### Related Formula k = A e^-E_mathrma/RT$k = A e^{-E_{\mathrm{a}}/RT}$ textRate propto frac1E_mathrma$\text{Rate} \propto \frac{1}{E_{\mathrm{a}}}$ ### Core Logic Let us break down each step of the mechanism: 1. **Step 1**: mathrmA rightarrow mathrmB$\mathrm{A} \rightarrow \mathrm{B}$ is **slow**. - Being the rate-determining step, it must have the **highest activation energy barrier** (E_mathrma1$E_{\mathrm{a1}}$). - Since Delta H = +mathrmve$\Delta H = +\mathrm{ve}$ (endothermic), the energy level of intermediate state B must be **higher** than the reactant state A. 2. **Step 2**: mathrmB rightarrow mathrmC$\mathrm{B} \rightarrow \mathrm{C}$ is **fast**. - It has a much **lower activation energy barrier** (E_mathrma2$E_{\mathrm{a2}}$). - Since Delta H = -mathrmve$\Delta H = -\mathrm{ve}$ (exothermic), the energy level of intermediate C is **lower** than state B. 3. **Step 3**: mathrmC rightarrow mathrmD$\mathrm{C} \rightarrow \mathrm{D}$ is **fast**. - It has a **low activation energy barrier** (E_mathrma3$E_{\mathrm{a3}}$). - Since Delta H = -mathrmve$\Delta H = -\mathrm{ve}$ (exothermic), the energy level of final state D is **lower** than state C. 4. **Net Reaction**: Exothermic with "net evolution of heat". - The potential energy of the final product D is **lower** than the initial potential energy of reactant A. ### Step 1: Check Potential Energy Profile Evaluating the transition states and relative energy levels in **Graph (1)**: - The first peak (transition state 1) is clearly the highest (E_mathrma1 > E_mathrma2, E_mathrma3$E_{\mathrm{a1}} > E_{\mathrm{a2}}, E_{\mathrm{a3}}$) implies$\implies$ Step 1 is the slowest. - The intermediate B is higher in energy than A. - Intermediates C and product D are progressively lower in energy. - Product D has lower energy than reactant A (net exothermic).

This perfectly corresponds to **Graph (1)**. ### Pattern Recognition Kinetics shortcut: Slow step = tallest peak. Exothermic step = drop in energy levels of products/intermediates. Endothermic step = climb in energy levels. Use these rules to visually scan energy profiles in under 5 seconds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula t_1/2 = fracln 2k$t_{1/2} = \frac{\ln 2}{k}$ t = frac1k lnleft(fracA_0A_tright)$t = \frac{1}{k} \ln\left(\frac{A_0}{A_t}\right)$ ### Core Logic Let's first analyze the order of the reaction using the concentration-time coordinates from the graph: - At t=5~mathrms$t=5~\mathrm{s}$, concentration [A] = 40~mathrmg~L^-1$[A] = 40~\mathrm{g~L^{-1}}$ - At t=15~mathrms$t=15~\mathrm{s}$, concentration [A] = 20~mathrmg~L^-1$[A] = 20~\mathrm{g~L^{-1}}$ Notice that the concentration drops to exactly half of its value (40 rightarrow 20$40 \rightarrow 20$) over a time interval of Delta t = 15 - 5 = 10~mathrms$\Delta t = 15 - 5 = 10~\mathrm{s}$. - At t=25~mathrms$t=25~\mathrm{s}$, concentration [A] = 10~mathrmg~L^-1$[A] = 10~\mathrm{g~L^{-1}}$ Again, the concentration drops to half (20 rightarrow 10$20 \rightarrow 10$) in another interval of Delta t = 25 - 15 = 10~mathrms$\Delta t = 25 - 15 = 10~\mathrm{s}$. Since the half-life (t_1/2$t_{1/2}$) is constant and independent of the initial concentration, this reaction follows **first-order kinetics**. ### Step 1: Calculate the Rate Constant (k) The half-life of the reaction is t_1/2 = 10~mathrms$t_{1/2} = 10~\mathrm{s}$. k = fracln 210 = frac2.303 log 210 = frac2.303 times 0.301010 approx 0.0693~mathrms^-1$k = \frac{\ln 2}{10} = \frac{2.303 \log 2}{10} = \frac{2.303 \times 0.3010}{10} \approx 0.0693~\mathrm{s^{-1}}$ ### Step 2: Calculate the Time for Decay to 2.5 g/L Given initial concentration A_0 = 50~mathrmg~L^-1$A_0 = 50~\mathrm{g~L^{-1}}$ and target concentration A_t = 2.5~mathrmg~L^-1$A_t = 2.5~\mathrm{g~L^{-1}}$: t = frac2.303k logleft(fracA_0A_tright)$t = \frac{2.303}{k} \log\left(\frac{A_0}{A_t}\right)$ t = frac2.303frac2.303 log 210 logleft(frac502.5right)$t = \frac{2.303}{\frac{2.303 \log 2}{10}} \log\left(\frac{50}{2.5}\right)$ t = frac10log 2 log(20)$t = \frac{10}{\log 2} \log(20)$ t = 10 times fraclog(10) + log(2)log(2)$t = 10 \times \frac{\log(10) + \log(2)}{\log(2)}$ t = 10 times left( frac1 + 0.30100.3010 right)$t = 10 \times \left( \frac{1 + 0.3010}{0.3010} \right)$ t = 10 times left( frac1.30100.3010 right) approx 43.22~mathrms$t = 10 \times \left( \frac{1.3010}{0.3010} \right) \approx 43.22~\mathrm{s}$ Rounding to the nearest integer gives **43** seconds. ### Pattern Recognition Shortcut trick: If t_1/2 = 10~mathrms$t_{1/2} = 10~\mathrm{s}$, any concentration drop of 2^n$2^n$ times takes n times t_1/2$n \times t_{1/2}$ seconds. Here, frac502.5 = 20$\frac{50}{2.5} = 20$. Since 2^4 = 16$2^4 = 16$ (takes 40 s) and 2^5 = 32$2^5 = 32$ (takes 50 s), a drop of 20 times must take slightly over 40 seconds. This confirms our calculation of 43 seconds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula Half-life equation for a Zero-Order reaction layout: t_1/2 = frac[mathrmA]_02K$t_{1/2} = \frac{[\mathrm{A}]_0}{2K}$ Integrated rate law configuration expression: [mathrmA]_t = -Kt + [mathrmA]_0$[\mathrm{A}]_t = -Kt + [\mathrm{A}]_0$ ### Core Logic Let's extract kinetic constants step-by-step: 1. The given plot displays a perfect linear variation passing through the origin: t_1/2 propto [mathrmA]_0$t_{1/2} \propto [\mathrm{A}]_0$. This confirms the transformation process follows **Zero-Order kinetics**. 2. The visual slope equation evaluates as: textSlope = frac12K = 76.92 implies K = frac12 times 76.92 = 0.0065mathrm~molcdot L^-1cdot min^-1$\text{Slope} = \frac{1}{2K} = 76.92 \implies K = \frac{1}{2 \times 76.92} = 0.0065\mathrm{~mol\cdot L^{-1}\cdot min^{-1}}$ ### Step 1: Compute Concentration at t = 10 min Apply the values to the integrated rate law formula track (t = 10mathrm~min$t = 10\mathrm{~min}$, [mathrmA]_0 = 2.5mathrm~molcdot L^-1$[\mathrm{A}]_0 = 2.5\mathrm{~mol\cdot L^{-1}}$): [mathrmA]_10 = -left(frac12 times 76.92right) times 10 + 2.5$[\mathrm{A}]_{10} = -\left(\frac{1}{2 \times 76.92}\right) \times 10 + 2.5$ [mathrmA]_10 = -0.0650 + 2.5 = 2.435mathrm~molcdot L^-1 = 2435 times 10^-3mathrm~molcdot L^-1$[\mathrm{A}]_{10} = -0.0650 + 2.5 = 2.435\mathrm{~mol\cdot L^{-1}} = 2435 \times 10^{-3}\mathrm{~mol\cdot L^{-1}}$ Hence, the target value for the blank field is 2435. ### Pattern Recognition Always identify the reaction order first by inspecting the axes layout: a linear plot of t_1/2$t_{1/2}$ versus [A]_0$[A]_0$ uniquely identifies zero-order behavior, whereas a horizontal flat line implies a first-order path. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula Arrhenius equation is given by: k = A e^-fracE_aR T$k = A e^{-\frac{E_a}{R T}}$ where: - k$k$ is the rate constant - A$A$ is the pre-exponential factor (frequency factor) - E_a$E_a$ is the activation energy ### Core Logic Evaluate each statement: - **A**: Arrhenius equation is an empirical relation that works well for both elementary and complex homogeneous reactions rightarrow$\rightarrow$ *Incorrect*. - **B**: Since the exponential term e^-E_a/RT$e^{-E_a/RT}$ is dimensionless, the pre-exponential factor A$A$ has the exact same unit as the rate constant k$k$ rightarrow$\rightarrow$ *Correct*. - **C**: For low E_a$E_a$, the term e^-E_a/RT$e^{-E_a/RT}$ is large, giving a high rate constant k$k$ and a fast reaction rightarrow$\rightarrow$ *Correct*. - **D**: A$A$ and E_a$E_a$ are assumed to be independent of temperature over a narrow range rightarrow$\rightarrow$ *Incorrect*. - **E**: A$A$ and E_a$E_a$ remain independent parameters of the system, not interdependent rightarrow$\rightarrow$ *Incorrect*. ### Step 1: Select correct statements Statements B and C are correct, matching Option (3). ### Pattern Recognition The exponential factor e^-E_a/RT$e^{-E_a/RT}$ represents the fraction of collisions with energy greater than the activation barrier. As E_a$E_a$ decreases, this fraction grows exponentially, explaining why low-activation pathways (like catalyzed reactions) run much faster. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula textBefore: fracdNdt = k_1 N implies N = N_0 e^k_1 t$\text{Before: } \frac{dN}{dt} = k_1 N \implies N = N_0 e^{k_1 t}$ textAfter: -fracdNdt = k_2 N^2 implies frac1N - frac1N_0 = k_2 t$\text{After: } -\frac{dN}{dt} = k_2 N^2 \implies \frac{1}{N} - \frac{1}{N_0} = k_2 t$ ### Core Logic Let's analyze the kinetics for the two stages: 1. **Before applying medicine**: - Bacterial growth follows 1^textst$1^{\text{st}}$ order kinetics: fracdNdt = k_1 N$\frac{dN}{dt} = k_1 N$. - Integrating this yields: N(t) = N_0 e^k_1 t$N(t) = N_0 e^{k_1 t}$. - The graph of fracNN_0$\frac{N}{N_0}$ vs t$t$ is a rising exponential curve starting from 1$1$ (since at t=0, fracNN_0 = 1$t=0, \frac{N}{N_0} = 1$). 2. **After applying medicine**: - Reductive rate is proportional to the square of existing bacteria (2^textnd$2^{\text{nd}}$ order decay): -fracdNdt = k_2 N^2 implies fracdNN^2 = -k_2 dt$-\frac{dN}{dt} = k_2 N^2 \implies \frac{dN}{N^2} = -k_2 dt$ - Integrating this yields: -frac1N = -k_2 t + C implies frac1N = k_2 t + frac1N_0$-\frac{1}{N} = -k_2 t + C \implies \frac{1}{N} = k_2 t + \frac{1}{N_0}$ N(t) = fracN_01 + N_0 k_2 t$N(t) = \frac{N_0}{1 + N_0 k_2 t}$ - A plot of N$N$ vs t$t$ or fracNN_0$\frac{N}{N_0}$ vs t$t$ for this decay is a hyperbolic curve decreasing gradually. - Option B correctly matches the exponential growth before medicine and the hyperbolic decay after medicine. ### Pattern Recognition First-order growth is an exponential curve (N_0 e^kt$N_0 e^{kt}$), while second-order decay behaves as a rational hyperbolic relationship (1 / (1 + bt)$1 / (1 + bt)$). Option B shows the exact transition from an exponential rise to a hyperbolic decay curve. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics