A pure first-order gas-phase decomposition takes place at constant volume. The initial pressure is . — Chemical Kinetics Chemistry Question
Question
A pure first-order gas-phase decomposition $A(g) \rightarrow B(g) + C(g)$ takes place at constant volume. The initial pressure is $P_0$. Let $P_t$ represent the total pressure of the completely closed system at time $t$. Which of the following is the correct integrated rate law?
Answer: B
💡 Solution & Explanation
At time $t$, $P_A = P_0 - x$, $P_B = x$, $P_C = x$. Total pressure $P_t = P_0 + x \Rightarrow x = P_t - P_0$. The remaining pressure of reactant A is $P_A = P_0 - x = P_0 - (P_t - P_0) = 2P_0 - P_t$. 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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