Reaction mathrmA(g) rightarrow 2mathrmB(g) + mathrmC(g)$\mathrm{A(g)} \rightarrow 2\mathrm{B(g)} + \mathrm{C(g)}$ is a first order reaction. It was started with pure A.

t / textmin$t / \text{min}$	Pressure of system at time t / textmm Hg$t / \text{mm Hg}$
10	160
infty$\infty$	240

Which of the following options is incorrect?

### Related Formula For a first-order integrated rate law expression: [A]_t = [A]_0 e^-k_A t quad textwhere k = fracln 2t_1/2$$[A]_t = [A]_0 e^{-k_A t} quad \text{where } k = \frac{ln 2}{t_{1/2}}$$ ### Core Logic We require the instantaneous concentration to be identical at time t$t$: [A]_t = [B]_t implies [A]_0 e^-k_A t = [B]_0 e^-k_B t$$[A]_t = [B]_t implies [A]_0 e^{-k_A t} = [B]_0 e^{-k_B t}$$ frac[A]_0[B]_0 = e^(k_A - k_B)t$$\frac{[A]_0}{[B]_0} = e^{(k_A - k_B)t}$$ ### Step 1: Substituting Parameters Substitute [A]_0 = 8[B]_0$[A]_0 = 8[B]_0$ and express rate constants in terms of half-lives: 8 = e^(k_A - k_B)t implies ln 8 = (k_A - k_B)t$$8 = e^{(k_A - k_B)t} implies ln 8 = (k_A - k_B)t$$ 3ln 2 = ln 2 left( frac1(t_1/2)_A - frac1(t_1/2)_B ight) times t$$3ln 2 = ln 2 \left( \frac{1}{(t_{1/2})_A} - \frac{1}{(t_{1/2})_B} ight) \times t$$ 3 = left( frac110 - frac140 ight) times t implies 3 = frac340 times t implies t = 40text min.$$3 = \left( \frac{1}{10} - \frac{1}{40} ight) \times t implies 3 = \frac{3}{40} \times t implies t = 40\text{ min.}$$ ### Pattern Recognition Shortcut: Express the concentration drop using half-life indices: [A]_t = frac[A]_02^t/10 = frac8[B]_02^t/10$$[A]_t = \frac{[A]_0}{2^{t/10}} = \frac{8[B]_0}{2^{t/10}}$$ [B]_t = frac[B]_02^t/40$$[B]_t = \frac{[B]_0}{2^{t/40}}$$ Equating both: 8 cdot 2^-t/10 = 2^-t/40 implies 2^3 = 2^fract10 - fract40 implies 3 = frac3t40 implies t = 40text min.$8 \cdot 2^{-t/10} = 2^{-t/40} implies 2^3 = 2^{\frac{t}{10} - \frac{t}{40}} implies 3 = \frac{3t}{40} implies t = 40\text{ min.}$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### 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 r = k [A]^n [B]^m$$r = k [A]^n [B]^m$$ ### Core Logic Let the initial rate relation be: r_1 = k [A]^n [B]^m$$r_1 = k [A]^n [B]^m$$ When concentration parameters shift ([A]' = 2[A]$[A]' = 2[A]$ and [B]' = frac[B]2$[B]' = \frac{[B]}{2}$): r_2 = k (2[A])^n left(frac[B]2right)^m = k cdot 2^n [A]^n cdot 2^-m [B]^m$$r_2 = k (2[A])^n \left(\frac{[B]}{2}\right)^m = k \cdot 2^n [A]^n \cdot 2^{-m} [B]^m$$ r_2 = 2^(n-m) cdot left(k [A]^n [B]^mright) = 2^(n-m) cdot r_1$$r_2 = 2^{(n-m)} \cdot \left(k [A]^n [B]^m\right) = 2^{(n-m)} \cdot r_1$$ Taking the ratio yields: fracr_2r_1 = 2^(n-m)$$\frac{r_2}{r_1} = 2^{(n-m)}$$ ### Pattern Recognition Powers simplify cleanly via exponent rules: doubling a base scales the expression by 2^n$2^n$, while halving scales it by 2^-m$2^{-m}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics