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 k = frac2.303t logleft(fracP_0P_Aright)$$k = \frac{2.303}{t} \log\left(\frac{P_0}{P_A}\right)$$ ### Core Logic For the reaction: mathrmA(g) rightarrow 2mathrmB(g) + mathrmC(g)$\mathrm{A(g)} \rightarrow 2\mathrm{B(g)} + \mathrm{C(g)}$ - At t=0$t=0$, pressure of A = P_0$A = P_0$, while B = 0$B = 0$ and C = 0$C = 0$. - At t=infty$t=\infty$, A$A$ is completely consumed, leaving 2P_0$2P_0$ of B$B$ and P_0$P_0$ of C$C$. P_infty = 3P_0 = 240text mm Hg implies P_0 = 80text mm Hg$$P_{\infty} = 3P_0 = 240\text{ mm Hg} \implies P_0 = 80\text{ mm Hg}$$ This confirms option (A) is correct. At any time t$t$, pressure of A = P_0 - x$A = P_0 - x$, B = 2x$B = 2x$, C = x$C = x$. P_t = P_0 + 2x = 80 + 2x$$P_t = P_0 + 2x = 80 + 2x$$ At t=10text min$t=10\text{ min}$, P_10 = 160text mm Hg$P_{10} = 160\text{ mm Hg}$: 80 + 2x = 160 implies x = 40text mm Hg$$80 + 2x = 160 \implies x = 40\text{ mm Hg}$$ Thus, partial pressure of A$A$ after 10text min$10\text{ min}$ is: P_A = P_0 - x = 80 - 40 = 40text mm Hg$$P_A = P_0 - x = 80 - 40 = 40\text{ mm Hg}$$ This confirms option (D) is correct. Now, calculate the rate constant k$k$: k = frac110 lnleft(frac8040right) = fracln 210 = 0.0693text min^-1$$k = \frac{1}{10} \ln\left(\frac{80}{40}\right) = \frac{\ln 2}{10} = 0.0693\text{ min}^{-1}$$ Therefore, option (C) which states k = 1.693text min^-1$k = 1.693\text{ min}^{-1}$ is incorrect. ### Pattern Recognition At t=infty$t=\infty$, the total pressure is 3$3$ times the initial pressure of A$A$. So, P_0 = P_infty / 3 = 80text mm Hg$P_0 = P_\infty / 3 = 80\text{ mm Hg}$. Half-life t_1/2 = 10text min$t_{1/2} = 10\text{ min}$ since P_A$P_A$ drops from 80$80$ to 40$40$ in 10text min$10\text{ min}$. Thus, k = 0.693 / 10 = 0.0693text min^-1$k = 0.693 / 10 = 0.0693\text{ min}^{-1}$. ### 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

### 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