### 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-1 ight) [mathrmA2]$\frac{k_1}{k{-1}} = \frac{[\mathrm{A}]^2}{[\mathrm{A}2]} \implies [\mathrm{A}]^2 = \left(\frac{k_1}{k{-1}} ight) [\mathrm{A}2]$ [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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### 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

### Related Formula K = A e^-E_a/RT$K = A e^{-E_a/RT}$ ### Core Logic Given the relationship between the rate constants: K = fracK_1 cdot K_2K_3$K = \frac{K_1 \cdot K_2}{K_3}$ Substituting the Arrhenius equation for each rate constant: A cdot e^-E_a/RT = fracA_1 cdot e^-E_a1/RT cdot A_2 cdot e^-E_a2/RTA_3 cdot e^-E_a3/RT$A \cdot e^{-E_a/RT} = \frac{A_1 \cdot e^{-E_{a1}/RT} \cdot A_2 \cdot e^{-E_{a2}/RT}}{A_3 \cdot e^{-E_{a3}/RT}}$ Combining the exponential terms using rules of exponents: A cdot e^-E_a/RT = left(fracA_1 cdot A_2A_3right) cdot e^frac-(E_a1 + E_a2 - E_a3)RT$A \cdot e^{-E_a/RT} = \left(\frac{A_1 \cdot A_2}{A_3}\right) \cdot e^{\frac{-(E_{a1} + E_{a2} - E_{a3})}{RT}}$ ### Step 1: Equating Activation Energies By comparing the powers of e$e$ on both sides, the overall activation energy E_a$E_a$ is related to the individual steps as follows: E_a = E_a1 + E_a2 - E_a3$E_a = E_{a1} + E_{a2} - E_{a3}$ Substitute the given values (E_a1 = 40$E_{a1} = 40$, E_a2 = 50$E_{a2} = 50$, E_a3 = 60$E_{a3} = 60$ kJ/mol): E_a = 40 + 50 - 60$E_a = 40 + 50 - 60$ E_a = 90 - 60$E_a = 90 - 60$ E_a = 30 text kJ/mol$E_a = 30 \text{ kJ/mol}$ ### Pattern Recognition When rate constants are multiplied or divided (K = K_1^a K_2^b / K_3^c$K = K_1^a K_2^b / K_3^c$), the corresponding overall activation energy follows the linear combination of the exponents: E_a = a E_a1 + b E_a2 - c E_a3$E_a = a E_{a1} + b E_{a2} - c E_{a3}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics