For a given reaction mathrmRrightarrow mathrmP,mathrmt_1 / 2$\mathrm{R}\rightarrow \mathrm{P},\mathrm{t}_{1 / 2}$ is related to [mathrmA]_0$[\mathrm{A}]_0$ as given in table :

[A]_0 / mathrmmol\,L^-1$[A]_0 / \mathrm{mol\,L^{-1}}$	t_1/2 / mathrmmin$t_{1/2} / \mathrm{min}$
0.100	200
0.025	100

Given: log 2 = 0.30$\log 2 = 0.30$ Which of the following is true? A. The order of the reaction is frac12$\frac{1}{2}$ . B. If [mathrmA]_0$[\mathrm{A}]_0$ is 1mathrmM$1\mathrm{M}$ , then mathrmt_1/2$\mathrm{t}_{1/2}$ is 200sqrt10$200\sqrt{10}$ min C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 mathrmM$0.100 \mathrm{M}$ to 0.500 mathrmM$0.500 \mathrm{M}$ . D. t_1 / 2$t_{1 / 2}$ is 800 mathrm~min$800 \mathrm{~min}$ for [mathrmA]_0 = 1.6 mathrmM$[\mathrm{A}]_0 = 1.6 \mathrm{M}$ Choose the correct answer from the options given below:

### Related Formula ln k = ln A - fracE_aRT$$\ln k = \ln A - \frac{E_a}{RT}$$ ### Core Logic For Reaction 1: ln k_1 = ln A - fracE_1RT$\ln k_1 = \ln A - \frac{E_1}{RT}$ For Reaction 2: ln k_2 = ln A - fracE_2RT$\ln k_2 = \ln A - \frac{E_2}{RT}$ (Pre-exponential factor A$A$ is the same). Subtracting the first from the second: ln k_2 - ln k_1 = -fracE_2RT - left(-fracE_1RTright)$$\ln k_2 - \ln k_1 = -\frac{E_2}{RT} - \left(-\frac{E_1}{RT}\right)$$ ln left(frack_2k_1right) = fracE_1 - E_2RT$$\ln \left(\frac{k_2}{k_1}\right) = \frac{E_1 - E_2}{RT}$$ Given that E_1$E_1$ exceeds E_2$E_2$ by 20text kJ mol^-1$20\text{ kJ mol}^{-1}$, E_1 - E_2 = 20000text J mol^-1$E_1 - E_2 = 20000\text{ J mol}^{-1}$. T = 300text K$T = 300\text{ K}$, R = 8.3text J K^-1text mol^-1$R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}$. ln left(frack_2k_1right) = frac200008.3 times 300 = frac2008.3 times 3 = frac20024.9$$\ln \left(\frac{k_2}{k_1}\right) = \frac{20000}{8.3 \times 300} = \frac{200}{8.3 \times 3} = \frac{200}{24.9}$$ ln left(frack_2k_1right) = 8.032$$\ln \left(\frac{k_2}{k_1}\right) = 8.032$$ Rounding off to nearest integer gives 8. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### 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