Chemical Equilibrium – JEE Main ChemistryPractice Questions & PYQs

The equilibrium constant $K_p$ for gas-phase reactions depends on the partial pressures of the products and reactants raised to the power of their stoichiometric coefficients. At 1990 K and 1 atm pressure, the mixture contains an equal number of $\text{Cl}_2$ molecules and $\text{Cl}$ atoms, which implies the mole fraction for each species is 0.5. Given the total pressure, the partial pressure for each component is 0.5 atm. For the reaction $\text{Cl}_2(g) \rightleftharpoons 2\text{Cl}(g)$, the expression for the equilibrium constant is:

$K_p = \frac{(P_{\text{Cl}})^2}{P_{\text{Cl}_2}}$

Substituting the partial pressures into this formula yields:

$K_p = \frac{(0.5)^2}{0.5} = 0.5$

Since this value is equivalent to $5 \times 10^{-1}$, the value of $x$ is 5.

Equilibrium reactions involving gases are governed by the relationship between the total number of moles of gaseous species and the system's total pressure, volume, and temperature. By analyzing the dissociation of the reactant $\mathrm{AB_2}$, we can determine the extent of the reaction using the Ideal Gas Law.

Starting with 1 mole of $\mathrm{AB_2}$ and letting $\alpha$ represent the degree of dissociation, the equilibrium mixture consists of $1-\alpha$ moles of $\mathrm{AB_2}$, $\alpha$ moles of $\mathrm{A}$, and $2\alpha$ moles of $\mathrm{B}$. The total number of gaseous moles present at equilibrium is $1+2\alpha$. Applying the Ideal Gas Law $PV = n_{\text{total}}RT$, with pressure $P = 1.9\ \text{atm}$, volume $V = 25\ \text{L}$, temperature $T = 300\ \text{K}$, and gas constant $R = 0.082\ \text{L}\ \text{atm}\ \text{mol}^{-1}\ \text{K}^{-1}$, the equation $1+2\alpha = \frac{PV}{RT}$ results in $1+2\alpha = \frac{1.9 \times 25}{0.082 \times 300} = 1.93$. Solving this leads to $\alpha = 0.465$.

The equilibrium constant $K_p$ is calculated using the partial pressures of the components, where the partial pressure of each gas is its mole fraction multiplied by the total pressure. The resulting expression for $K_p$ is $K_p = \frac{4\alpha^3 P^2}{(1+2\alpha)^2(1-\alpha)}$. Inserting the values $\alpha = 0.465$ and $P = 1.9\ \text{atm}$ results in $K_p = \frac{4 \times (0.465)^3 \times (1.9)^2}{(1.93)^2 \times (1 - 0.465)} = 0.7285$. This value is equivalent to $72.85 \times 10^{-2}$, which allows us to determine that $x = 73$.

The law of mass action defines the equilibrium constant $K_c$ as the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to its stoichiometric coefficient. For the reaction $A + B \rightleftharpoons C + D$, the expression is $K_c = \frac{[C][D]}{[A][B]}$. Given an initial concentration of 1 M for every component, the system shifts to equilibrium, causing reactant concentrations to decrease by $x$ and product concentrations to increase by $x$, resulting in equilibrium concentrations of $[A] = 1-x$, $[B] = 1-x$, $[C] = 1+x$, and $[D] = 1+x$.

Substituting these values into the equilibrium expression gives:
$100 = \frac{(1+x)(1+x)}{(1-x)(1-x)} = \left(\frac{1+x}{1-x}\right)^2$
Taking the square root of both sides yields $\frac{1+x}{1-x} = 10$. Solving this algebraic equation leads to $1+x = 10 - 10x$, which simplifies to $11x = 9$, resulting in $x = \frac{9}{11}$. The equilibrium concentration of $D$ is $1+x = 1 + \frac{9}{11} = \frac{20}{11} \approx 1.818$. Expressing this as $181.8 \times 10^{-2}$, the value rounded to the nearest integer is $182 \times 10^{-2}$ M.

Complex ion formation is governed by the equilibrium constant of the reaction, which defines the relationship between the concentrations of the metal ion, the ligand, and the complex at equilibrium. Given the formation constant $K_f = 1.0 \times 10^8$ for $[\mathrm{Ag}(\mathrm{NH}_3)_2]^{+}$ is large, the reaction $\mathrm{Ag}^{+} + 2\mathrm{NH}_3 \rightleftharpoons [\mathrm{Ag}(\mathrm{NH}_3)_2]^{+}$ effectively shifts to the right, meaning we can assume the concentration of the complex $[\mathrm{Ag}(\mathrm{NH}_3)_2]^{+}$ is essentially equal to the initial concentration of $\mathrm{Ag}^{+}$ ions, which is 0.80 M.

Using the equilibrium expression K_f = \frac{[\mathrm{Ag}(\mathrm{NH}_3)_2]^{+}}{[\mathrm{Ag}^{+}]\[\mathrm{NH}_3]^2}$, we substitute the target concentration of free silver ions ($5.0 × 10^{-8}$M) and the known formation constant to solve for the remaining free ammonia concentration. This gives$1.0 × 10^8 = \frac{0.80}{(5.0 × 10^{-8})[\mathrm{NH}_3]^2}$, which simplifies to$[\mathrm{NH}_3]^2 = 0.16$, resulting in a free ammonia concentration of 0.40 M. Because each mole of complex requires two moles of ammonia, we also account for the 1.6 M of ammonia consumed to form the complex from the 0.80 M of silver ions. Adding the free and consumed ammonia concentrations results in a total ammonia concentration of 2.0 M, and applying this to the 2 L solution volume gives a requirement of 4 moles.

The equilibrium state for the decomposition of solid ammonium hydrosulphide is governed by the concentrations of its gaseous products, described by the reaction $\mathrm{NH}_4\,\mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_3(g) + \mathrm{H}_2\,\mathrm{S}(g)$. With a molar mass of $51\,\text{g/mol}$, the initial $5.1\,\text{g}$ sample corresponds to $0.1\,\text{mol}$ of the solid. Since $20\,\%$ of this solid decomposes, $0.02\,\text{mol}$ of both $\mathrm{NH}_3$ and $\mathrm{H}_2\,\mathrm{S}$ are liberated into the $2\,\text{L}$ flask, creating a concentration of $0.01\,\text{M}$ for each gas.

The equilibrium constant $K_c$ is the product of these concentrations, calculated as $(0.01)(0.01) = 10^{-4}$. To determine $K_p$, we use the relation $K_p = K_c(RT)^{\Delta n}$, where $\Delta n = 2$ represents the change in moles of gas during the reaction. Using a temperature of $300\,\text{K}$ and the gas constant $R = 0.082\,\text{L}\,\text{atm}\,\text{K}^{-1}\,\text{mol}^{-1}$, the calculation $K_p = 10^{-4} \times (0.082 \times 300)^2$ yields approximately $0.06$. Given that $K_p = x \times 10^{-2}$, this result confirms that the integer value of $x$ is $6$.

Chemical equilibrium is established when the rate of the forward reaction equals the rate of the reverse reaction, ensuring that the concentrations of the species remain constant over time. By equating the forward rate $4.8 \times 10^{-5} [\mathrm{PtCl}_4]^{2-}$ to the reverse rate $2.4 \times 10^{-3} [\mathrm{Pt}(\mathrm{H}_2\mathrm{O})\mathrm{Cl}_3]^- [\mathrm{Cl}^-]$, the mathematical relationship between the reactant and product concentrations at equilibrium is defined. Solving for the equilibrium constant $K_c$ requires determining the ratio of the reactant concentration to the product concentrations, expressed as K_c = \frac{[\mathrm{PtCl}_4]^{2-}}{[\mathrm{Pt}(\mathrm{H}_2\mathrm{O})\mathrm{Cl}_3]^- \[\mathrm{Cl}^-]$}$. Substituting the given rate coefficients into this ratio yields$\frac{2.4 \times 10^{-3}}{4.8 \times 10^{-5}}$, which simplifies to 50.

The dissociation of a weak acid in a solution containing a strong acid is significantly suppressed because the excess hydrogen ions shift the equilibrium position to the left, which is a classic manifestation of the common ion effect. Because hydrochloric acid is a strong electrolyte, it fully dissociates, creating an initial hydrogen ion concentration of $0.1 \text{ M}$ that remains practically unchanged by the minimal dissociation of the weak acid. The weak acid equilibrium is represented by K_{a} = \frac{[\mathrm{H}^{+}]\$[\mathrm{A}^{-}]$}{[\mathrm{HA}]}$, where the concentrations at equilibrium can be approximated based on the assumption that the degree of dissociation$\alpha$is much less than$1$.

Given the dominance of the strong acid, we can set $[\mathrm{H}^{+}] \approx 0.1 \text{ M}$ and $[\mathrm{HA}] \approx 0.01 \text{ M}$, while the concentration of the conjugate base is $[\mathrm{A}^{-}] = 0.01 \alpha$. Substituting these values into the equilibrium constant expression yields $2.0 \times 10^{-6} = \frac{(0.1)(0.01 \alpha)}{0.01}$. Upon canceling the concentration terms for the weak acid, the equation simplifies to $0.1 \alpha = 2.0 \times 10^{-6}$, which leads to a degree of dissociation of $\alpha = 2.0 \times 10^{-5}$. Consequently, the required integer value is $2$.

The mixture of NH4Cl and NH3 forms a basic buffer, where the hydroxide ion concentration is determined by the Henderson-Hasselbalch equation. The total volume of the solution is 7 mL, found by summing the 5 mL of NH4Cl solution and the 2 mL of NH3 solution.

The concentration of each component in the mixture is calculated by dividing the millimoles of the substance by the total volume. Dividing the 0.252 millimoles of NH4Cl by the 7 mL total volume gives a concentration of 0.036 M, and dividing the 0.042 millimoles of NH3 by the 7 mL total volume gives a concentration of 0.006 M. The Henderson-Hasselbalch equation is:
$\mathrm{pOH} = \mathrm{p}K_b + \log \frac{[\text{Salt}]}{[\text{Base}]}$
The $\mathrm{p}K_b$ is the negative logarithm of 1.8 × 10⁻⁵, which is approximately 4.74. Substituting this value leads to the calculation:
$\mathrm{pOH} = 4.74 + \log \left( \frac{0.036}{0.006} \right) = 4.74 + \log(6) \approx 5.52$
Because $[\mathrm{OH}^{-}]$ is $10^{-\mathrm{pOH}}$, the hydroxide concentration is $10^{-5.52}$, which results in approximately $2.9 \times 10^{-6}$ M. In the form $x \times 10^{-6}$ M, the value of $x$ is 3.

The equilibrium constant relates the concentrations of reactants and products once a system has reached a stable state, following the expression $K = \frac{[C]^2}{[A][B]}$. Starting with an initial concentration of 1 M for each species, we let the change in concentration be $y$, causing reactants to decrease and the product to increase by $2y$ based on the stoichiometry. This results in equilibrium concentrations of $[A] = 1 - y$, $[B] = 1 - y$, and $[C] = 1 + 2y$.

Substituting these into the expression gives the equation:

$100 = \frac{(1 + 2y)^2}{(1 - y)(1 - y)}$

Taking the square root of both sides results in $10 = \frac{1 + 2y}{1 - y}$. Solving for $y$ involves multiplying 10 by the denominator to get $10 - 10y = 1 + 2y$, which simplifies to $12y = 9$, resulting in $y = 0.75$. Finally, substituting this back into the expression for the equilibrium concentration of $C$ yields $[C] = 1 + 2(0.75) = 2.5 \text{ M}$. As 2.5 M is equivalent to 25 times 10 to the power of -1 M, the value of $x$ is 25.

The relationship between the standard Gibbs free energy change and the equilibrium constant $K_{p}$ is governed by the equation $\Delta G^{\circ} = -RT \ln K_{p}$. Given the conversion factor $\ln 10 = 2.3$, this expression is rewritten as $\Delta G^{\circ} = -2.3 RT \log_{10} K_{p}$. Using the provided values where $\Delta G^{\circ} = 25200 \text{ J mol}^{-1}$, $R = 8.3 \text{ J mol}^{-1} \text{ K}^{-1}$, and $T = 400 \text{ K}$, the equation becomes $25200 = -2.3 \times 8.3 \times 400 \times \log_{10} K_{p}$.

Solving this for $\log_{10} K_{p}$ leads to $\frac{-25200}{2.3 \times 8.3 \times 400}$, which simplifies to $-3.3$. We can express $K_{p}$ as $10^{-3.3}$, or $10^{-3} \times 10^{-0.3}$. Applying the given antilog value of $10^{-0.3} = 0.501$, we find that $K_{p} = 0.501 \times 10^{-3} = 5.01 \times 10^{-4}$.

To determine the equilibrium constant in terms of concentration, $K_{c}$, we utilize the conversion formula $K_{p} = K_{c}(RT)^{\Delta n_{g}}$. For the reaction $2 \text{A}(g) \rightleftharpoons \text{A}_{2}(g)$, the change in the number of gaseous moles is $\Delta n_{g} = 1 - 2 = -1$. Rearranging the formula gives $K_{c} = K_{p} \times (RT)^{1}$. Substituting the known values, we calculate $K_{c} = (5.01 \times 10^{-4}) \times 8.3 \times 400$. Calculating the product of $8.3$ and $400$ gives $3320$, and multiplying this by $5.01 \times 10^{-4}$ yields $1.66332$. Expressing this result as $166 \times 10^{-2}$, the value rounded to the nearest integer is $166$.

Chemical equilibrium is determined by the balance between forward and reverse reaction rates, described quantitatively by the equilibrium constant $K_c$. In the dissociation of $\mathrm{PCl}_5$ into $\mathrm{PCl}_3$ and $\mathrm{Cl}_2$, the relationship is expressed as $K_c = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]}$. Given that the reaction takes place in a $1\text{ L}$ container, the concentration of each species is numerically equivalent to its mole count. We define $x$ as the moles of $\mathrm{PCl}_5$ that dissociate, meaning at equilibrium there are $3.0 - x$ moles of $\mathrm{PCl}_5$ remaining, while $\mathrm{PCl}_3$ and $\mathrm{Cl}_2$ are each present in $x$ moles.

Substituting these quantities into the equilibrium expression leads to $\frac{x^2}{3.0 - x} = 1.844$. Rearranging this relationship results in the quadratic equation $x^2 + 1.844x - 5.532 = 0$. Applying the quadratic formula to solve for the positive root gives $x \approx 1.604$. Subtracting this dissociation amount from the initial $3.0$ moles yields the final quantity of reactant, which is $1.396$ moles. Expressing this as an integer multiplied by $10^{-3}$ corresponds to $1396 \times 10^{-3}$, which rounds to the integer $1400$.

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 difference between the \sum of the stoichiometric coefficients of the gaseous products and the gaseous reactants. For the decomposition reaction $\mathrm{N_2O_4}(g) \rightleftharpoons 2\,\mathrm{NO_2}(g)$, the calculation for the change \in moles is $\Delta n_g = 2 - 1 = 1$.

Substituting the given values $K_p = 600.1$, $K_c = 20.4$, and $R = 0.0831$ into the relationship results in the equation $600.1 = 20.4 \times (0.0831 \times T)^1$ Solving this expression for $T$ requires dividing 600.1 by the product of 20.4 and 0.0831. This calculation yields a temperature of approximately 354 K.

The equilibrium constant for a reversible gas-phase reaction is determined by the ratio of product partial pressures to reactant partial pressures, with each term raised to the power of its stoichiometric coefficient. For the reaction $2 \mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{g})$, the equilibrium expression is $K_P = \frac{(P_{\mathrm{SO}_3})^2}{(P_{\mathrm{SO}_2})^2 \times P_{\mathrm{O}_2}}$. Before calculating, it is necessary to express all pressures in the same units; since the sulfur oxides are given in kilopascals, the oxygen partial pressure of $530 \, \text{Pa}$ should be converted to $0.53 \, \text{kPa}$. Substituting these values into the equilibrium expression gives $K_P = \frac{43^2}{45^2 \times 0.53} = \frac{1849}{2025 \times 0.53} = \frac{1849}{1073.25} \approx 1.7228$. Expressing this result in the requested format of $X \times 10^{-2}$ requires adjusting the decimal point, which yields $172.28 \times 10^{-2}$, and rounding to the nearest integer gives 172.

The equilibrium constant for a heterogeneous reaction excludes pure solids, as their activities are defined as unity. For the reaction $\mathrm{A}(\mathrm{s}) \rightleftharpoons \mathrm{M}(\mathrm{s}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g})$, the expression for the equilibrium constant simplifies to $K_p = (P_{\mathrm{O}_2})^{1/2}$. By substituting the given value of 4 into the equation $4 = (P_{\mathrm{O}_2})^{1/2}$ and squaring both sides, we determine that the partial pressure of oxygen at equilibrium is 16 atm.

The relationship between the standard Gibbs free energy change and the equilibrium constant of a chemical reaction is defined by the equation
$\Delta G^\circ = -RT \ln K_p$
This principle connects the thermodynamic potential of a reaction to the extent to which it proceeds at equilibrium.

Substituting the provided values $K_p = 100$ and $T = 300\ \text{K}$ into the equation gives $\Delta G^\circ = -R \times 300 \times \ln 100$. Because $100$ is equivalent to $10^2$, the logarithmic term simplifies to $2 \ln 10$. Utilizing the provided value $\ln 10 = 2.3$, the value of $\ln 100$ becomes $2 \times 2.3 = 4.6$.

Multiplying these factors yields $\Delta G^\circ = -R \times 300 \times 4.6$, which results in a value of $-1380 R$. Since the problem defines the free energy change as $-x R$, matching the expression indicates that $x$ is 1380.

The equilibrium constants $K_P$ and $K_C$ for a gaseous reaction are linked by the equation $K_P = K_C(RT)^{\Delta n}$. The value of $\Delta n$ is determined by calculating the change in the number of gaseous moles, which for this reaction is $2 - 1 = 1$. By rearranging the relationship to solve for $K_C$, the formula becomes $K_C = \frac{K_P}{RT}$. Substituting $K_P = 47.9$, $R = 0.083$, and $T = 288$ into this formula gives
$K_C = \frac{47.9}{0.083 \times 288}$
which evaluates to approximately 2.003, and the nearest integer is 2.

The relationship between the acidity and basicity of an aqueous solution is governed by the self-ionization constant of water, $K_w$, which is defined as [\mathrm{H}_3\mathrm{O}^+]\[\mathrm{OH}^-]$ = 10^{-14}$at 298 K. Because$\mathrm{Ba}(\mathrm{OH})_2$is a strong base that dissociates completely in water, it releases two hydroxide ions for every formula unit dissolved. Starting with a 0.005 M concentration of$\mathrm{Ba}(\mathrm{OH})_2$, the hydroxide ion concentration is determined by the equation$[\mathrm{OH}^-] = 2 \times 0.005 = 0.01 = 10^{-2}$. Substituting this value into the water ionization expression as$[\mathrm{H}_3\mathrm{O}^+] = \frac{10^{-14}}{10^{-2}} = 10^{-12}$provides the concentration of hydronium ions in mol/L. This result, expressed as 1 multiplied by$10^{-12}$ mol/L, indicates that the required integer is 1.

The overall stability of a coordination complex formed in sequential stages is equal to the product of its individual stepwise formation constants, whereas the dissociation constant for the complete complex is the reciprocal of this overall formation constant. Multiplying the provided stepwise values for the formation of the complex yields the overall equilibrium constant:

$K_{overall} = 10^4 \times 1.58 \times 10^3 \times 5 \times 10^2 \times 10^2 = 7.9 \times 10^{11}$

The dissociation constant for $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$ is then determined by taking the inverse of this overall formation constant:

$K_{diss} = \frac{1}{7.9 \times 10^{11}} = 1.26 \times 10^{-12}$

Comparing this result to the expression $x \times 10^{-12}$, the value of x is 1.26.

When two consecutive chemical equilibrium reactions are added together to form a net overall process, the equilibrium constant of the combined reaction is the product of the equilibrium constants of the individual steps. For the processes $A \rightleftharpoons B + C$ and $B + C \rightleftharpoons P$, the respective equilibrium constants are $K_{\mathrm{eq}}^{(1)} = \frac{[B][C]}{[A]}$ and $K_{\mathrm{eq}}^{(2)} = \frac{[P]}{[B][C]}$. Multiplying these two constants causes the intermediate species $[B]$ and $[C]$ to appear in the numerator and denominator and cancel out, which yields the net equilibrium constant expression $K_{\mathrm{eq}}^{(1)} \times K_{\mathrm{eq}}^{(2)} = \frac{[P]}{[A]}$ for the reaction $A \rightleftharpoons P$.

The equilibrium constant $K$ for a reversible reaction $A \rightleftharpoons B$ quantifies the ratio of product concentration to reactant concentration at equilibrium, expressed as $K = \frac{[B]}{[A]}$. By counting the species in the diagram, there are 11 particles representing product B and 6 particles representing reactant A. Dividing the count of the products by the reactants results in $11/6$, which simplifies to approximately 2.