Decomposition of A is a first order reaction at T(K) and is given by textA(textg) rightarrow textB(textg) + textC(textg)$\text{A}(\text{g}) \rightarrow \text{B}(\text{g}) + \text{C}(\text{g})$. In a closed 1 L vessel, 1 bar textA(textg)$\text{A}(\text{g})$ is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in textmin^-1$\text{min}^{-1}$) of the reaction? (log 2 = 0.3)$(\log 2 = 0.3)$

### 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 = frac2.303t logleft(fracP_02P_0 - P_texttotalright)$$k = \frac{2.303}{t} \log\left(\frac{P_0}{2P_0 - P_{\text{total}}}\right)$$ or equivalent first-order expression. ### Core Logic For textA(textg) rightarrow textB(textg) + textC(textg)$\text{A}(\text{g}) \rightarrow \text{B}(\text{g}) + \text{C}(\text{g})$: - Initial pressure: P_0 = 1 text bar$P_0 = 1 \text{ bar}$ - At time t = 100 text min$t = 100 \text{ min}$, pressure of A remaining = 1 - P$= 1 - P$, pressures of B and C = P$= P$. - Total pressure P_texttotal = 1 - P + P + P = 1 + P = 1.5 text bar implies P = 0.5 text bar$P_{\text{total}} = 1 - P + P + P = 1 + P = 1.5 \text{ bar} \implies P = 0.5 \text{ bar}$. Remaining pressure of A = 1 - 0.5 = 0.5 text bar$= 1 - 0.5 = 0.5 \text{ bar}$. ### Step 1: Calculating Rate Constant k = frac1100 lnleft(frac10.5right) = frac0.693100 = 6.9 times 10^-3 text min^-1$$k = \frac{1}{100} \ln\left(\frac{1}{0.5}\right) = \frac{0.693}{100} = 6.9 \times 10^{-3} \text{ min}^{-1}$$ ### Pattern Recognition Sees: gaseous phase first-order kinetics with total pressure data. Trap: Confusing partial pressure of reactant with total pressure in rate expressions. ### 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