The temperature at which the rate constants of the given below two gaseous reactions become equal is ____ K. (Nearest integer). X rightarrow Y quad k_1 = 10^6e^frac-30000T$$X \rightarrow Y \quad k_{1} = 10^{6}e^{\frac{-30000}{T}}$$ P rightarrow Q quad k_2 = 10^4e^frac-24000T$$P \rightarrow Q \quad k_{2} = 10^{4}e^{\frac{-24000}{T}}$$ Given: ln 10 = 2.303$\ln 10 = 2.303$

### Related Formula ln k = ln A - fracE_aRT$$\ln k = \ln A - \frac{E_a}{RT}$$ ### Core Logic For Reaction 1: ln k_1 = ln A - fracE_1RT$\ln k_1 = \ln A - \frac{E_1}{RT}$ For Reaction 2: ln k_2 = ln A - fracE_2RT$\ln k_2 = \ln A - \frac{E_2}{RT}$ (Pre-exponential factor A$A$ is the same). Subtracting the first from the second: ln k_2 - ln k_1 = -fracE_2RT - left(-fracE_1RTright)$$\ln k_2 - \ln k_1 = -\frac{E_2}{RT} - \left(-\frac{E_1}{RT}\right)$$ ln left(frack_2k_1right) = fracE_1 - E_2RT$$\ln \left(\frac{k_2}{k_1}\right) = \frac{E_1 - E_2}{RT}$$ Given that E_1$E_1$ exceeds E_2$E_2$ by 20text kJ mol^-1$20\text{ kJ mol}^{-1}$, E_1 - E_2 = 20000text J mol^-1$E_1 - E_2 = 20000\text{ J mol}^{-1}$. T = 300text K$T = 300\text{ K}$, R = 8.3text J K^-1text mol^-1$R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}$. ln left(frack_2k_1right) = frac200008.3 times 300 = frac2008.3 times 3 = frac20024.9$$\ln \left(\frac{k_2}{k_1}\right) = \frac{20000}{8.3 \times 300} = \frac{200}{8.3 \times 3} = \frac{200}{24.9}$$ ln left(frack_2k_1right) = 8.032$$\ln \left(\frac{k_2}{k_1}\right) = 8.032$$ Rounding off to nearest integer gives 8. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula k = frac2.303t logleft(fracP_02P_0 - P_texttotalright)$$k = \frac{2.303}{t} \log\left(\frac{P_0}{2P_0 - P_{\text{total}}}\right)$$ or equivalent first-order expression. ### Core Logic For textA(textg) rightarrow textB(textg) + textC(textg)$\text{A}(\text{g}) \rightarrow \text{B}(\text{g}) + \text{C}(\text{g})$: - Initial pressure: P_0 = 1 text bar$P_0 = 1 \text{ bar}$ - At time t = 100 text min$t = 100 \text{ min}$, pressure of A remaining = 1 - P$= 1 - P$, pressures of B and C = P$= P$. - Total pressure P_texttotal = 1 - P + P + P = 1 + P = 1.5 text bar implies P = 0.5 text bar$P_{\text{total}} = 1 - P + P + P = 1 + P = 1.5 \text{ bar} \implies P = 0.5 \text{ bar}$. Remaining pressure of A = 1 - 0.5 = 0.5 text bar$= 1 - 0.5 = 0.5 \text{ bar}$. ### Step 1: Calculating Rate Constant k = frac1100 lnleft(frac10.5right) = frac0.693100 = 6.9 times 10^-3 text min^-1$$k = \frac{1}{100} \ln\left(\frac{1}{0.5}\right) = \frac{0.693}{100} = 6.9 \times 10^{-3} \text{ min}^{-1}$$ ### Pattern Recognition Sees: gaseous phase first-order kinetics with total pressure data. Trap: Confusing partial pressure of reactant with total pressure in rate expressions. ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula textRate = k[A]$$\text{Rate} = k[A]$$ Where [A]$[A]$ is the molar concentration of reactant A. ### Core Logic For a first order reaction, the rate is directly proportional to the concentration of the reactant, not the total volume or the total number of moles. In Run 1: [A] = 10text M$[A] = 10\text{ M}$. textRate_1 = k(10)$\text{Rate}_1 = k(10)$. In Run 2: [A] = 10text M$[A] = 10\text{ M}$. textRate_2 = k(10)$\text{Rate}_2 = k(10)$. (Volume increased, but molarity is identical). In Run 3: 100 mL of 10 M solution is diluted with 100 mL water. New volume is 200 mL. M_1 V_1 = M_2 V_2 implies 10 times 100 = M_2 times 200 implies M_2 = 5text M$M_1 V_1 = M_2 V_2 \implies 10 \times 100 = M_2 \times 200 \implies M_2 = 5\text{ M}$. [A] = 5text M$[A] = 5\text{ M}$. textRate_3 = k(5)$\text{Rate}_3 = k(5)$. ### Step 1: Final Conclusion Therefore, textRate_3 < textRate_1 = textRate_2$\text{Rate}_3 < \text{Rate}_1 = \text{Rate}_2$. ### Pattern Recognition Rate laws depend exclusively on molar concentration (M). Diluting the solution decreases rate, while just taking a larger volume of the same stock solution keeps the rate identical. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

### Related Formula k = A e^-E_a/RT$$k = A e^{-E_a/RT}$$ ln k = ln A - fracE_aRT implies log k = log A - fracE_a2.303 R T$$\ln k = \ln A - \frac{E_a}{RT} \implies \log k = \log A - \frac{E_a}{2.303 R T}$$ ### Core Logic Statement (A): BANNED - e^-E_a/RT$e^{-E_a/RT}$ represents fraction of molecules with energy ge E_a$\ge E_a$ (not less). Statement (B): CORRECT - Lower activation energy E_a$E_a$ increases the rate constant k$k$, speeding up the reaction. Statement (C): CORRECT - For most reactions, a 10^circtextC$10^\circ\text{C}$ rise in temperature doubles the rate coefficient. Statement (D): INCORRECT - Plot of log k$\log k$ vs 1/T$1/T$ has slope equal to -fracE_a2.303 R$-\frac{E_a}{2.303 R}$ (the factor 2.303$2.303$ is missing). ### Pattern Recognition Sees: Arrhenius statements. Shortcut: Watch for missing 2.303$2.303$ in log$\log$ slope equation and 'less than' vs 'greater than' in exponential fraction definition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics