(\mathrm{g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})}$, the calculation yields
$\Delta n_g = 2 - 1 = 1$
which confirms that $K_p$ and $K_c$ are not equal. In all other provided reactions, there are two moles of gaseous reactants and two moles of gaseous products, resulting in $\Delta n_g = 0$.

When analyzing equilibrium constants, modifying a chemical equation changes its associated constant in predictable ways. Reversing a reaction transforms the equilibrium constant into its reciprocal, while multiplying the entire equation by a specific factor requires raising the constant to the power of that factor.

Starting with the reaction $3 A_2 + B_2 \rightleftharpoons 2 A_3 B$ with constant $K_1$, reversing it yields $2 A_3 B \rightleftharpoons 3 A_2 + B_2$ with a constant of $\frac{1}{K_1}$. To reach the second given equilibrium of $A_3 B \rightleftharpoons \frac{3}{2} A_2 + \frac{1}{2} B_2$, every coefficient in this reversed reaction must be multiplied by $\frac{1}{2}$. Applying this factor means raising the constant to the power of $\frac{1}{2}$, which results in $K_2 = (\frac{1}{K_1})^{1/2}$, simplifying to $K_2 = \frac{1}{\sqrt{K_1}}$.

Solubility is governed by the solubility product constant, $K_{sp}$, which represents the equilibrium between the solid salt and its dissolved ions. In pure water, the dissolution of $\text{AgCl}$ follows the equilibrium $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$, leading to the expression $K_{sp} =$[\text{Ag}^+]$[\text{Cl}^-]$. Since the molar concentration of both silver and chloride ions must be equal to the solubility$s$, the equation becomes$K_{sp} = s^2$, so$10^{-10} = s^2$, which indicates that the solubility in water is$10^{-5}$ M.

When dissolved in a $0.1$ M potassium chloride solution, the common ion effect significantly alters the solubility. Because potassium chloride is a strong electrolyte, the chloride ion concentration is essentially provided by the salt, making $[\text{Cl}^-] \approx 0.1$ M. If the new solubility is $s'$, the solubility product expression becomes $K_{sp} = s' \times 0.1$, or $10^{-10} = s' \times 0.1$. Solving for $s'$ yields a solubility of $10^{-9}$ M. The ratio of the solubility in the potassium chloride solution to the solubility in water is $10^{-9} / 10^{-5}$, which results in $10^{-4}$.

Le Chatelier's principle states that equilibrium systems respond to environmental stress by shifting in a direction that alleviates that stress, which dictates the specific conditions necessary to drive the exothermic reaction $\mathrm{N}_2(\mathrm{g})+3\mathrm{H}_2(\mathrm{g})\rightleftharpoons2\mathrm{NH}_3(\mathrm{g})$ forward. Because the reaction releases energy, indicated by $\Delta H = -Q$, cooler temperatures favor the products, while the reduction in gas volume-moving from four moles on the left to two moles on the right-means that higher pressure effectively pushes the system forward. Finally, increasing the concentration of a reactant such as $\mathrm{H}_2$ creates additional stress that the system offsets by shifting production in the forward direction.

Titrations involving a weak acid, such as oxalic acid, and a strong base like sodium hydroxide require an indicator that changes color within the basic pH range to accurately detect the equivalence point. Phenolphthalein is specifically chosen for this reaction because it remains colourless in the acidic environment of the flask but transitions to a distinct pink color immediately as the solution becomes alkaline upon the addition of a slight excess of $NaOH$. This clear visual shift indicates that the neutralization reaction between the $C_2H_2O_4$ and the base has reached completion.

Chemical equilibrium is governed by the principle that the equilibrium constant $K_c$ remains fixed for a specific temperature regardless of the starting concentrations. Based on the provided equilibrium state for the reaction $2NO \rightleftharpoons N_2 + O_2$, the constant is calculated using the expression $K_c = \frac{[N_2][O_2]}{[NO]^2}$, resulting in a value of 1.607.

Starting with 0.1 mol/L of $NO$, the dissociation process involves two moles of $NO$ yielding one mole each of $N_2$ and $O_2$. Expressing the concentrations at equilibrium in terms of the dissociation degree $\alpha$ provides the relations $[NO] = 0.1(1-\alpha)$ and $[N_2] = [O_2] = 0.05\alpha$. Substituting these into the equilibrium expression gives $K_c = \frac{(0.05\alpha)^2}{(0.1(1-\alpha))^2}$, which is set equal to 1.607.

Taking the square root of both sides simplifies the expression to $\frac{0.05\alpha}{0.1(1-\alpha)} = \sqrt{1.607}$. This results in $\frac{\alpha}{2(1-\alpha)}$ equaling approximately 1.27. Rearranging the terms yields $\frac{\alpha}{1-\alpha} = 2.54$, which simplifies to $3.54\alpha = 2.54$. Solving for the degree of dissociation results in 0.717.

The direction in which a reversible chemical reaction proceeds is determined by the relationship between the reaction quotient $Q_c$ and the equilibrium constant $K_c$. By calculating the current ratio of product concentrations to reactant concentrations, one can assess whether the system is at equilibrium or needs to shift to balance the reaction.

For the reaction $2 A \rightleftharpoons B + C$, the reaction quotient is defined by the expression $Q_c = \frac{[B][C]}{[A]^2}$. Substituting the given concentration values of $2 × 10^{-3}$ M for all species into this equation results \in $Q_c = \frac{(2 \times 10^{-3})(2 \times 10^{-3})}{(2 \times 10^{-3})^2}$, which simplifies to $Q_c = 1$. Since this calculated value is greater than the provided equilibrium constant $K_c = 4 × 10^{-3}$, the system currently contains a higher concentration of products relative to reactants than is permissible at equilibrium, compelling the reaction to shift in the backward direction to restore balance.

The dissociation constant for a weak acid is determined by the relation $K_a = C \alpha^2$, where $C$ represents the concentration and $\alpha$ the degree of dissociation. Converting the 1% dissociation to a decimal gives an $\alpha$ value of 0.01. Substituting the concentration of 0.1 mol L^-1 and the dissociation value into the formula results in $K_a = 0.1 \times (0.01)^2$. Squaring 0.01 provides 0.0001, which leads to the final dissociation constant of $K_a = 1 \times 10^{-5}$.

To determine the standard Gibbs free energy change, we utilize the relationship between free energy and the equilibrium constant, expressed as $\Delta G^{\circ} = -RT \ln K_{eq}$. For practical calculations, this is converted to $\Delta G^{\circ} = -2.303 RT \log K_{eq}$ by using the conversion factor between natural and common logarithms.

Begin by calculating the equilibrium constant using the law of mass action, which gives $K_{eq} = \frac{[C][D]}{[A][B]}$. Substituting the equilibrium concentrations of 10, 6, 2, and 3 into this expression results in $K_{eq} = \frac{10 \times 6}{2 \times 3} = 10$. Apply the values $R = 2 \text{ cal/mol K}$ and $T = 300 \text{ K}$ to the equation $\Delta G^{\circ} = -2.303 \times 2 \times 300 \times \log(10)$. Since $\log(10) = 1$, the calculation simplifies to $\Delta G^{\circ} = -2.303 \times 600$, which results in $-1381.8 \text{ cal}$.

An acidic buffer maintains a stable pH by containing a weak acid alongside its conjugate base, which is supplied through the acid's salt with a strong base. When these components are mixed, the weak acid serves to neutralize added hydroxide ions while the conjugate base effectively neutralizes added protons, a relationship governed by the Henderson-Hasselbalch equation $pH = pK_a + \log \frac{[Salt]}{[Acid]}$. This combination prevents drastic changes in acidity, confirming that the essential requirement for forming an acidic buffer is the pairing of a weak acid with its corresponding salt derived from a strong base.

The combination of a weak acid and the salt of its conjugate base forms an acidic buffer, where the pH is defined by the Henderson-Hasselbalch equation:
$\text{pH} = \text{p}K_{a} + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$
Since the mixture contains equal volumes of 0.1 M sodium acetate and 0.01 M acetic acid, the volume components cancel out, and the concentration ratio simplifies to 0.1 divided by 0.01. Substituting the given pKa of 4.57 into the equation results in 4.57 plus log(10), yielding a final value of $\text{pH} = 5.57$.

The law of mass action allows us to relate the equilibrium concentrations of reactants and products through the equilibrium constant, where concentrations are raised to the power of their stoichiometric coefficients. For the transformation of oxygen into ozone, the expression is defined as $K_c = \frac{[\mathrm{O}_{3}]^2}{[\mathrm{O}_{2}]^3}$.

Rearranging this relationship to solve for the concentration of ozone gives $[\mathrm{O}_{3}]^2 = K_c \times [\mathrm{O}_{2}]^3$. Substituting the given equilibrium constant of $3.0 \times 10^{-59}$ and the oxygen concentration of $0.040 \text{ M}$ into the equation, we calculate $[\mathrm{O}_{3}]^2 = 3.0 \times 10^{-59} \times (0.040)^3$, which equals $1.9 \times 10^{-63}$, or alternatively expressed as $19 \times 10^{-64}$. Taking the square root of this value yields an equilibrium concentration of ozone equal to $4.38 \times 10^{-32} \text{ M}$.

The acidity of a solution depends on the concentration of hydrogen ions, which for a weak acid is found by multiplying the initial molar concentration by the degree of ionization. For a 0.01 M acetic acid solution that is 1% ionized, the degree of ionization corresponds to 0.01. The concentration of hydrogen ions is calculated using the equation $[H^+] = 0.01 \times 0.01 = 10^{-4}$ M. Since pH is defined as the negative logarithm of the hydrogen ion concentration, the calculation $\text{pH} = -\log_{10}(10^{-4})$ yields a result of 4.

The relationship between the equilibrium constants $K_p$ and $K_c$ is defined by the expression $K_p = K_c(RT)^{\Delta n_g}$, where $\Delta n_g$ represents the change in the number of moles of gaseous products minus gaseous reactants. For the reaction $\mathrm{CO}_2(\mathrm{g}) + \mathrm{C}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})$, we exclude the solid carbon and find that the net change in gaseous moles is $\Delta n_g = 2 - 1 = 1$. By substituting the provided values into the equation $3.0 = K_c(0.083 \times 1000)^1$, we calculate the constant $K_c$ to be $K_c = 3.6 \times 10^{-2}$.

Dimethylammonium acetate forms from the reaction of a weak acid, acetic acid, with a weak base, dimethylamine, meaning its aqueous pH is determined by the balance between their relative ionization tendencies. The acidity of this specific type of salt is defined by the relationship:
$\text{pH} = 7 + \frac{1}{2}(\text{p}K_a - \text{p}K_b)$
Calculating the difference between the $\text{p}K_a$ of 4.77 and the $\text{p}K_b$ of 3.27 yields 1.50. Dividing this value by 2 results in 0.75, and combining this with the baseline of 7 gives a solution pH of 7.75.

The common ion effect describes the reduction in solubility of a sparingly soluble salt when it dissolves in a solution already containing one of its constituent ions. In this scenario, the dissociation equilibrium $\mathrm{Ni(OH)_2} \rightleftharpoons \mathrm{Ni^{2+}} + 2\mathrm{OH^-}$ is suppressed by the presence of $\mathrm{NaOH}$, which acts as a strong electrolyte and provides a high concentration of hydroxide ions. Since the concentration of $\mathrm{NaOH}$ is 0.1 M, the total hydroxide ion concentration in the solution is approximately 0.1 M, as the additional contribution of ions from the nickel salt is negligible.

Using the provided ionic product value of 2 × 10^-15, the equilibrium expression is $K_{sp} =$[\mathrm{Ni^{2+}}]$$[\mathrm{OH^-}]^2$. Representing the solubility of the nickel hydroxide as $s$, the concentration of nickel ions is $s$ and the concentration of hydroxide ions is 0.1 M. Substituting these values into the equilibrium equation gives $2 × 10^{-15} = s(0.1)^2$. This simplifies to $2 × 10^{-15} = s(0.01)$, which results \in a solubility $s$ of 2 × 10^-13 M.

The relationship between the standard Gibbs free energy change and the equilibrium constant for a reversible reaction is defined by the thermodynamic equation $\Delta_rG^{\theta} = -RT \ln K_c$. In this expression, $R$ represents the universal gas constant, $T$ denotes the absolute temperature in Kelvin, and $K_c$ is the equilibrium constant of the reaction. Substituting the known values into this relationship, where $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$, $T = 300\,\text{K}$, and $K_c = 2\times10^{13}$, yields the expression $\Delta_rG^{\theta} = -8.314\,\text{J mol}^{-1}\text{K}^{-1} \times 300\,\text{K} \times \ln(2\times10^{13})$. This resulting calculation provides the standard Gibbs free energy change for the hydrolysis of sucrose at the specified temperature.

A basic buffer is formed by the combination of a weak base and its conjugate salt, which is precisely the result of partially neutralizing a weak base with a strong acid. When 100 mL of 0.1 M $\mathrm{HCl}$ is mixed with 200 mL of 0.1 M $\mathrm{NH}_4\mathrm{OH}$, the acid reacts completely with a portion of the base according to the following equation:

$\mathrm{HCl} + \mathrm{NH}_4\mathrm{OH} \rightarrow \mathrm{NH}_4\mathrm{Cl} + \mathrm{H}_2\mathrm{O}$

The $\mathrm{HCl}$ provides 10 mmol of substance, while the $\mathrm{NH}_4\mathrm{OH}$ provides 20 mmol. As the limiting reactant, the entire 10 mmol of $\mathrm{HCl}$ is consumed, which results in the formation of 10 mmol of the salt $\mathrm{NH}_4\mathrm{Cl}$ while simultaneously leaving 10 mmol of the weak base $\mathrm{NH}_4\mathrm{OH}$ unreacted. Because the final mixture contains both a weak base and its corresponding salt, it satisfies the requirements to function as a basic buffer.