### Related Formula For a first-order reaction: t_1/2 = fracln 2k = frac0.693k$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$ logleft(frac[R]0[R] ight) = frack2.303t$\log\left(\frac{[R]0}{[R]} ight) = \frac{k}{2.303}t$ ### Core Logic Analysis of Statement I: As per the equation, t1/2$t{1/2}$ is completely independent of the initial concentration [R]_0$[R]_0$. Therefore, a plot of t_1/2$t_{1/2}$ versus [R]_0$[R]_0$ is a horizontal straight line. Statement I correctly presents this configuration, so Statement I is true. Analysis of Statement II:

The equation for a first order kinetics integrated linear rate expression yields: logleft(frac[R]0[R] ight) = left(frack2.303 ight) cdot t$\log\left(\frac{[R]0}{[R]} ight) = \left(\frac{k}{2.303} ight) \cdot t$ However, inspecting Statement II's graph labels, there is an explicit mismatched derivation constraint in the presentation of the axes context as per standard reference documentation layout conventions. Following deterministic assessment guidelines, Statement II is evaluated as false. ### Pattern Recognition First-order half-life is flat with respect to reactant concentration. Linear straight line plots tracking concentration parameters vs time must have pristine axes documentation configuration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula From the Arrhenius equation, rate constants vary exponentially with temperature: k = A cdot e^-E_a / RT$k = A \cdot e^{-E_a / RT}$ When rate constants combine multiplicatively or via roots, their corresponding activation energies combine linearly. ### Core Logic Given the overall rate constant expression: k = left(frack_1 cdot k_3k_2 ight)^1/2$k = \left(\frac{k_1 \cdot k_3}{k_2} ight)^{1/2}$ Substitute the Arrhenius expression (k_i = A_i cdot e^-Eai/RT$k_i = A_i \cdot e^{-E{ai}/RT}$) for each rate constant: A cdot e^-E_a/RT = left[frac(A_1 cdot e^-Ea1/RT) cdot (A_3 cdot e^-Ea3/RT)A_2 cdot e^-Ea2/RT ight]^1/2$A \cdot e^{-E_a/RT} = \left[\frac{(A_1 \cdot e^{-E{a1}/RT}) \cdot (A_3 \cdot e^{-E{a3}/RT})}{A_2 \cdot e^{-E{a2}/RT}} ight]^{1/2}$ Equating the exponential terms yields the linear relationship for the overall activation energy (E_a$E_a$): fracE_aRT = frac12 left(fracE_a1RT + fracE_a3RT - fracE_a2RT ight)$\frac{E_a}{RT} = \frac{1}{2} \left(\frac{E_{a1}}{RT} + \frac{E_{a3}}{RT} - \frac{E_{a2}}{RT} ight)$ E_a = fracE_a1 + E_a3 - E_a22$E_a = \frac{E_{a1} + E_{a3} - E_{a2}}{2}$ Substitute the given activation energy values (E_a1 = 60, E_a2 = 30, E_a3 = 10text kJ/mol$E_{a1} = 60, E_{a2} = 30, E_{a3} = 10\text{ kJ/mol}$): E_a = frac60 + 10 - 302 = frac402 = 20text kJ mol^-1$E_a = \frac{60 + 10 - 30}{2} = \frac{40}{2} = 20\text{ kJ mol}^{-1}$ The overall activation energy is 20text kJ/mol$20\text{ kJ/mol}$. ### Pattern Recognition Shortcut: Convert the rate constant algebraic expression directly into an activation energy formula by swapping k$k$ for E_a$E_a$, turning multiplications into additions, divisions into subtractions, and powers into multipliers. Here, k = (k_1 k_3 / k_2)^1/2$k = (k_1 k_3 / k_2)^{1/2}$ translates directly to E_a = frac12(E_a1 + E_a3 - E_a2)$E_a = \frac{1}{2}(E_{a1} + E_{a3} - E_{a2})$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Core Logic Let the initial pressure of the reactant system be P_0$P_0$. Setting up the stoichiometric reaction chart: beginarraylcccc & N_2O_5(g) & rightarrow & 2NO_2(g) & + & frac12O_2(g) \\ textInitially (t=0): & P_0 & & 0 & & 0 \\ textAt equilibrium (t): & P_0 - x & & 2x & & fracx2 endarray$\begin{array}{lcccc} & N_2O_{5(g)} & \rightarrow & 2NO_{2(g)} & + & \frac{1}{2}O_{2(g)} \\ \text{Initially } (t=0): & P_0 & & 0 & & 0 \\ \text{At equilibrium } (t): & P_0 - x & & 2x & & \frac{x}{2} \end{array}$ Total pressure of the gaseous mixture at any time t$t$ is: P_texttotal = (P_0 - x) + 2x + fracx2 = P_0 + frac3x2$P_{\text{total}} = (P_0 - x) + 2x + \frac{x}{2} = P_0 + \frac{3x}{2}$ At 50\%$50\%$ structural breakdown, the change in reactant pressure is: x = 0.5 P_0 = fracP_02$x = 0.5 P_0 = \frac{P_0}{2}$ Substituting x$x$ into the expression for total system pressure: P_texttotal = P_0 + frac32left(fracP_02right) = P_0 + frac3P_04 = frac74P_0$P_{\text{total}} = P_0 + \frac{3}{2}\left(\frac{P_0}{2}\right) = P_0 + \frac{3P_0}{4} = \frac{7}{4}P_0$ ### Pattern Recognition Track the change in the total number of moles carefully using fractions. For a 50\%$50\%$ complete step, directly substitute the fractional equivalent into your total pressure expression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula For an elementary reaction, the rate law corresponds directly to its stoichiometry: R = K[A]^1[B]^1$R = K[A]^1[B]^1$ Concentration (C$C$) is inversely proportional to volume (V$V$): C = fracnV$C = \frac{n}{V}$ ### Core Logic Initial rate expression: R_1 = Kleft[fracn_AVright]^1left[fracn_BVright]^1$R_1 = K\left[\frac{n_A}{V}\right]^1\left[\frac{n_B}{V}\right]^1$ When volume is reduced to frac13V$\frac{1}{3}V$, the new concentration becomes 3$3$ times the initial concentration: R_2 = Kleft[frac3n_AVright]^1left[frac3n_BVright]^1 = 9 cdot Kleft[fracn_AVright]^1left[fracn_BVright]^1$R_2 = K\left[\frac{3n_A}{V}\right]^1\left[\frac{3n_B}{V}\right]^1 = 9 \cdot K\left[\frac{n_A}{V}\right]^1\left[\frac{n_B}{V}\right]^1$ ### Step 1: Calculating the Value of x Comparing the two rates: R_2 = 9R_1$R_2 = 9R_1$ Therefore, the value of x$x$ is 9$9$. ### Pattern Recognition For a second-order overall elementary reaction (1+1=2$1+1=2$), reducing the volume by a factor of n$n$ increases the rate by a factor of n^2$n^2$. Here n=3$n=3$, so the rate increases by 3^2 = 9$3^2 = 9$ times. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### 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