Masterclass: 25 Solved JEE Advanced Numericals on Chemical Kinetics
Conquer the rates of reactions! This exhaustive guide features complex multi-step problems on the Arrhenius equation, sequential/parallel reactions, steady-state approximations, and advanced integrated rate laws. Click "View Solution" to reveal the answers.
These numericals combine rigorous calculus, collision theory, and reaction mechanisms. Before proceeding, ensure you have mastered the core theories from our fundamental guides in the Class 12 Chemistry Master Hub.
View Solution
Step 1: Write the expression for the overall rate of reaction
According to stoichiometry, the overall rate is defined by dividing individual rates by their stoichiometric coefficients:
$\text{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = +\frac{1}{4} \frac{d[NO_2]}{dt} = +\frac{d[O_2]}{dt}$
Step 2: Calculate Overall Rate
We are given $-\frac{d[N_2O_5]}{dt} = 2.4 \times 10^{-4}$.
$\text{Overall Rate} = \frac{1}{2} (2.4 \times 10^{-4}) = 1.2 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$.
Step 3: Calculate Rate of Appearance of $NO_2$
$\frac{1}{4} \frac{d[NO_2]}{dt} = \text{Overall Rate} = 1.2 \times 10^{-4}$
$\frac{d[NO_2]}{dt} = 4 \times 1.2 \times 10^{-4} = 4.8 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$.
View Solution
Step 1: Write the integrated rate law for first-order
$t = \frac{2.303}{k} \log_{10} \frac{[A_0]}{[A_t]}$
Step 2: Calculate $t_{99.9\%}$
For $99.9\%$ completion, the amount left is $[A_t] = [A_0] - 0.999[A_0] = 0.001[A_0]$.
$t_{99.9\%} = \frac{2.303}{k} \log_{10} \frac{[A_0]}{0.001[A_0]}$
$t_{99.9\%} = \frac{2.303}{k} \log_{10} (1000) = \frac{2.303}{k} \times 3 = \frac{6.909}{k}$.
Step 3: Relate to Half-Life
We know $t_{1/2} = \frac{0.693}{k}$.
Let us find the ratio: $\frac{t_{99.9\%}}{t_{1/2}} = \frac{6.909 / k}{0.693 / k} = \frac{6.909}{0.693} \approx 9.96 \approx 10$.
View Solution
Step 1: Write the two-temperature Arrhenius equation
$\log_{10} \left( \frac{k_2}{k_1} \right) = \frac{{E_a}}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$
Step 2: Identify variables
$k_2 / k_1 = 4$
$T_1 = 293 \text{ K}$, $T_2 = 313 \text{ K}$
$T_2 - T_1 = 20 \text{ K}$
Step 3: Solve for ${E_a}$
$\log_{10} (4) = \frac{{E_a}}{2.303 \times 8.314} \left( \frac{20}{293 \times 313} \right)$
$0.602 = \frac{{E_a}}{19.147} \left( \frac{20}{91709} \right)$
$0.602 = \frac{20 {E_a}}{1755952}$
${E_a} = \frac{0.602 \times 1755952}{20} = 52854 \text{ J mol}^{-1}$.
View Solution
Step 1: Recall Zero-Order properties
For a zero-order reaction, the rate is constant. The amount reacted is directly proportional to time. Integrated rate law: $x = kt$ (where $x$ is the amount reacted).
Step 2: Setup proportionality
$50\%$ completion takes $20 \text{ min}$.
Therefore, $1\%$ completion takes $20 / 50 = 0.4 \text{ min}$.
Step 3: Calculate for $80\%$
Time for $80\%$ = $80 \times 0.4 \text{ min} = 32 \text{ minutes}$.
Exp 1: $[A]=0.1$, $[B]=0.1$, $\text{Rate}=2 \times 10^{-3}$
Exp 2: $[A]=0.1$, $[B]=0.2$, $\text{Rate}=8 \times 10^{-3}$
Exp 3: $[A]=0.2$, $[B]=0.1$, $\text{Rate}=2 \times 10^{-3}$
Determine the rate law and the overall order of the reaction.
View Solution
Step 1: Assume a general rate law
$\text{Rate} = k[A]^x[B]^y$
Step 2: Find order with respect to B ($y$)
Compare Exp 1 and Exp 2 (where $[A]$ is constant at $0.1$):
$\frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[B]_2}{[B]_1}\right)^y$
$\frac{8 \times 10^{-3}}{2 \times 10^{-3}} = \left(\frac{0.2}{0.1}\right)^y \implies 4 = 2^y \implies y = 2$.
Step 3: Find order with respect to A ($x$)
Compare Exp 1 and Exp 3 (where $[B]$ is constant at $0.1$):
$\frac{\text{Rate}_3}{\text{Rate}_1} = \left(\frac{[A]_3}{[A]_1}\right)^x$
$\frac{2 \times 10^{-3}}{2 \times 10^{-3}} = \left(\frac{0.2}{0.1}\right)^x \implies 1 = 2^x \implies x = 0$.
Step 4: Finalize
Overall order = $x + y = 0 + 2 = 2$.
View Solution
Step 1: Formula for optical rotation kinetics
For a first-order reaction monitored by optical rotation:
$k = \frac{2.303}{t} \log_{10} \frac{r_0 - r_{\infty}}{r_t - r_{\infty}}$
Step 2: Substitute values
$r_0 - r_{\infty} = 20 - (-4) = 24^{\circ}$ (This is proportional to initial concentration $[A_0]$).
$r_t - r_{\infty} = 8 - (-4) = 12^{\circ}$ (This is proportional to remaining concentration $[A_t]$).
Step 3: Calculate $k$
$k = \frac{2.303}{100} \log_{10} \left( \frac{24}{12} \right)$
$k = \frac{2.303}{100} \log_{10} (2)$
$k = \frac{2.303 \times 0.301}{100} = 0.00693 \text{ min}^{-1}$.
View Solution
Step 1: Setup Pressure ICE Table
Initial ($t=0$): $A = P_i$, $B = 0$, $C = 0$. Total = $P_i = 100$.
At time $t$: $A = P_i - p$, $B = 2p$, $C = p$.
Total Pressure ($P_t$) = $(P_i - p) + 2p + p = P_i + 2p$.
Step 2: Find $p$ (pressure reacted)
$150 = 100 + 2p \implies 2p = 50 \implies p = 25 \text{ mmHg}$.
Step 3: Find Partial Pressure of A remaining ($P_A$)
$P_A = P_i - p = 100 - 25 = 75 \text{ mmHg}$.
Step 4: Apply first-order gas equation
$k = \frac{2.303}{t} \log_{10} \frac{P_i}{P_A}$
$k = \frac{2.303}{20} \log_{10} \frac{100}{75} = \frac{2.303}{20} \log_{10} (1.333)$
$k = \frac{2.303 \times 0.125}{20} = 0.0144 \text{ min}^{-1}$.
View Solution
Step 1: Write Arrhenius expressions
Uncatalyzed: $\ln k_{uncat} = \ln A - \frac{{E_{a1}}}{RT}$
Catalyzed: $\ln k_{cat} = \ln A - \frac{{E_{a2}}}{RT}$
Step 2: Subtract to find the ratio
$\ln \left(\frac{k_{cat}}{k_{uncat}}\right) = \frac{{E_{a1}} - {E_{a2}}}{RT}$
Converting to base 10:
$\log_{10} \left(\frac{k_{cat}}{k_{uncat}}\right) = \frac{{E_{a1}} - {E_{a2}}}{2.303 RT}$
Step 3: Substitute and Calculate
$\Delta {E_a} = 60,000 - 40,000 = 20,000 \text{ J mol}^{-1}$.
$\log_{10} (\text{Ratio}) = \frac{20000}{2.303 \times 8.314 \times 300} = \frac{20000}{5744.1} = 3.48$
$\text{Ratio} = 10^{3.48} \approx 3020$.
Step 1: $O_3 \rightleftharpoons O_2 + O$ (Fast equilibrium, constants $k_1$ forward, $k_{-1}$ reverse)
Step 2: $O_3 + O \xrightarrow{k_2} 2O_2$ (Slow)
Derive the overall rate law for the disappearance of $O_3$ using the Steady State Approximation for the intermediate atomic Oxygen ($O$).
View Solution
Step 1: Identify the intermediate
Atomic oxygen ($O$) is the highly reactive intermediate. Apply SSA: $\frac{d[O]}{dt} = 0$.
Step 2: Setup SSA equation
Formation of $O$: $k_1[O_3]$.
Consumption of $O$: $k_{-1}[O_2][O] + k_2[O_3][O]$.
$k_1[O_3] - k_{-1}[O_2][O] - k_2[O_3][O] = 0$
Solve for $[O]$: $[O] = \frac{k_1[O_3]}{k_{-1}[O_2] + k_2[O_3]}$
Step 3: Substitute into overall rate
The rate of the overall reaction is defined by the slow step: $\text{Rate} = k_2[O_3][O]$.
$\text{Rate} = k_2[O_3] \left( \frac{k_1[O_3]}{k_{-1}[O_2] + k_2[O_3]} \right) = \frac{k_1 k_2 [O_3]^2}{k_{-1}[O_2] + k_2[O_3]}$.
Step 4: Apply limiting condition
Since Step 2 is very slow, $k_2[O_3] \ll k_{-1}[O_2]$. The denominator simplifies to $k_{-1}[O_2]$.
$\text{Rate} = \frac{k_1 k_2 [O_3]^2}{k_{-1} [O_2]} = k_{obs} \frac{[O_3]^2}{[O_2]}$.
View Solution
Step 1: Understand Temperature Coefficient ($Q_{10}$)
The temperature coefficient tells us the factor by which the rate increases for every $10^{\circ}\text{C}$ (or $10 \text{ K}$) rise in temperature. Here, it doubles ($Q_{10} = 2$).
Step 2: Calculate the number of $10^{\circ}$ intervals
$\Delta T = 80 - 20 = 60^{\circ}\text{C}$.
Number of intervals ($n$) = $60 / 10 = 6$.
Step 3: Calculate the final rate
$\text{Final Rate} = \text{Initial Rate} \times (Q_{10})^n$
$\text{Final Rate} = X \times (2)^6 = X \times 64$.
Path 1: $A \rightarrow B$ with $k_1 = 3 \times 10^{-3} \text{ s}^{-1}$
Path 2: $A \rightarrow C$ with $k_2 = 1 \times 10^{-3} \text{ s}^{-1}$
Calculate the overall half-life of A and the percentage yield of product B.
View Solution
Step 1: Overall Rate Constant ($k_{overall}$)
For parallel first-order paths, the overall rate of disappearance of A is the sum of the individual rates.
$k_{overall} = k_1 + k_2 = (3 + 1) \times 10^{-3} = 4 \times 10^{-3} \text{ s}^{-1}$.
Step 2: Calculate Half-Life
$t_{1/2} = \frac{0.693}{k_{overall}} = \frac{0.693}{4 \times 10^{-3}} = 173.25 \text{ seconds}$.
Step 3: Calculate Percentage Yield
The yield of a specific product is determined by the ratio of its rate constant to the overall rate constant.
$\% \text{ Yield of B} = \frac{k_1}{k_1 + k_2} \times 100 = \frac{3}{4} \times 100 = 75\%$.
View Solution
Step 1: Recognize the requirement
The concentration dropping from $0.4 \text{ M}$ to $0.2 \text{ M}$ is exactly one half-life ($t_{1/2}$).
Step 2: Use Second-Order Half-Life Formula
Unlike first-order reactions, the half-life of a second-order reaction depends inversely on the initial concentration.
$t_{1/2} = \frac{1}{k[A_0]}$
Step 3: Calculation
$t_{1/2} = \frac{1}{(5.0 \times 10^{-2}) \times 0.4} = \frac{1}{0.02} = 50 \text{ seconds}$.
View Solution
Step 1: Equate their Arrhenius equations
$k_1 = k_2 \implies A_1 e^{-{E_{a1}}/RT} = A_2 e^{-{E_{a2}}/RT}$
Step 2: Take natural logarithm on both sides
$\ln A_1 - \frac{{E_{a1}}}{RT} = \ln A_2 - \frac{{E_{a2}}}{RT}$
Step 3: Rearrange to solve for $T$
$\frac{{E_{a2}} - {E_{a1}}}{RT} = \ln A_2 - \ln A_1 = \ln \left( \frac{A_2}{A_1} \right)$
$T = \frac{{E_{a2}} - {E_{a1}}}{R \ln(A_2/A_1)}$
Step 4: Substitute values
$\Delta {E_a} = 60000 - 50000 = 10000 \text{ J/mol}$.
$\ln(A_2/A_1) = \ln(10^{12}/10^{10}) = \ln(10^2) = 2 \times 2.303 = 4.606$.
$T = \frac{10000}{8.314 \times 4.606} = \frac{10000}{38.29} \approx 261.4 \text{ K}$.
View Solution
Step 1: Relationship between $\Delta H$ and ${E_a}$
The enthalpy change of a reaction is strictly the difference between the activation energies of the forward and backward paths.
$\Delta H = {E_{a(forward)}} - {E_{a(backward)}}$
Step 2: Substitute and Solve
$-45 = 30 - {E_{a(backward)}}$
${E_{a(backward)}} = 30 - (-45) = 75 \text{ kJ mol}^{-1}$.
Concept Check: This makes physical sense. An exothermic reaction drops into a deeper potential energy well, so the backward reaction must climb out of that deep well, requiring a massive activation energy.
View Solution
Step 1: Recall the generalized half-life formula
For any $n$-th order reaction (except $n=1$), $t_{1/2} \propto \frac{1}{[A_0]^{n-1}}$.
Step 2: Setup ratio
$\frac{(t_{1/2})_2}{(t_{1/2})_1} = \left( \frac{[A_0]_1}{[A_0]_2} \right)^{n-1}$
Step 3: Substitute the given conditions
Let initial concentration be $a$, so $[A_0]_2 = 2a$.
The new half-life is $1/4$th, so ratio is $1/4$.
$1/4 = \left( \frac{a}{2a} \right)^{n-1}$
$(1/2)^2 = (1/2)^{n-1}$
Step 4: Solve for $n$
Equating the exponents: $2 = n - 1 \implies n = 3$.
View Solution
Step 1: Understand Radioactive Decay
Radioactive decay rigidly follows first-order kinetics. The activity ($A$) is directly proportional to the amount of radioactive nuclei present.
Initial activity $A_0 = 16.0$, Current activity $A_t = 4.0$.
Step 2: Mental Check (Half-Life logic)
The activity dropped from 16 to 8 (1 half-life), then from 8 to 4 (a 2nd half-life). Exactly 2 half-lives have passed.
Step 3: Rigorous Calculation (using decay constant $\lambda$)
$\lambda = \frac{0.693}{5730} = 1.21 \times 10^{-4} \text{ yr}^{-1}$.
$t = \frac{2.303}{\lambda} \log_{10} \frac{A_0}{A_t} = \frac{2.303}{1.21 \times 10^{-4}} \log_{10} \left( \frac{16}{4} \right)$
$t = \frac{2.303}{1.21 \times 10^{-4}} \times \log_{10}(4) = \frac{2.303 \times 0.602}{1.21 \times 10^{-4}} \approx 11460 \text{ years}$.
View Solution
Step 1: Define Overall Rate Constant
$k_{obs} = k_1 + k_2$
Step 2: Apply Arrhenius definition
The rigorous definition of activation energy is ${E_a} = RT^2 \frac{d(\ln k)}{dT}$.
Applying this to $k_{obs}$:
${E_{eff}} = RT^2 \frac{d}{dT} \ln(k_1 + k_2) = RT^2 \frac{1}{k_1 + k_2} \frac{d(k_1 + k_2)}{dT}$
Step 3: Differentiate individual constants
From Arrhenius, $\frac{dk_1}{dT} = k_1 \frac{{E_1}}{RT^2}$ and $\frac{dk_2}{dT} = k_2 \frac{{E_2}}{RT^2}$.
Step 4: Substitute back
${E_{eff}} = \frac{RT^2}{k_1 + k_2} \left[ k_1 \frac{{E_1}}{RT^2} + k_2 \frac{{E_2}}{RT^2} \right]$
The $RT^2$ terms cancel out brilliantly.
1. $Cl_2 \rightleftharpoons 2Cl$ (Fast equilibrium, $K_c$)
2. $NO + Cl \xrightarrow{k_2} NOCl$ (Slow, Rate determining step)
Derive the rate law. What is the order with respect to $Cl_2$?
View Solution
Step 1: Rate determining step
The rate is governed by the slow step: $\text{Rate} = k_2[NO][Cl]$.
However, $[Cl]$ is an intermediate and cannot appear in the final rate law.
Step 2: Use the pre-equilibrium condition
From Step 1, $K_c = \frac{[Cl]^2}{[Cl_2]} \implies [Cl]^2 = K_c[Cl_2] \implies [Cl] = (K_c)^{1/2}[Cl_2]^{1/2}$.
Step 3: Substitute into rate law
$\text{Rate} = k_2 [NO] \left( (K_c)^{1/2} [Cl_2]^{1/2} \right)$
$\text{Rate} = k_{obs} [NO]^1 [Cl_2]^{1/2}$
View Solution
Step 1: Calculate Equilibrium Concentration
At equilibrium, Rate forward = Rate backward. $k_f [A_{eq}] = k_b [B_{eq}]$.
We know $[A_{eq}] + [B_{eq}] = [A_0] = 1$. So $[B_{eq}] = 1 - [A_{eq}]$.
$10^{-2} [A_{eq}] = 3 \times 10^{-2} (1 - [A_{eq}])$
$[A_{eq}] = 3 - 3[A_{eq}] \implies 4[A_{eq}] = 3 \implies [A_{eq}] = 0.75 \text{ M}$.
Step 2: Calculate relaxation time (Half-life of approach)
For a reversible first-order system, the approach to equilibrium behaves like a simple first-order reaction with an effective rate constant $k_{eff} = k_f + k_b$.
$k_{eff} = 10^{-2} + 3 \times 10^{-2} = 4 \times 10^{-2} \text{ s}^{-1}$.
$t_{50\% \text{ to eq}} = \frac{0.693}{k_{eff}} = \frac{0.693}{0.04} = 17.325 \text{ seconds}$.
View Solution
Step 1: Define theoretical maximum
From $k = A e^{-{E_a}/RT}$, the theoretical maximum rate constant (at $T = \infty$ where $e^0 = 1$) is exactly $A$.
The problem asks for the temperature where $k = A/2$.
Step 2: Setup Arrhenius
$A/2 = A e^{-{E_a}/RT}$
$1/2 = e^{-{E_a}/RT}$
Step 3: Solve for T
Take natural log of both sides: $\ln(1/2) = -{E_a}/RT \implies -\ln 2 = -{E_a}/RT$
$\ln 2 = {E_a}/RT \implies T = \frac{{E_a}}{R \ln 2}$.
$T = \frac{50000}{8.314 \times 0.693} = \frac{50000}{5.761} \approx 8696 \text{ K}$.
View Solution
Step 1: Definitions
Half life $t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}$.
Mean life ($\tau$) is the sum of lifetimes of all atoms divided by total atoms, derived via calculus as $\tau = \frac{1}{k}$.
Step 2: Compare
$\tau = \frac{1}{k} = 1.44 \times \frac{0.693}{k} = 1.44 \times t_{1/2}$.
Since $1.44 > 1$, average life is greater than half-life.
View Solution
Step 1: Use the $t_{max}$ formula
From solving the complex differential equations for consecutive kinetics, the time of maximum intermediate concentration is given by:
$t_{max} = \frac{\ln(k_1/k_2)}{k_1 - k_2}$
Step 2: Substitute and solve
$t_{max} = \frac{\ln(2/1)}{2 - 1} = \frac{\ln 2}{1} = 0.693 \text{ seconds}$.
View Solution
Step 1: Understand the Titration Data
At $t=0$, the volume $V_0$ corresponds to the acid catalyst alone.
At $t=\infty$, $V_{\infty}$ corresponds to the catalyst + total acetic acid produced (which relates to initial ester $[A_0]$).
At $t$, $V_t$ corresponds to catalyst + acetic acid produced so far.
Step 2: Proportionalities
$[A_0] \propto (V_{\infty} - V_0) = 30 - 10 = 20$.
$[A_t] \propto (V_{\infty} - V_t) = 30 - 15 = 15$.
Step 3: Calculate $k$
$k = \frac{2.303}{t} \log_{10} \frac{V_{\infty} - V_0}{V_{\infty} - V_t}$
$k = \frac{2.303}{10} \log_{10} \left(\frac{20}{15}\right) = \frac{2.303}{10} \log_{10} (1.333) = 0.0287 \text{ min}^{-1}$.
View Solution
Step 1: Setup Calculus
$-\frac{d[A]}{dt} = k[A]^{3/2} \implies [A]^{-3/2} d[A] = -k dt$
Step 2: Integrate
$\int_{[A_0]}^{[A_t]} [A]^{-1.5} d[A] = -k \int_0^t dt$
$\left[ \frac{[A]^{-0.5}}{-0.5} \right] = -kt \implies -2 \left( \frac{1}{\sqrt{[A_t]}} - \frac{1}{\sqrt{[A_0]}} \right) = -kt$
$kt = 2 \left( \frac{1}{\sqrt{[A_t]}} - \frac{1}{\sqrt{[A_0]}} \right)$
Step 3: Calculate time
$0.1 \times t = 2 \left( \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{4}} \right) = 2 \left( 1 - 0.5 \right) = 2(0.5) = 1.0$
$t = 1.0 / 0.1 = 10 \text{ seconds}$.
View Solution
Step 1: Match with Arrhenius equation
$\ln k = -\frac{{E_a}}{R} \left(\frac{1}{T}\right) + \ln A$.
This is in the form $y = mx + c$, where $y = \ln k$ and $x = 1/T$.
Step 2: Extract ${E_a}$ from Slope
Slope ($m$) = $-{E_a}/R$.
$-12000 = -{E_a} / 8.314 \implies {E_a} = 12000 \times 8.314 = 99768 \text{ J/mol}$.
Step 3: Extract $A$ from Intercept
Intercept ($c$) = $\ln A = 25$.
$A = e^{25}$.
Mastering the Rates of Chemistry
Congratulations on tackling these 25 highly advanced numericals on Chemical Kinetics. While thermodynamics tells us whether a reaction is possible, kinetics governs whether it happens in a microsecond or a millennium. From navigating complex Arrhenius temperature shifts to applying the Steady-State Approximation on fleeting intermediates, you now possess the mathematical rigorousness demanded by top-tier engineering and medical entrance exams. Keep refining your integration skills, and always watch your units!
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