Masterclass: 25 Solved JEE Advanced Numericals on Chemical Equilibrium
Conquer reversible reactions! This exhaustive guide features highly complex multi-step problems on Simultaneous Equilibria, Degree of Dissociation, Le Chatelier's Principle, and Thermodynamic Equilibrium Constants. Click "View Solution" to reveal the step-by-step breakdown.
Chemical Equilibrium demands a flawless understanding of stoichiometry and algebraic manipulation (the legendary ICE table). Before attempting these rigorous problems, ensure you clearly understand the difference between Reaction Quotient ($Q_c$) and Equilibrium Constant ($K_c$), the relationship $K_p = K_c(RT)^{\Delta n_g}$, and the precise application of Le Chatelier's Principle.
View Solution
Step 1: Write the $K_p$ and $K_c$ relationship
The fundamental equation linking the two constants is:
$K_p = K_c(RT)^{\Delta n_g}$
Step 2: Calculate $\Delta n_g$
$\Delta n_g = (\text{Moles of gaseous products}) - (\text{Moles of gaseous reactants})$
$\Delta n_g = 2 - (2 + 1) = 2 - 3 = -1$.
Step 3: Substitute known values and solve for $T$
$11.4 = 278 \times (0.0821 \times T)^{-1}$
$11.4 = \frac{278}{0.0821 \times T}$
Rearranging to isolate $T$:
$T = \frac{278}{11.4 \times 0.0821}$
$T = \frac{278}{0.93594}$
$T = 297.02 \text{ K}$.
View Solution
Step 1: Calculate Theoretical Vapor Density ($D$)
Molar mass of $PCl_5 = 31 + 5(35.5) = 31 + 177.5 = 208.5 \text{ g mol}^{-1}$.
Theoretical vapor density ($D$) = $\frac{\text{Molar Mass}}{2} = \frac{208.5}{2} = 104.25$.
Step 2: Understand the Dissociation Formula
For a reaction $A \rightleftharpoons nB$, the degree of dissociation $\alpha$ is related to vapor densities by:
$\alpha = \frac{D - d}{d(n - 1)}$
Where $D$ is theoretical vapor density, and $d$ is observed vapor density.
Step 3: Substitute values and calculate $\alpha$
For this specific reaction, 1 mole of reactant yields 2 moles of products, so $n = 2$.
Observed vapor density ($d$) = $57.9$.
$\alpha = \frac{104.25 - 57.9}{57.9(2 - 1)}$
$\alpha = \frac{46.35}{57.9 \times 1}$
$\alpha = 0.80$.
View Solution
Step 1: Write the fundamental thermodynamic equation
$\Delta G^{\circ} = -RT \ln K_p = -2.303 RT \log_{10} K_p$.
Step 2: Check Units Carefully
$\Delta G^{\circ} = -33.2 \text{ kJ mol}^{-1} = -33200 \text{ J mol}^{-1}$.
Step 3: Substitute and solve for $\log_{10} K_p$
$-33200 = -2.303 \times 8.314 \times 298 \times \log_{10} K_p$
$-33200 = -5705.8 \times \log_{10} K_p$
$\log_{10} K_p = \frac{-33200}{-5705.8} = 5.818$.
Step 4: Find the Antilog
$K_p = 10^{5.818} = 10^{0.818} \times 10^5 \approx 6.58 \times 10^5$.
Since standard states were based on partial pressures in atmospheres, the constant is implicitly $K_p$.
View Solution
Step 1: Setup the ICE Table using Partial Pressures
Let the equilibrium partial pressure of $CO_2$ be $P$.
According to the 2:1 stoichiometry, the partial pressure of $NH_3$ must be twice that of $CO_2$, so $P_{NH_3} = 2P$.
(The solid $NH_2COONH_4$ does not exert partial pressure).
Step 2: Relate to Total Pressure
$P_{\text{total}} = P_{NH_3} + P_{CO_2}$
$0.3 \text{ atm} = 2P + P = 3P$
$P = 0.1 \text{ atm}$.
Step 3: Determine Individual Partial Pressures
$P_{CO_2} = 0.1 \text{ atm}$
$P_{NH_3} = 2(0.1) = 0.2 \text{ atm}$.
Step 4: Calculate $K_p$
$K_p = (P_{NH_3})^2 \times (P_{CO_2})^1$
$K_p = (0.2)^2 \times (0.1) = 0.04 \times 0.1 = 0.004 \text{ atm}^3$.
View Solution
Step 1: Calculate Initial Concentrations
$[H_2] = \frac{0.4 \text{ mol}}{2.0 \text{ L}} = 0.2 \text{ M}$
$[I_2] = \frac{0.4 \text{ mol}}{2.0 \text{ L}} = 0.2 \text{ M}$
$[HI] = \frac{2.0 \text{ mol}}{2.0 \text{ L}} = 1.0 \text{ M}$
Step 2: Calculate the Reaction Quotient ($Q_c$)
$Q_c = \frac{[HI]^2}{[H_2][I_2]}$
$Q_c = \frac{(1.0)^2}{(0.2)(0.2)} = \frac{1.0}{0.04} = 25.0$.
Step 3: Compare $Q_c$ with $K_c$
We are given $K_c = 50.0$.
Since $Q_c < K_c$ ($25.0 < 50.0$), the ratio of products to reactants is currently too low. The system is not at equilibrium.
Step 4: Conclusion
To reach equilibrium, the system must increase the numerator (products) and decrease the denominator (reactants). Therefore, the reaction will proceed in the forward direction.
View Solution
Step 1: Set up the ICE Table (in Molarity)
Initial $[CO] = 1.0 / 10 = 0.1 \text{ M}$. Initial $[H_2O] = 0.1 \text{ M}$.
$CO \quad + \quad H_2O \rightleftharpoons \quad CO_2 \quad + \quad H_2$
I: $0.1$ $\quad\quad$ $0.1$ $\quad\quad\quad$ $0$ $\quad\quad\quad$ $0$
C: $-x$ $\quad\quad$ $-x$ $\quad\quad\quad$ $+x$ $\quad\quad\quad$ $+x$
E: $0.1-x$ $\quad$ $0.1-x$ $\quad\quad$ $x$ $\quad\quad\quad$ $x$
Step 2: Write $K_c$ expression and substitute
$K_c = \frac{[CO_2][H_2]}{[CO][H_2O]} = 4.0$
$4.0 = \frac{(x)(x)}{(0.1 - x)(0.1 - x)} = \frac{x^2}{(0.1 - x)^2}$
Step 3: Solve the perfect square
Take the square root of both sides (since concentrations must be positive):
$2.0 = \frac{x}{0.1 - x}$
$2.0(0.1 - x) = x$
$0.2 - 2x = x$
$3x = 0.2 \implies x = \frac{0.2}{3} = 0.0667 \text{ M}$.
Step 4: Find final concentrations
$[CO_2] = [H_2] = x = 0.0667 \text{ M}$.
$[CO] = [H_2O] = 0.1 - 0.0667 = 0.0333 \text{ M}$.
View Solution
Step 1: Write the Van 't Hoff Equation
$\log_{10} \left( \frac{K_2}{K_1} \right) = \frac{\Delta H^{\circ}}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$
Step 2: Identify variables
$K_1 = 10^{-4}$, $T_1 = 300 \text{ K}$
$K_2 = 10^{-2}$, $T_2 = 400 \text{ K}$
Step 3: Substitute and evaluate
$\log_{10} \left( \frac{10^{-2}}{10^{-4}} \right) = \frac{\Delta H^{\circ}}{2.303 \times 8.314} \left( \frac{400 - 300}{300 \times 400} \right)$
$\log_{10} (10^2) = \frac{\Delta H^{\circ}}{19.147} \left( \frac{100}{120000} \right)$
$2 = \frac{\Delta H^{\circ}}{19.147} \left( \frac{1}{1200} \right)$
Step 4: Solve for $\Delta H^{\circ}$
$\Delta H^{\circ} = 2 \times 19.147 \times 1200$
$\Delta H^{\circ} = 45952.8 \text{ J mol}^{-1} = 45.95 \text{ kJ mol}^{-1}$.
Check: $K$ increased as $T$ increased, meaning the reaction is endothermic ($\Delta H^{\circ} > 0$). Our positive result aligns perfectly.
View Solution
Case (a): Addition of Inert Gas at Constant Volume
Adding Argon increases the total pressure of the system. However, because the volume is constant, the partial pressures of the reactive gases ($PCl_5, PCl_3, Cl_2$) do not change ($P_i = n_i RT / V$). Since the partial pressures are unchanged, the reaction quotient $Q_p$ remains equal to $K_p$.
Result: No shift in equilibrium. Degree of dissociation remains unchanged.
Case (b): Addition of Inert Gas at Constant Pressure
To keep the total pressure constant while adding Argon, the volume of the container must expand. This expansion dilutes the reactive gases, causing their individual partial pressures to drop.
Mathematically, let total pressure be $P$. $K_p = \frac{(x_{PCl_3}P)(x_{Cl_2}P)}{(x_{PCl_5}P)} = \frac{n_{PCl_3} n_{Cl_2}}{n_{PCl_5} n_{\text{total}}} P$.
Adding Argon drastically increases $n_{\text{total}}$. To keep $K_p$ constant, the numerator ($n_{PCl_3} n_{Cl_2}$) must increase.
Result: The equilibrium shifts forward (towards more moles of gas). The degree of dissociation increases.
View Solution
Step 1: Moles at Equilibrium (ICE Table)
$N_2 \quad + \quad 3H_2 \rightleftharpoons \quad 2NH_3$
I: $1$ $\quad\quad\quad$ $3$ $\quad\quad\quad$ $0$
C: $-x$ $\quad\quad$ $-3x$ $\quad\quad$ $+2x$
E: $1-x$ $\quad$ $3-3x$ $\quad\quad$ $2x$
Step 2: Total Moles at Equilibrium
$n_{\text{total}} = (1 - x) + (3 - 3x) + 2x = 4 - 2x$.
Step 3: Calculate Mole Fractions ($X_i$)
$X_{N_2} = \frac{1 - x}{4 - 2x}$
$X_{H_2} = \frac{3 - 3x}{4 - 2x} = \frac{3(1 - x)}{4 - 2x}$
$X_{NH_3} = \frac{2x}{4 - 2x}$
Step 4: Write $K_p$ Expression
$K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} = \frac{(X_{NH_3} P)^2}{(X_{N_2} P)(X_{H_2} P)^3} = \frac{(X_{NH_3})^2}{(X_{N_2})(X_{H_2})^3 P^2}$
Step 5: Substitute and Simplify
$K_p = \frac{\left( \frac{2x}{4-2x} \right)^2}{\left( \frac{1-x}{4-2x} \right) \left( \frac{3(1-x)}{4-2x} \right)^3 P^2}$
$K_p = \frac{4x^2 / (4-2x)^2}{[27(1-x)^4 / (4-2x)^4] P^2}$
$K_p = \frac{4x^2 (4-2x)^2}{27(1-x)^4 P^2}$.
View Solution
Step 1: Setup Moles
Start with 1 mole of $N_2O_4$.
At equilibrium: Moles of $N_2O_4 = 1 - \alpha$. Moles of $NO_2 = 2\alpha$.
Total moles = $1 - \alpha + 2\alpha = 1 + \alpha$.
Step 2: Setup Partial Pressures
$P_{N_2O_4} = \left( \frac{1-\alpha}{1+\alpha} \right) P$
$P_{NO_2} = \left( \frac{2\alpha}{1+\alpha} \right) P$
Step 3: Construct $K_p$
$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{\left[ \frac{2\alpha}{1+\alpha} P \right]^2}{\left( \frac{1-\alpha}{1+\alpha} \right) P} = \frac{4\alpha^2 P^2 / (1+\alpha)^2}{(1-\alpha) P / (1+\alpha)}$
$K_p = \frac{4\alpha^2 P}{(1+\alpha)(1-\alpha)} = \frac{4\alpha^2 P}{1 - \alpha^2}$.
Step 4: Apply Approximation
If $\alpha$ is very small, $\alpha \ll 1$, then $1 - \alpha^2 \approx 1$.
The equation simplifies to $K_p \approx 4\alpha^2 P$.
Rearranging for $\alpha$:
$\alpha^2 = \frac{K_p}{4P} \implies \alpha = \frac{\sqrt{K_p}}{2} \frac{1}{\sqrt{P}}$.
$A_{(s)} \rightleftharpoons C_{(g)} + D_{(g)} \quad (K_{p1} = 400 \text{ atm}^2)$
$B_{(s)} \rightleftharpoons C_{(g)} + E_{(g)} \quad (K_{p2} = 900 \text{ atm}^2)$
If both solids are placed in an evacuated container and allowed to reach equilibrium simultaneously, calculate the total equilibrium pressure.
View Solution
Step 1: Understand the common ion effect in gases
Gas $C$ is produced by both reactions. This creates a "common gas effect" where the partial pressure of $C$ suppresses the dissociation of both solids.
Step 2: Assign Variables
Let $P_1$ be the partial pressure of $D$ created by Reaction 1. Thus, Reaction 1 also creates $P_1$ of gas $C$.
Let $P_2$ be the partial pressure of $E$ created by Reaction 2. Thus, Reaction 2 also creates $P_2$ of gas $C$.
At simultaneous equilibrium, the total partial pressure of $C$ everywhere in the flask is $(P_1 + P_2)$.
Step 3: Write $K_p$ equations
$K_{p1} = P_C \times P_D = (P_1 + P_2) \times P_1 = 400$
$K_{p2} = P_C \times P_E = (P_1 + P_2) \times P_2 = 900$
Step 4: Solve the simultaneous equations
Add the two equations together:
$(P_1 + P_2)P_1 + (P_1 + P_2)P_2 = 400 + 900$
Factor out $(P_1 + P_2)$:
$(P_1 + P_2)(P_1 + P_2) = 1300$
$(P_1 + P_2)^2 = 1300$
$P_1 + P_2 = \sqrt{1300} \approx 36.05 \text{ atm}$. (This is the total pressure of gas $C$).
Step 5: Calculate Total Pressure
Total Pressure $= P_C + P_D + P_E = (P_1 + P_2) + P_1 + P_2 = 2(P_1 + P_2)$.
Total Pressure $= 2 \times 36.05 = 72.1 \text{ atm}$.
(1) $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)} \quad K_1 = 10^{-4}$
(2) $N_{2(g)} + 2O_{2(g)} \rightleftharpoons 2NO_{2(g)} \quad K_2 = 10^{-6}$
Calculate the equilibrium constant $K_3$ for the reaction: $NO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons NO_{2(g)}$.
View Solution
Step 1: Analyze the target equation
Target: $NO + \frac{1}{2}O_2 \rightleftharpoons NO_2$.
We need to build this equation using equations (1) and (2).
Step 2: Manipulate given equations
We need $NO$ on the reactant side. Equation (1) has $2NO$ on the product side. Therefore, reverse equation (1) and divide by 2.
Reversing (1) gives $K = 1/K_1$. Dividing by 2 takes the square root: $K'_1 = (1/K_1)^{1/2} = (K_1)^{-1/2}$.
Reaction A: $NO \rightleftharpoons \frac{1}{2}N_2 + \frac{1}{2}O_2 \quad K_A = (10^{-4})^{-1/2} = 10^2 = 100$.
We need $NO_2$ on the product side. Equation (2) has $2NO_2$ on the product side. Divide equation (2) by 2.
Dividing by 2 takes the square root: $K'_2 = (K_2)^{1/2}$.
Reaction B: $\frac{1}{2}N_2 + O_2 \rightleftharpoons NO_2 \quad K_B = (10^{-6})^{1/2} = 10^{-3}$.
Step 3: Add manipulated equations
Add Reaction A and Reaction B:
LHS: $NO + \frac{1}{2}N_2 + O_2$
RHS: $\frac{1}{2}N_2 + \frac{1}{2}O_2 + NO_2$
Net: $NO + \frac{1}{2}O_2 \rightleftharpoons NO_2$ (This perfectly matches the target!).
Step 4: Calculate final $K$
When equations are added, their equilibrium constants are multiplied.
$K_3 = K_A \times K_B = 10^2 \times 10^{-3} = 10^{-1} = 0.1$.
View Solution
Step 1: Write the combined dissolution equation
Reaction 1: $AgCl_{(s)} \rightleftharpoons Ag^+ + Cl^- \quad (K_{sp} = 10^{-10})$
Reaction 2: $Ag^+ + 2NH_3 \rightleftharpoons [Ag(NH_3)_2]^+ \quad (K_f = 10^8)$
Net Reaction: $AgCl_{(s)} + 2NH_3 \rightleftharpoons [Ag(NH_3)_2]^+ + Cl^-$.
Step 2: Calculate the net equilibrium constant ($K_{net}$)
Adding reactions multiplies constants: $K_{net} = K_{sp} \times K_f = 10^{-10} \times 10^8 = 10^{-2} = 0.01$.
Step 3: Setup ICE Table for the net reaction
Let solubility be $S$.
Initial: $NH_3 = 1.0 \text{ M}$. Products $= 0$.
Change: $NH_3 = -2S$ (since 2 moles are consumed per mole of AgCl dissolved). Products $= +S$.
Equilibrium: $NH_3 = 1.0 - 2S$. Complex $= S$. $Cl^- = S$.
Step 4: Solve for S
$K_{net} = \frac{[\text{Complex}][Cl^-]}{[NH_3]^2}$
$0.01 = \frac{S \times S}{(1.0 - 2S)^2} = \frac{S^2}{(1.0 - 2S)^2}$
Take the square root of both sides:
$0.1 = \frac{S}{1.0 - 2S}$
$0.1 - 0.2S = S$
$1.2S = 0.1 \implies S = \frac{0.1}{1.2} = \frac{1}{12} \approx 0.0833 \text{ M}$.
View Solution
Step 1: Setup Moles and Mole Fractions
Start with 1 mole of $A$. At equilibrium: $n_A = 1 - \alpha$, $n_B = 2\alpha$.
Total moles $= 1 - \alpha + 2\alpha = 1 + \alpha$.
Mole fractions: $X_A = \frac{1-\alpha}{1+\alpha}$, $X_B = \frac{2\alpha}{1+\alpha}$.
Step 2: Setup Partial Pressures
$P_A = X_A \times P = \frac{1-\alpha}{1+\alpha} P$
$P_B = X_B \times P = \frac{2\alpha}{1+\alpha} P$
Step 3: Write $K_p$ and substitute
$K_p = \frac{(P_B)^2}{P_A} = \frac{\left( \frac{2\alpha}{1+\alpha} P \right)^2}{\frac{1-\alpha}{1+\alpha} P}$
$K_p = \frac{4\alpha^2 P^2 / (1+\alpha)^2}{(1-\alpha)P / (1+\alpha)} = \frac{4\alpha^2 P}{(1+\alpha)(1-\alpha)} = \frac{4\alpha^2 P}{1 - \alpha^2}$.
Step 4: Algebraically isolate $\alpha$
$K_p (1 - \alpha^2) = 4\alpha^2 P$
$K_p - K_p \alpha^2 = 4P \alpha^2$
$K_p = \alpha^2 (4P + K_p)$
$\alpha^2 = \frac{K_p}{4P + K_p}$
$\alpha = \sqrt{\frac{K_p}{K_p + 4P}}$. (Proved).
View Solution
Step 1: Find Mole Fractions in the bulk equilibrium mixture
Average Molar Mass $M_{\text{avg}} = X_{NO_2}(46) + X_{N_2O_4}(92)$.
Since $X_{N_2O_4} = 1 - X_{NO_2}$:
$69 = 46X_{NO_2} + 92(1 - X_{NO_2})$
$69 = 46X_{NO_2} + 92 - 92X_{NO_2} = 92 - 46X_{NO_2}$
$46X_{NO_2} = 23 \implies X_{NO_2} = 0.5$.
Thus, $X_{N_2O_4} = 0.5$. The mixture is equimolar.
Step 2: Apply Graham's Law of Effusion
The rate of effusion ($r$) of a gas is proportional to its partial pressure and inversely proportional to the square root of its molar mass: $r \propto \frac{P}{\sqrt{M}}$.
Since $X_{NO_2} = X_{N_2O_4}$, their partial pressures are equal. Let's find the ratio of their moles in the effused mixture ($n'$):
$\frac{n'_{NO_2}}{n'_{N_2O_4}} = \frac{P_{NO_2} / \sqrt{46}}{P_{N_2O_4} / \sqrt{92}} = \frac{\sqrt{92}}{\sqrt{46}} = \sqrt{2} \approx 1.414$.
Step 3: Calculate the new Mole Fractions in the effused gas
Let $n'_{N_2O_4} = 1$, then $n'_{NO_2} = 1.414$. Total moles $= 2.414$.
New $X'_{NO_2} = \frac{1.414}{2.414} = 0.586$.
New $X'_{N_2O_4} = \frac{1}{2.414} = 0.414$.
Step 4: Calculate the apparent molar mass of the effused mixture
$M'_{\text{avg}} = (0.586 \times 46) + (0.414 \times 92) = 26.956 + 38.088 = 65.04 \text{ g/mol}$.
Insight: The effused gas is lighter than the bulk gas because the lighter $NO_2$ molecules effuse faster!
View Solution
Step 1: Calculate $K_c$
$K_c$ is constant. Using initial equilibrium data:
$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(0.4)(0.4)}{0.2} = \frac{0.16}{0.2} = 0.8 \text{ M}$.
Step 2: Find Moles at Initial Equilibrium
Moles = Molarity $\times$ Volume ($10 \text{ L}$).
$n_{PCl_5} = 0.2 \times 10 = 2 \text{ moles}$.
$n_{PCl_3} = 0.4 \times 10 = 4 \text{ moles}$.
$n_{Cl_2} = 0.4 \times 10 = 4 \text{ moles}$.
Step 3: Setup the Shift
Let the new volume be $V$. Le Chatelier's Principle says expanding the volume shifts the reaction forward to produce more gas molecules. Let $y$ moles of $PCl_5$ dissociate further.
New moles: $n'_{PCl_5} = 2 - y$, $n'_{PCl_3} = 4 + y$, $n'_{Cl_2} = 4 + y$.
Step 4: Use new concentration to find $y$
We are given new $[PCl_5] = 0.05 \text{ M}$.
$[PCl_5] = \frac{2 - y}{V} = 0.05 \implies 2 - y = 0.05V \implies y = 2 - 0.05V$.
Step 5: Apply $K_c$ to the new state
New concentrations: $[PCl_3] = \frac{4+y}{V}$, $[Cl_2] = \frac{4+y}{V}$.
$K_c = \frac{([PCl_3])([Cl_2])}{[PCl_5]} \implies 0.8 = \frac{\left(\frac{4+y}{V}\right)^2}{0.05}$
$0.8 \times 0.05 = \left(\frac{4+y}{V}\right)^2 \implies 0.04 = \left(\frac{4+y}{V}\right)^2$.
Taking square root: $0.2 = \frac{4+y}{V} \implies 4 + y = 0.2V$.
Step 6: Solve system of equations
We have: $y = 2 - 0.05V$ and $y = 0.2V - 4$.
$2 - 0.05V = 0.2V - 4$
$6 = 0.25V \implies V = \frac{6}{0.25} = 24 \text{ L}$.
View Solution
Step 1: Analyze the $K_p$ expression
Since hydrates are solids, they do not appear in the $K_p$ expression.
$K_p = (P_{H_2O})^2$
We are given $K_p = 1.0 \times 10^{-4}$.
$(P_{H_2O})^2 = 10^{-4} \implies P_{H_2O} = \sqrt{10^{-4}} = 10^{-2} \text{ atm} = 0.01 \text{ atm}$.
Step 2: Convert to consistent units
Convert the equilibrium vapor pressure of water from atm to torr.
$P_{H_2O} = 0.01 \text{ atm} \times 760 \text{ torr/atm} = 7.6 \text{ torr}$.
This is the actual partial pressure of water vapor in the air maintained by the hydrate equilibrium.
Step 3: Calculate Relative Humidity (RH)
Relative Humidity is defined as the ratio of the actual partial pressure of water vapor to the saturated vapor pressure of pure water at that temperature, expressed as a percentage.
$RH = \left( \frac{P_{actual}}{P_{pure}} \right) \times 100$
$RH = \left( \frac{7.6}{24.0} \right) \times 100 = 0.3166 \times 100 = 31.66\%$.
View Solution
Step 1: Setup ICE Table in Reverse
Since we start with products, the reaction must proceed backward to reach equilibrium.
$A \quad + \quad B \rightleftharpoons \quad C \quad + \quad D$
I: $0$ $\quad\quad\quad$ $0$ $\quad\quad\quad$ $2.0$ $\quad\quad$ $2.0$
C: $+x$ $\quad\quad$ $+x$ $\quad\quad$ $-x$ $\quad\quad$ $-x$
E: $x$ $\quad\quad\quad$ $x$ $\quad\quad\quad$ $2.0-x$ $\quad$ $2.0-x$
Step 2: Apply $K_c$ Expression
$K_c = \frac{[C][D]}{[A][B]} = 9.0$
$9.0 = \frac{(2.0 - x)(2.0 - x)}{(x)(x)} = \frac{(2.0 - x)^2}{x^2}$
Step 3: Solve Perfect Square
Take the square root of both sides:
$3.0 = \frac{2.0 - x}{x}$
$3.0x = 2.0 - x$
$4.0x = 2.0 \implies x = 0.5 \text{ M}$.
Step 4: Find Concentration of A
$[A] = x = 0.5 \text{ M}$.
View Solution
Step 1: Setup ICE Table
$2H_2S \rightleftharpoons 2H_2 + S_2$
I: $0.1$ $\quad\quad\quad$ $0$ $\quad\quad$ $0$
C: $-2x$ $\quad\quad$ $+2x$ $\quad$ $+x$
E: $0.1-2x$ $\quad$ $2x$ $\quad\quad$ $x$
Step 2: Write $K_c$ expression
$K_c = \frac{[H_2]^2[S_2]}{[H_2S]^2} = 1.0 \times 10^{-7}$
$10^{-7} = \frac{(2x)^2(x)}{(0.1 - 2x)^2} = \frac{4x^3}{(0.1 - 2x)^2}$
Step 3: Apply Chemical Approximation
Since $K_c$ is extremely small ($10^{-7}$), the reaction hardly proceeds forward. Therefore, $2x$ is negligibly small compared to $0.1$.
Assume $0.1 - 2x \approx 0.1$.
The equation simplifies to:
$10^{-7} = \frac{4x^3}{(0.1)^2} = \frac{4x^3}{0.01}$
$10^{-7} \times 0.01 = 4x^3 \implies 10^{-9} = 4x^3$
$x^3 = 0.25 \times 10^{-9} = 250 \times 10^{-12}$.
Step 4: Solve for $x$
$x = (250)^{1/3} \times 10^{-4}$.
Since $6^3 = 216$ and $7^3 = 343$, $(250)^{1/3} \approx 6.3$.
$x = 6.3 \times 10^{-4} \text{ M}$.
Check approximation: $2x \approx 0.00126$, which is $1.26\%$ of $0.1$. The $< 5\%$ rule holds, so the approximation is valid.
View Solution
Step 1: Understand Large K_c implications
A $K_c$ of $10^8$ means the reaction goes virtually to $100\%$ completion. Standard ICE tables with $-x$ will fail because $(1.0-x)$ will round to zero, crashing the math. We must treat it as a limiting reagent problem first, then let it "bounce back" slightly to equilibrium.
Step 2: Drive reaction to 100% completion
$A$ is the limiting reagent (1 mole vs 2 moles of B).
Assume all $A$ reacts. $A$ becomes $0$. $B$ becomes $2.0 - 1.0 = 1.0 \text{ M}$. $C$ produced = $1.0 \text{ M}$.
Step 3: "Bounce Back" ICE Table (Reverse reaction)
Now let a minuscule amount $y$ of $C$ decompose back to satisfy the equilibrium constant.
$A \quad + \quad B \rightleftharpoons \quad C$
I: $0$ $\quad\quad\quad$ $1.0$ $\quad\quad$ $1.0$
C: $+y$ $\quad\quad$ $+y$ $\quad\quad$ $-y$
E: $y$ $\quad\quad\quad$ $1.0+y$ $\quad$ $1.0-y$
Step 4: Solve for $y$ using approximations
Since the forward reaction is highly favored, the reverse reaction is practically nonexistent. Thus $y$ is extremely small ($y \ll 1.0$).
$1.0 + y \approx 1.0$
$1.0 - y \approx 1.0$
$K_c = \frac{[C]}{[A][B]} = \frac{1.0}{(y)(1.0)} = \frac{1}{y}$
Step 5: Calculate
$10^8 = \frac{1}{y} \implies y = 10^{-8} \text{ M}$.
View Solution
Step 1: Understand the Equilibrium state
For the reaction $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$, the equilibrium constant $K_p = P_{H_2O}$.
The problem asks for the temperature where $P_{H_2O} = 1 \text{ atm}$. Thus, we need the temperature where $K_p = 1$.
Step 2: Apply Thermodynamics
$\Delta G^{\circ} = -RT \ln K_p$.
If $K_p = 1$, then $\ln(1) = 0$, which means $\Delta G^{\circ} = 0$.
Step 3: Relate to Enthalpy and Entropy
$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$
Set $\Delta G^{\circ} = 0$:
$0 = \Delta H^{\circ} - T\Delta S^{\circ} \implies T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$
Step 4: Calculation
Ensure units match (convert $\Delta H$ to Joules): $\Delta H^{\circ} = 40660 \text{ J mol}^{-1}$.
$T = \frac{40660}{108.8} = 373.7 \text{ K}$.
View Solution
Step 1: Understand Thermodynamic invariance
A catalyst lowers the activation energy barrier for both the forward and reverse reactions by the exact same amount. It does NOT change the relative energy levels of reactants and products ($\Delta G^{\circ}$ is constant). Therefore, the equilibrium constant $K_c$ remains strictly unchanged.
Step 2: New $K_c$
$K_c$ is still exactly 10.
Step 3: Kinetic Relationship
We know $K_c = \frac{k_f}{k_b}$.
For $K_c$ to remain constant while $k_f$ increases by a factor of 1000, $k_b$ must also increase by the exact same factor of 1000.
View Solution
Step 1: Check Reaction Direction via $Q_c$
$Q_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} = \frac{0.5}{(0.1)(0.1)} = \frac{0.5}{0.01} = 50$.
Since $Q_c > K_c$ ($50 > 10$), the reaction must shift backward (to the left) to reach equilibrium.
Step 2: Setup ICE Table (Backward Shift)
Let $x$ be the amount of $PCl_5$ that decomposes.
$PCl_3 \quad + \quad Cl_2 \rightleftharpoons \quad PCl_5$
I: $0.1$ $\quad\quad\quad$ $0.1$ $\quad\quad\quad$ $0.5$
C: $+x$ $\quad\quad$ $+x$ $\quad\quad\quad$ $-x$
E: $0.1+x$ $\quad$ $0.1+x$ $\quad\quad$ $0.5-x$
Step 3: Apply $K_c$ and solve quadratic
$10 = \frac{0.5 - x}{(0.1 + x)(0.1 + x)} = \frac{0.5 - x}{(0.1 + x)^2}$
$10(0.01 + 0.2x + x^2) = 0.5 - x$
$0.1 + 2x + 10x^2 = 0.5 - x$
$10x^2 + 3x - 0.4 = 0$.
Step 4: Quadratic Formula
$x = \frac{-3 \pm \sqrt{3^2 - 4(10)(-0.4)}}{2(10)} = \frac{-3 \pm \sqrt{9 + 16}}{20} = \frac{-3 \pm \sqrt{25}}{20}$
$x = \frac{-3 \pm 5}{20}$.
We must take the positive root for physical reality: $x = \frac{2}{20} = 0.1 \text{ M}$.
Step 5: Final Concentration
$[PCl_3] = 0.1 + x = 0.1 + 0.1 = 0.2 \text{ M}$.
View Solution
Step 1: Find Initial $K_c$
$K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.2/1)^2}{(0.1/1)} = \frac{0.04}{0.1} = 0.4 \text{ M}$.
Step 2: Apply Volume Change (Le Chatelier)
Expanding the volume decreases pressure, shifting the equilibrium to the right (side with more gas moles). Let $x$ moles of $N_2O_4$ dissociate.
Step 3: Setup New ICE Table
Moles before shift: $N_2O_4 = 0.1$, $NO_2 = 0.2$.
Moles after shift: $N_2O_4 = 0.1 - x$, $NO_2 = 0.2 + 2x$.
New Volume $= 10 \text{ L}$.
Step 4: Solve for $x$ using $K_c$
$K_c = \frac{([NO_2]/10)^2}{[N_2O_4]/10} = \frac{(0.2+2x)^2 / 100}{(0.1-x) / 10} = \frac{(0.2+2x)^2}{10(0.1-x)}$
We know $K_c = 0.4$, so:
$0.4 = \frac{(0.2+2x)^2}{1 - 10x} \implies 0.4(1 - 10x) = (0.2+2x)^2$
$0.4 - 4x = 0.04 + 0.8x + 4x^2$
$4x^2 + 4.8x - 0.36 = 0$
Divide by 4: $x^2 + 1.2x - 0.09 = 0$.
Step 5: Quadratic Formula
$x = \frac{-1.2 \pm \sqrt{(1.2)^2 - 4(1)(-0.09)}}{2} = \frac{-1.2 \pm \sqrt{1.44 + 0.36}}{2} = \frac{-1.2 \pm \sqrt{1.80}}{2}$
$\sqrt{1.80} \approx 1.341$.
$x = \frac{-1.2 + 1.341}{2} = \frac{0.141}{2} \approx 0.0705 \text{ moles}$.
Step 6: New Moles of $NO_2$
$n_{NO_2} = 0.2 + 2x = 0.2 + 2(0.0705) = 0.2 + 0.141 = 0.341 \text{ moles}$.
View Solution
Step 1: Setup the Reverse ICE Table
We are starting with the product $HD$ and letting it decompose backward.
Let $2x$ be the amount of $HD$ that disproportionates to avoid fractions.
$H_2 \quad + \quad D_2 \rightleftharpoons \quad 2HD$
I: $0$ $\quad\quad\quad$ $0$ $\quad\quad\quad$ $1.0$
C: $+x$ $\quad\quad$ $+x$ $\quad\quad\quad$ $-2x$
E: $x$ $\quad\quad\quad$ $x$ $\quad\quad\quad$ $1.0-2x$
Step 2: Apply $K_c$
$K_c = \frac{[HD]^2}{[H_2][D_2]} = 4.0$
Since volume $V$ cancels out (as $\Delta n = 0$):
$4.0 = \frac{(1.0 - 2x)^2}{(x)(x)} = \frac{(1.0 - 2x)^2}{x^2}$
Step 3: Solve the Perfect Square
Take the square root of both sides:
$2.0 = \frac{1.0 - 2x}{x}$
$2.0x = 1.0 - 2x$
$4.0x = 1.0 \implies x = 0.25 \text{ moles}$.
Step 4: Calculate Percentage Disproportionation
The amount of $HD$ that reacted is $2x = 2(0.25) = 0.50 \text{ moles}$.
Initial amount of $HD$ was $1.0 \text{ mole}$.
Percentage $= \left( \frac{0.50}{1.0} \right) \times 100 = 50\%$.
Insight: At statistical equilibrium, a 1:1 mixture of $H$ and $D$ atoms distributes randomly into $H_2$, $D_2$, and $HD$ in a 1:1:2 ratio. Thus, half the atoms end up in $HD$, and the other half split between $H_2$ and $D_2$!
Mastering the Dynamics of Reversibility
Congratulations on completing these 25 highly advanced numericals on Chemical Equilibrium. By conquering simultaneous equilibria, reverse ICE tables, the van't Hoff thermodynamic link, and the mathematical nuance of Le Chatelier's volume expansions, you have built the ultimate foundation for all of physical chemistry. Remember that in equilibrium, mastering the algebra of perfect squares and quadratic approximations is just as important as knowing the chemistry!
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