### Related Formula textRate of Reaction (ROR) = -frac12 fracDelta [mathrmN_2mathrmO_5]Delta t = frac14 fracDelta [mathrmNO_2]Delta t$\text{Rate of Reaction (ROR)} = -\frac{1}{2} \frac{\Delta [\mathrm{N}_2\mathrm{O}_5]}{\Delta t} = \frac{1}{4} \frac{\Delta [\mathrm{NO}_2]}{\Delta t}$ ### Core Logic First, find the rate of disappearance of mathrmN_2mathrmO_5$\mathrm{N}_2\mathrm{O}_5$. -fracDelta [mathrmN_2mathrmO_5]Delta t = - frac(2.75 - 3)30 = frac0.2530 mathrmmol \, L^-1 \, min^-1$-\frac{\Delta [\mathrm{N}_2\mathrm{O}_5]}{\Delta t} = - \frac{(2.75 - 3)}{30} = \frac{0.25}{30} \mathrm{mol \, L^{-1} \, min^{-1}}$ Now, equate it to the general Rate of Reaction: textROR = -frac12 fracDelta [mathrmN_2mathrmO_5]Delta t = frac12 left(frac0.2530right) = frac0.12530 = frac1240 mathrmmol \, L^-1 \, min^-1$\text{ROR} = -\frac{1}{2} \frac{\Delta [\mathrm{N}_2\mathrm{O}_5]}{\Delta t} = \frac{1}{2} \left(\frac{0.25}{30}\right) = \frac{0.125}{30} = \frac{1}{240} \mathrm{mol \, L^{-1} \, min^{-1}}$ ### Step 1: Calculate the Rate of Formation of mathrmNO_2$\mathrm{NO}_2$ Rate of formation of mathrmNO_2 = fracDelta [mathrmNO_2]Delta t = 4 times textROR$\mathrm{NO}_2 = \frac{\Delta [\mathrm{NO}_2]}{\Delta t} = 4 \times \text{ROR}$ = 4 times frac1240 = frac160 mathrmmol \, L^-1 \, min^-1$= 4 \times \frac{1}{240} = \frac{1}{60} \mathrm{mol \, L^{-1} \, min^{-1}}$ Convert this to scientific notation to find x$x$: frac160 approx 0.01666 = 16.66 times 10^-3 mathrmmol \, L^-1 \, min^-1$\frac{1}{60} \approx 0.01666 = 16.66 \times 10^{-3} \mathrm{mol \, L^{-1} \, min^{-1}}$ Rounding to the nearest integer, we get x = 17$x = 17$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula Rate = k[A]$Rate = k[A]$ [A] = [A]_0 e^-kt$[A] = [A]_0 e^{-kt}$ t_1/2 = fracln 2k$t_{1/2} = \frac{\ln 2}{k}$ ### Core Logic For a first order reaction, rate is directly proportional to concentration. R_1 = k[A]_10 = k[A]_0 e^-k(10 times 60)$R_1 = k[A]_{10} = k[A]_0 e^{-k(10 \times 60)}$ R_2 = k[A]_20 = k[A]_0 e^-k(20 times 60)$R_2 = k[A]_{20} = k[A]_0 e^{-k(20 \times 60)}$ ### Step 1: Setting up equations 0.04 = k[A]_0 e^-600k dots (1)$0.04 = k[A]_0 e^{-600k} \dots (1)$ 0.03 = k[A]_0 e^-1200k dots (2)$0.03 = k[A]_0 e^{-1200k} \dots (2)$ ### Step 2: Solving for k Dividing equation (1) by (2): frac0.040.03 = frace^-600ke^-1200k$\frac{0.04}{0.03} = \frac{e^{-600k}}{e^{-1200k}}$ frac43 = e^600k$\frac{4}{3} = e^{600k}$ Take natural log on both sides: lnleft(frac43right) = 600k$\ln\left(\frac{4}{3}\right) = 600k$ k = fracln(4/3)600 text s^-1$k = \frac{\ln(4/3)}{600} \text{ s}^{-1}$ ### Step 3: Calculating half life t_1/2 = fracln 2k = fracln 2fracln(4/3)600 = frac600 ln 2ln 4 - ln 3 text seconds$t_{1/2} = \frac{\ln 2}{k} = \frac{\ln 2}{\frac{\ln(4/3)}{600}} = \frac{600 \ln 2}{\ln 4 - \ln 3} \text{ seconds}$ Convert to minutes by dividing by 60: t_1/2 = frac10 ln 2ln 4 - ln 3 text minutes$t_{1/2} = \frac{10 \ln 2}{\ln 4 - \ln 3} \text{ minutes}$ Substitute log values (since ln x = 2.303 log x$\ln x = 2.303 \log x$, the 2.303 cancels out): t_1/2 = 10 times fraclog 2log 4 - log 3 text minutes$t_{1/2} = 10 \times \frac{\log 2}{\log 4 - \log 3} \text{ minutes}$ t_1/2 = 10 times frac0.30102(0.3010) - 0.4771$t_{1/2} = 10 \times \frac{0.3010}{2(0.3010) - 0.4771}$ t_1/2 = 10 times frac0.30100.6020 - 0.4771 = 10 times frac0.30100.1249$t_{1/2} = 10 \times \frac{0.3010}{0.6020 - 0.4771} = 10 \times \frac{0.3010}{0.1249}$ t_1/2 = 24.099 approx 24 text minutes$t_{1/2} = 24.099 \approx 24 \text{ minutes}$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Core Logic Consider a general gas phase reaction: A rightarrow B + C$A \rightarrow B + C$ Initial (t=0$t=0$): quad P_i quadquad 0 quadquad 0$\quad P_i \quad\quad 0 \quad\quad 0$ At time t$t$: quad P_i - x quadquad x quadquad x$\quad P_i - x \quad\quad x \quad\quad x$ Total pressure at time t$t$: P_t = (P_i - x) + x + x = P_i + x$P_t = (P_i - x) + x + x = P_i + x$ x = P_t - P_i$x = P_t - P_i$ Partial pressure of A at time t$t$ (P_A$P_A$): P_A = P_i - x$P_A = P_i - x$ P_A = P_i - (P_t - P_i) = 2P_i - P_t$P_A = P_i - (P_t - P_i) = 2P_i - P_t$ For a first-order reaction: k = frac2.303t log fracP_0P_t$k = \frac{2.303}{t} \log \frac{P_0}{P_t}$ Here, P_0 = P_i$P_0 = P_i$ and the pressure of the reactant at time t$t$ is P_A$P_A$. k = frac2.303t log fracP_i2P_i - P_t$k = \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t}$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics