A pure first-order gas-phase decomposition takes place at constant volume. If is the initial pressur — Chemical Kinetics Chemistry Question
Question
A pure first-order gas-phase decomposition $A(g) \rightarrow B(g) + C(g)$ takes place at constant volume. If $P_0$ is the initial pressure and $P_t$ is the total system pressure at time $t$, which is the correct integrated rate law?
Answer: B
💡 Solution & Explanation
At time $t$, pressure of A is $P_0 - x$, B is $x$, C is $x$. Total pressure $P_t = P_0 + x$, so $x = P_t - P_0$. Remaining pressure of A is $P_A = P_0 - x = 2P_0 - P_t$. Substituting this into the master first-order equation gives $k = \frac{2.303}{t} \log\left(\frac{P_0}{2P_0 - P_t}\right)$.
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