Drug X becomes ineffective after 50%$50\%$ decomposition. The original concentration of drug in a bottle was 16 mg/mL$16\text{ mg/mL}$ which becomes 4 mg/mL$4\text{ mg/mL}$ in 12 months. The expiry time of the drug in months is (Assume that the decomposition of the drug follows first order kinetics). (1) 12 (2) 2 (3) 3 (4) 6

### Related Formula ln k = ln A - (Eₐ)/(RT)$$\ln k = \ln A - \frac{E_a}{RT}$$ ### Core Logic For Reaction 1: ln k₁ = ln A - (E₁)/(RT)$\ln k_1 = \ln A - \frac{E_1}{RT}$ For Reaction 2: ln k₂ = ln A - (E₂)/(RT)$\ln k_2 = \ln A - \frac{E_2}{RT}$ (Pre-exponential factor A$A$ is the same). Subtracting the first from the second: ln k₂ - ln k₁ = -(E₂)/(RT) - (-(E₁)/(RT))$$\ln k_2 - \ln k_1 = -\frac{E_2}{RT} - \left(-\frac{E_1}{RT}\right)$$ ln ((k₂)/(k₁)) = (E₁ - E₂)/(RT)$$\ln \left(\frac{k_2}{k_1}\right) = \frac{E_1 - E_2}{RT}$$ Given that E₁$E_1$ exceeds E₂$E_2$ by 20 kJ mol⁻¹$20\text{ kJ mol}^{-1}$, E₁ - E₂ = 20000 J mol⁻¹$E_1 - E_2 = 20000\text{ J mol}^{-1}$. T = 300 K$T = 300\text{ K}$, R = 8.3 J K⁻¹ mol⁻¹$R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}$. ln ((k₂)/(k₁)) = (20000)/(8.3 × 300) = (200)/(8.3 × 3) = (200)/(24.9)$$\ln \left(\frac{k_2}{k_1}\right) = \frac{20000}{8.3 \times 300} = \frac{200}{8.3 \times 3} = \frac{200}{24.9}$$ ln ((k₂)/(k₁)) = 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 = (2.303)/(t) ( P₀2P₀ - Ptotal)$$k = \frac{2.303}{t} \log\left(\frac{P_0}{2P_0 - P_{\text{total}}}\right)$$ or equivalent first-order expression. ### Core Logic For A(g) arrow B(g) + C(g)$\text{A}(\text{g}) \rightarrow \text{B}(\text{g}) + \text{C}(\text{g})$: - Initial pressure: P₀ = 1 bar$P_0 = 1 \text{ bar}$ - At time t = 100 min$t = 100 \text{ min}$, pressure of A remaining = 1 - P$= 1 - P$, pressures of B and C = P$= P$. - Total pressure Ptotal = 1 - P + P + P = 1 + P = 1.5 bar P = 0.5 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 bar$= 1 - 0.5 = 0.5 \text{ bar}$. ### Step 1: Calculating Rate Constant k = (1)/(100) ln((1)/(0.5)) = (0.693)/(100) = 6.9 × 10⁻³ 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 Rate = k[A]$$\text{Rate} = k[A]$$ Where [A]$[A]$ is the molar concentration of reactant A. ### Core Logic For a first order reaction, the rate is directly proportional to the concentration of the reactant, not the total volume or the total number of moles. In Run 1: [A] = 10 M$[A] = 10\text{ M}$. Rate₁ = k(10)$\text{Rate}_1 = k(10)$. In Run 2: [A] = 10 M$[A] = 10\text{ M}$. Rate₂ = k(10)$\text{Rate}_2 = k(10)$. (Volume increased, but molarity is identical). In Run 3: 100 mL of 10 M solution is diluted with 100 mL water. New volume is 200 mL. M₁ V₁ = M₂ V₂ 10 × 100 = M₂ × 200 M₂ = 5 M$M_1 V_1 = M_2 V_2 \implies 10 \times 100 = M_2 \times 200 \implies M_2 = 5\text{ M}$. [A] = 5 M$[A] = 5\text{ M}$. Rate₃ = k(5)$\text{Rate}_3 = k(5)$. ### Step 1: Final Conclusion Therefore, Rate₃ < Rate₁ = Rate₂$\text{Rate}_3 < \text{Rate}_1 = \text{Rate}_2$. ### Pattern Recognition Rate laws depend exclusively on molar concentration (M). Diluting the solution decreases rate, while just taking a larger volume of the same stock solution keeps the rate identical. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula k = A e-Eₐ/RT$$k = A e^{-E_a/RT}$$ ### Core Logic Equate the two rate constants: k₁ = k₂$k_{1} = k_{2}$ 10⁶e(-30000)/(T) = 10⁴e(-24000)/(T)$$10^{6}e^{\frac{-30000}{T}} = 10^{4}e^{\frac{-24000}{T}}$$ Divide both sides by 10⁴$10^4$: 10²e(-30000)/(T) = e(-24000)/(T)$$10^{2}e^{\frac{-30000}{T}} = e^{\frac{-24000}{T}}$$ Divide both sides by e(-30000)/(T)$e^{\frac{-30000}{T}}$: 100 = e(-24000)/(T)e(-30000)/(T)$$100 = \frac{e^{\frac{-24000}{T}}}{e^{\frac{-30000}{T}}}$$ 100 = e(6000)/(T)$$100 = e^{\frac{6000}{T}}$$ ### Step 1: Solve for T Take the natural logarithm (ln$\ln$) on both sides: ln(100) = (6000)/(T)$$\ln(100) = \frac{6000}{T}$$ 2 ln(10) = (6000)/(T)$$2 \ln(10) = \frac{6000}{T}$$ Substitute ln 10 = 2.303$\ln 10 = 2.303$: 2 × 2.303 = (6000)/(T)$$2 \times 2.303 = \frac{6000}{T}$$ 4.606 = (6000)/(T)$$4.606 = \frac{6000}{T}$$ T = (6000)/(4.606) = 1302.64 K$$T = \frac{6000}{4.606} = 1302.64\text{ K}$$ ### Step 2: Rounding Nearest integer is 1303. ### Pattern Recognition Simple exponential equating. Group powers of 10 on one side and exponentials on the other, then apply natural log. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula k = A e-Eₐ/RT$$k = A e^{-E_a/RT}$$ ln k = ln A - (Eₐ)/(RT) k = A - (Eₐ)/(2.303 R T)$$\ln k = \ln A - \frac{E_a}{RT} \implies \log k = \log A - \frac{E_a}{2.303 R T}$$ ### Core Logic Statement (A): BANNED - e-Eₐ/RT$e^{-E_a/RT}$ represents fraction of molecules with energy ≥ Eₐ$\ge E_a$ (not less). Statement (B): CORRECT - Lower activation energy Eₐ$E_a$ increases the rate constant k$k$, speeding up the reaction. Statement (C): CORRECT - For most reactions, a 10^$10^\circ\text{C}$ rise in temperature doubles the rate coefficient. Statement (D): INCORRECT - Plot of k$\log k$ vs 1/T$1/T$ has slope equal to -(Eₐ)/(2.303 R)$-\frac{E_a}{2.303 R}$ (the factor 2.303$2.303$ is missing). ### Pattern Recognition Sees: Arrhenius statements. Shortcut: Watch for missing 2.303$2.303$ in $\log$ slope equation and 'less than' vs 'greater than' in exponential fraction definition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics