Decomposition of A is a first order reaction at T(K) and is given by textA(textg) rightarrow textB(textg) + textC(textg)$\text{A}(\text{g}) \rightarrow \text{B}(\text{g}) + \text{C}(\text{g})$. In a closed 1 L vessel, 1 bar textA(textg)$\text{A}(\text{g})$ is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in textmin^-1$\text{min}^{-1}$) of the reaction? (log 2 = 0.3)$(\log 2 = 0.3)$

### 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