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 = frac1t ln fracP_0P_0 - x $$k = \frac{1}{t} \ln \frac{P_0}{P_0 - x} $$ where P_0$P_0$ is the initial pressure of reactant textA$\text{A}$, and x$x$ is the change in pressure at time t$t$. ### Core Logic Let's establish the ice table for total pressure calculation: beginarrayrccc & textA(g) & ightarrow & textB(g) & + & textC(g) \ textAt t=0: & P_0 & & 0 & & 0 \ textAt t=t: & P_0 - x & & x & & x \ textAt t=infty: & 0 & & P_0 & & P_0 endarray $$\begin{array}{rccc} & \text{A}(g) & ightarrow & \text{B}(g) & + & \text{C}(g) \ \text{At } t=0: & P_0 & & 0 & & 0 \ \text{At } t=t: & P_0 - x & & x & & x \ \text{At } t=\infty: & 0 & & P_0 & & P_0 \end{array} $$ From the data given at t = infty$t = \infty$: P_infty = P_0 + P_0 = 2P_0 implies P_0 = fracP_infty2 $$P_\infty = P_0 + P_0 = 2P_0 \implies P_0 = \frac{P_\infty}{2} $$ From the data given at time t$t$: P_t = (P_0 - x) + x + x = P_0 + x $$P_t = (P_0 - x) + x + x = P_0 + x $$ x = P_t - P_0 = P_t - fracP_infty2 $$x = P_t - P_0 = P_t - \frac{P_\infty}{2} $$ ### Step 1: Algebraic Substitution Now, compute the amount of reactant remaining at time t$t$: P_0 - x = fracP_infty2 - left(P_t - fracP_infty2 ight) = P_infty - P_t$$P_0 - x = \frac{P_\infty}{2} - \left(P_t - \frac{P_\infty}{2} ight) = P_\infty - P_t$$ Substitute P_0$P_0$ and (P_0 - x)$(P_0 - x)$ back into the primary kinetic expression: k = frac1t ln fracfracP_infty2P_infty - P_t = frac1t ln fracP_infty2(P_infty - P_t) $$k = \frac{1}{t} \ln \frac{\frac{P_\infty}{2}}{P_\infty - P_t} = \frac{1}{t} \ln \frac{P_\infty}{2(P_\infty - P_t)} $$ ### Pattern Recognition For a standard gaseous decomposition textA ightarrow ntextB + mtextC$\text{A} ightarrow n\text{B} + m\text{C}$, tracking the infinite pressure P_infty$P_\infty$ offers a clean mapping to initial reactant amounts. Since 1 mole of gas generates 2 moles of product gas here, P_0$P_0$ is exactly half of P_infty$P_\infty$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### 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