Drug X becomes ineffective after 50\%$50\%$ decomposition. The original concentration of drug in a bottle was 16text mg/mL$16\text{ mg/mL}$ which becomes 4text 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 Exponential growth equation model: N = N_0 e^Kt$N = N_0 e^{Kt}$ Normalized configuration formula: fracNN_0 = e^Kt$$\frac{N}{N_0} = e^{Kt}$$ ### Core Logic Radioactive decay follows a decreasing exponential path (N = N_0 e^-lambda t$N = N_0 e^{-\lambda t}$). Conversely, cell culture growth functions via an *increasing* exponential pattern because the rate of growth is directly proportional to the current population size (dN/dt = KN$dN/dt = KN$). This results in an exponential curve that starts at fracNN_0 = 1$\frac{N}{N_0} = 1$ when t = 0$t = 0$ and curves sharply upward over time. ### Step 1: Finding the Matching Curve Plotting fracNN_0$\frac{N}{N_0}$ against time shows an upward-clinging exponential profile starting from 1$1$, which perfectly matches the curve in option (4).

### Pattern Recognition The expression e^Kt$e^{Kt}$ dictates an exponential increase. Ensure the curve starts from a non-zero value (1$1$) at t=0$t=0$, as fracN_0N_0 = 1$\frac{N_0}{N_0} = 1$, rather than starting from the origin (0$0$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula textRate = k cdot [textReactants]^textorder$$\text{Rate} = k \cdot [\text{Reactants}]^{\text{order}}$$ ### Core Logic The slowest elementary step controls the net kinetic pathway rate law : textRate = k_2[mathrmA][mathrmB2] quad dots textEquation (1)$$\text{Rate} = k_2[\mathrm{A}][\mathrm{B}2] \quad \dots \text{Equation (1)}$$ Since [mathrmA]$[\mathrm{A}]$ behaves as a transient intermediate species, replace it using the prior fast equilibrium step : frack_1k-1 = frac[mathrmA]^2[mathrmA2] implies [mathrmA]^2 = left(frack_1k-1right) [mathrmA2]$$\frac{k_1}{k{-1}} = \frac{[\mathrm{A}]^2}{[\mathrm{A}2]} \implies [\mathrm{A}]^2 = \left(\frac{k_1}{k{-1}}\right) [\mathrm{A}2]$$ [mathrmA] = sqrtfrack_1k-1 cdot [mathrmA2]^1/2$$[\mathrm{A}] = \sqrt{\frac{k_1}{k{-1}}} \cdot [\mathrm{A}2]^{1/2}$$ Substitute [mathrmA]$[\mathrm{A}]$ back into Equation (1) : textRate = k_2 sqrtfrack_1k-1 cdot [mathrmA2]^1/2[mathrmB2]$$\text{Rate} = k_2 \sqrt{\frac{k_1}{k{-1}}} \cdot [\mathrm{A}2]^{1/2}[\mathrm{B}2]$$ Sum of powers determining overall order: textOrder = frac12 + 1 = 1.5$$\text{Order} = \frac{1}{2} + 1 = 1.5$$ Hence, Option (1) is correct. ### Pattern Recognition Whenever a fast initial step dissociates a molecule into matching independent halves, it always injects a fractional order component of 0.5$0.5$ relative to that parent species.

### Related Formula N = fracN_02^n$$N = \frac{N_0}{2^n}$$ where n = fractt_1/2$n = \frac{t}{t_{1/2}}$ (number of half-lives). Alternatively, using the first-order decay formula: t = frac2.303lambda log left( fracN_0N_t right)$$t = \frac{2.303}{\lambda} \log \left( \frac{N_0}{N_t} \right)$$ where lambda = frac0.693t_1/2$\lambda = \frac{0.693}{t_{1/2}}$. ### Core Logic The atmospheric ratio of ^14mathrmC/^12mathrmC$^{14}\mathrm{C}/^{12}\mathrm{C}$ acts as the initial activity or amount (N_0$N_0$) when the tree was alive. The current ratio in the wood represents the amount left at time t$t$ (N_t$N_t$). Given that N_t = frac18 N_0$N_t = \frac{1}{8} N_0$. ### Step 1: Calculate Half-lives fracN_tN_0 = frac18$$\frac{N_t}{N_0} = \frac{1}{8}$$ left(frac12right)^n = frac18 = left(frac12right)^3$$ \left(\frac{1}{2}\right)^n = \frac{1}{8} = \left(\frac{1}{2}\right)^3$$ So, the number of half-lives passed, n = 3$n = 3$. ### Step 2: Calculate Age t = n times t_1/2$$t = n \times t_{1/2}$$ t = 3 times 5730 text years$$t = 3 \times 5730 \text{ years}$$ t = 17190 text years$$t = 17190 \text{ years}$$ ### Pattern Recognition Whenever the remaining fraction is a perfect power of 1/2$1/2$ (like 1/2, 1/4, 1/8, 1/16$1/2, 1/4, 1/8, 1/16$), just find the exponent n$n$ and multiply by t_1/2$t_{1/2}$. Here, 1/8 = (1/2)^3 rightarrow 3$1/8 = (1/2)^3 \rightarrow 3$ half-lives. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula k = frac0.693, t_1/2 quad textand quad t = frac2.303, k log_10 left(fraca, a-xright)$$k = \frac{0.693}, {t_{1/2}} \quad \text{and} \quad t = \frac{2.303}, {k} \log_{10} \left(\frac{a}, {a-x}\right)$$ ### Core Logic Given t_1/2 = 36text hours$t_{1/2} = 36\text{ hours}$, calculate the decay constant (k$k$): k = frac0.693, 36 = 0.01925text hr^-1$$k = \frac{0.693}, {36} = 0.01925\text{ hr}^{-1}$$ We want to find the fraction remaining after 1text day = 24text hours$1\text{ day} = 24\text{ hours}$: log_10 left(fraca, a-xright) = frack times t, 2.303 = frac0.01925 times 24, 2.303 = 0.2006$$\log_{10} \left(\frac{a}, {a-x}\right) = \frac{k \times t}, {2.303} = \frac{0.01925 \times 24}, {2.303} = 0.2006$$ ### Step 1: Antilog Application Taking the antilog on both sides: fraca, a-x = 1.587 implies textFraction remaining left(fraca-x, aright) = frac1, 1.587 approx 0.6301$$\frac{a}, {a-x} = 1.587 \implies \text{Fraction remaining } \left(\frac{a-x}, {a}\right) = \frac{1}, {1.587} \approx 0.6301$$ Expressing the remaining fraction in the requested format: 0.6301 = 63 times 10^-2$$0.6301 = 63 \times 10^{-2}$$ Thus, the required integer value is **63**. ### Pattern Recognition Ensure all time variables are in matching units (hours) before substituting values into first-order kinetic equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula Rate law relation expression: R = k[textHI]^n$$R = k[\text{HI}]^n$$ where n$n$ represents the overall reaction order indicator. ### Step 1: Set up ratios using data subsets Comparing data from experiment 1 and experiment 2: fracR_2R_1 = frac3.0 times 10^-37.5 times 10^-4 = left(frac0.010.005right)^n$$\frac{R_2}{R_1} = \frac{3.0 \times 10^{-3}}{7.5 \times 10^{-4}} = \left(\frac{0.01}{0.005}\right)^n$$ 4 = (2)^n$4 = (2)^n$ 2^2 = 2^n implies n = 2$$2^2 = 2^n \implies n = 2$$ ### Pattern Recognition Doubling concentration (0.005 rightarrow 0.01$0.005 \rightarrow 0.01$) increases the reaction rate by 4 times (7.5 times 10^-4 rightarrow 3.0 times 10^-3$7.5 \times 10^{-4} \rightarrow 3.0 \times 10^{-3}$). Hence, it is a clear second-order (2^2 = 4$2^2 = 4$) dynamic pattern. ### Chapter Mix Class 12 Chemistry: Chemical Kinetics