The integrated rate law equation for a first- | vedclass

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Question

(A) For the reaction $A(g) \rightarrow B(g) + C(g)$:      Time $t=0$    $P_i$    $0$    $0$        Time $t$    $P_i - x$    $x$    $x$  Total pressure

Vedclass · Accepted Answer

$k = \frac{2.303}{t} \log \frac{P_i}{2 P_i - P_t}$

Vedclass · Answer

(A) For the reaction $A(g) \rightarrow B(g) + C(g)$:      Time $t=0$    $P_i$    $0$    $0$        Time $t$    $P_i - x$    $x$    $x$  Total pressure at time $t$ is $P_t = (P_i - x) + x + x = P_i + x$.Therefore,$x = P_t - P_i$.The partial pressure of reactant $A$ at time $t$ is $P_A = P_i - x = P_i - (P_t - P_i) = 2 P_i - P_t$.Substituting into the first-order rate equation $k = \frac{2.303}{t} \log \frac{P_i}{P_A}$:$k = \frac{2.303}{t} \log \frac{P_i}{2 P_i - P_t}$.

The integrated rate law equation for a first-order gas-phase reaction $A(g) \rightarrow B(g) + C(g)$ is given by (where $P_i$ is the initial pressure and $P_t$ is the total pressure at time $t$):

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