For a first-order gaseous isothermal reaction formally defined as carried out in a vessel of strictl — Chemical Kinetics Chemistry Question
Question
For a first-order gaseous isothermal reaction formally defined as $A(g) \rightarrow B(g) + C(g)$ carried out in a vessel of strictly constant volume, if $P_0$ is the isolated initial pressure of $A$ and $P_t$ is the overarching total pressure of the mixed gases at time $t$, the integrated rate constant $k$ equates exactly to:
Answer: A
💡 Solution & Explanation
Initial condition: $P_A = P_0$, others 0. At time $t$: $P_A = P_0 - x$, $P_B = x$, $P_C = x$. Total pressure $P_t = (P_0 - x) + x + x = P_0 + x$. Consequently, $x = P_t - P_0$. The partial pressure of reactant $A$ at time $t$ is $P_A = P_0 - x = P_0 - (P_t - P_0) = 2P_0 - P_t$. Substituting this into the first-order gas equation gives $k = \frac{1}{t} \ln \left(\frac{P_0}{2P_0 - P_t}\right)$.
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