Consider the first-order gas-phase decomposit | vedclass

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(B) For the reaction: $A_{(g)} \longrightarrow B_{(g)} + C_{(g)}$At $t = 0$: Initial pressure of $A = P_i$,$P_B = 0$,$P_C = 0$.At time $t$: Let the de

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

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(B) For the reaction: $A_{(g)} \longrightarrow B_{(g)} + C_{(g)}$At $t = 0$: Initial pressure of $A = P_i$,$P_B = 0$,$P_C = 0$.At time $t$: Let the decrease in pressure of $A$ be $x$. Then,$P_A = P_i - x$,$P_B = x$,and $P_C = x$.The total pressure at time $t$ is $P_t = P_A + P_B + P_C = (P_i - x) + x + x = P_i + x$.From this,$x = P_t - P_i$.The partial pressure of $A$ at time $t$ is $P_A = P_i - x = P_i - (P_t - P_i) = 2P_i - P_t$.For a first-order reaction,$K = \frac{2.303}{t} \log \frac{P_{	ext{initial}}}{P_{	ext{final}}}$.Substituting the values,$K = \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t}$.

$S$.no.	Time	Total pressure in Pascal
$1.$	At the end of $10 \ min$	$300$
$2.$	After completion	$200$

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