### Related Formula log k = log A - fracE_a2.303 R T$\log k = \log A - \frac{E_a}{2.303 R T}$ textSlope of the line = -fracE_a2.303 R implies |textSlope| propto E_a$\text{Slope of the line} = -\frac{E_a}{2.303 R} \implies |\text{Slope}| \propto E_a$ ### Core Logic From the given graph, we look at the steepness (magnitude of the negative slope) of lines 1, 2, and 3: - Line 2 is the steepest, meaning it has the largest slope magnitude. - Line 1 has an intermediate slope. - Line 3 is the flattest, indicating the smallest slope magnitude. Since the activation energy E_a$E_a$ is directly proportional to the magnitude of this slope: |textSlope_2| > |textSlope_1| > |textSlope_3| implies E_a2 > E_a1 > E_a3$|\text{Slope}_2| > |\text{Slope}_1| > |\text{Slope}_3| \implies E_{a2} > E_{a1} > E_{a3}$ ### Pattern Recognition In Arrhenius coordinates, steepness equals barriers. A steeper line means the reaction rate is highly sensitive to temperature because it has a higher activation energy (E_a$E_a$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula t_1/2 = frac[A]_02k quad text(for Zero-Order रिएक्शन)$t_{1/2} = \frac{[A]_0}{2k} \quad \text{(for Zero-Order रिएक्शन)}$ t = frac[A]_0 - [A]_tk$t = \frac{[A]_0 - [A]_t}{k}$ ### Core Logic 1. Find the rate constant k$k$ using the given half-life parameters: 1 text hour = 60 text min = frac2.02k implies k = frac2.02 times 60 = frac160 mathrm~M cdot min^-1$1 \text{ hour} = 60 \text{ min} = \frac{2.0}{2k} \implies k = \frac{2.0}{2 \times 60} = \frac{1}{60} \mathrm{~M \cdot min^{-1}}$ 2. Calculate the time t$t$ to drop from 0.50 mathrm~molcdot L^-1$0.50 \mathrm{~mol\cdot L^{-1}}$ to 0.25 mathrm~molcdot L^-1$0.25 \mathrm{~mol\cdot L^{-1}}$: t = frac0.50 - 0.25k = frac0.25left(frac160right) = 0.25 times 60 = 15 text minutes$t = \frac{0.50 - 0.25}{k} = \frac{0.25}{\left(\frac{1}{60}\right)} = 0.25 \times 60 = 15 \text{ minutes}$ ### Pattern Recognition For zero-order systems, the rate of reaction is entirely independent of concentration. This means the time required to consume a specific quantity of reactant scales linearly with the concentration change (t = fracDelta Ck$t = \frac{\Delta C}{k}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula Delta H = E_a(f) - E_a(b)$\Delta H = E_{a(f)} - E_{a(b)}$ ### Core Logic A catalyst accelerates both forward and backward path steps symmetrically by carving a lower activation energy profile route. * Uncatalyzed values: Delta H = 180 - 200 = -20mathrm~kJ~mol^-1$\Delta H = 180 - 200 = -20\mathrm{~kJ~mol}^{-1}$. * Catalyzed values: E_a(f)' = 80mathrm~kJ~mol^-1$E_{a(f)}' = 80\mathrm{~kJ~mol}^{-1}$ and E_a(b)' = 100mathrm~kJ~mol^-1$E_{a(b)}' = 100\mathrm{~kJ~mol}^{-1}$, leading to Delta H' = 80 - 100 = -20mathrm~kJ~mol^-1$\Delta H' = 80 - 100 = -20\mathrm{~kJ~mol}^{-1}$. Thermodynamic parameters (Delta H$\Delta H$, Delta G$\Delta G$, Delta S$\Delta S$) depend strictly on the initial and final energy states of reactants and products, meaning they are completely unaltered by the presence of a catalyst. ### Pattern Recognition Catalysts alter only kinetic properties (rate, activation barriers). They have zero impact on equilibrium positions or thermodynamic state parameters like Delta G$\Delta G$ or Delta H$\Delta H$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### 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