Reaction mathrmA(g) rightarrow 2mathrmB(g) + mathrmC(g)$\mathrm{A(g)} \rightarrow 2\mathrm{B(g)} + \mathrm{C(g)}$ is a first order reaction. It was started with pure A.

t / textmin$t / \text{min}$	Pressure of system at time t / textmm Hg$t / \text{mm Hg}$
10	160
infty$\infty$	240

Which of the following options is incorrect?

### 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

### Related Formula For a first-order reaction: t = frac2.303k logleft(fracC_0C_tright)$$t = \frac{2.303}{k} \log\left(\frac{C_0}{C_t}\right)$$ Alternatively, using half-life t_50\%$t_{50\%}$: C_t = fracC_02^n$$C_t = \frac{C_0}{2^n}$$ where n$n$ is the number of half-lives (n = fractt_50\%$n = \frac{t}{t_{50\%}}$). ### Core Logic Step 1: Calculate t_1$t_1$ for decomposition to frac14$\frac{1}{4}$ of initial concentration: C_t = fracC_04 = fracC_02^2 implies n = 2$$C_t = \frac{C_0}{4} = \frac{C_0}{2^2} \implies n = 2$$ t_1 = 2 cdot t_50\%$$t_1 = 2 \cdot t_{50\%}$$ Step 2: Calculate t_2$t_2$ for decomposition to frac18$\frac{1}{8}$ of initial concentration: C_t = fracC_08 = fracC_02^3 implies n = 3$$C_t = \frac{C_0}{8} = \frac{C_0}{2^3} \implies n = 3$$ t_2 = 3 cdot t_50\%$$t_2 = 3 \cdot t_{50\%}$$ Step 3: Find the ratio fract_1t_2$\frac{t_1}{t_2}$: fract_1t_2 = frac2 cdot t_50\%3 cdot t_50\% = frac23$$\frac{t_1}{t_2} = \frac{2 \cdot t_{50\%}}{3 \cdot t_{50\%}} = \frac{2}{3}$$ ### Pattern Recognition For first-order kinetics, every step of concentration halving takes exactly one half-life (t_50\%$t_{50\%}$). textInitial xrightarrowt_50\% frac12 xrightarrowt_50\% frac14 quad text(Total 2 text half-lives)$\text{Initial} \xrightarrow{t_{50\%}} \frac{1}{2} \xrightarrow{t_{50\%}} \frac{1}{4} \quad \text{(Total } 2 \text{ half-lives)}$ frac14 xrightarrowt_50\% frac18 quad text(Total 3 text half-lives)$\frac{1}{4} \xrightarrow{t_{50\%}} \frac{1}{8} \quad \text{(Total } 3 \text{ half-lives)}$ Therefore, the ratio is simply the ratio of the number of half-lives: 2 : 3$2 : 3$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula N_t = N_0 left(frac12 ight)^n$$N_t = N_0 left(\frac{1}{2} ight)^n$$ ### Core Logic Let's track concentration reductions: 16text mg/mL xrightarrowt_1/2 8text mg/mL xrightarrowt_1/2 4text mg/mL$$16\text{ mg/mL} xrightarrow{t_{1/2}} 8\text{ mg/mL} xrightarrow{t_{1/2}} 4\text{ mg/mL}$$ This total progression constitutes exactly 2$2$ half-lives (n = 2$n = 2$). 2 cdot t_1/2 = 12text months implies t_1/2 = 6text months$$2 cdot t_{1/2} = 12\text{ months} implies t_{1/2} = 6\text{ months}$$ Since the drug becomes ineffective right after 50\%$50\%$ decomposition, its functional expiry limit is exactly 1$1$ half-life period. textExpiry time = t_1/2 = 6text months$$\text{Expiry time} = t_{1/2} = 6\text{ months}$$ ### Pattern Recognition For multi-step concentration halving, bypass complex integrated logarithmic rate expressions by directly applying integer half-life steps (16 ightarrow 8 ightarrow 4$16 ightarrow 8 ightarrow 4$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula The dependence of half-life on initial concentration is given by: t_1/2 propto frac1[A]_0^n-1$$t_{1/2} \propto \frac{1}{[A]_0^{n-1}}$$ ### Step 1: Finding the reaction order (n) Using the values provided: frac(t_1/2)_1(t_1/2)_2 = left( frac[A]_0,2[A]_0,1 right)^n-1$$\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{[A]_{0,2}}{[A]_{0,1}} \right)^{n-1}$$ frac200100 = left( frac0.0250.100 right)^n-1 Rightarrow 2 = left( frac14 right)^n-1$$\frac{200}{100} = \left( \frac{0.025}{0.100} \right)^{n-1} \Rightarrow 2 = \left( \frac{1}{4} \right)^{n-1}$$ 2 = 2^-2(n-1) Rightarrow 1 = -2n + 2 Rightarrow n = frac12$$2 = 2^{-2(n-1)} \Rightarrow 1 = -2n + 2 \Rightarrow n = \frac{1}{2}$$ Hence, statement A is correct. ### Step 2: Checking half-life at other concentrations Since n = frac12$n = \frac{1}{2}$, t_1/2 propto sqrt[A]_0$t_{1/2} \propto \sqrt{[A]_0}$. - For [A]_0 = 1\,mathrmM$[A]_0 = 1\,\mathrm{M}$: frac200t_1/2 = sqrtfrac0.11 Rightarrow t_1/2 = 200sqrt10\,mathrmmin$$\frac{200}{t_{1/2}} = \sqrt{\frac{0.1}{1}} \Rightarrow t_{1/2} = 200\sqrt{10}\,\mathrm{min}$$ Hence, statement B is correct. - For [A]_0 = 1.6\,mathrmM$[A]_0 = 1.6\,\mathrm{M}$: frac200t_1/2 = sqrtfrac0.11.6 = sqrtfrac116 = frac14 Rightarrow t_1/2 = 800\,mathrmmin$$\frac{200}{t_{1/2}} = \sqrt{\frac{0.1}{1.6}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \Rightarrow t_{1/2} = 800\,\mathrm{min}$$ Hence, statement D is correct. ### Pattern Recognition Sees: Half-life reducing as initial concentration decreases. Trap: Assuming all reactions are first or zero order without calculations. Shortcut: Reduction of [A]_0$[A]_0$ by 4 causes reduction of t_1/2$t_{1/2}$ by 2 rightarrow$\rightarrow$ indicates a square root dependence (t_1/2 propto sqrtA_0$t_{1/2} \propto \sqrt{A_0}$), which implies n = 0.5$n = 0.5$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics