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(D) For the reaction $A_{(g)} 	o 3B_{(g)}$:At $t = 0$: $P_A = P_0$,$P_B = 0$. Total pressure $P_0$.At $t = t$: $P_A = P_0 - x$,$P_B = 3x$. Total pres

Vedclass · Accepted Answer

$K = \frac{1}{t} \ln \left( \frac{2P_\infty}{3(P_\infty - P_t)} \right)$

Vedclass · Answer

(D) For the reaction $A_{(g)} 	o 3B_{(g)}$:At $t = 0$: $P_A = P_0$,$P_B = 0$. Total pressure $P_0$.At $t = t$: $P_A = P_0 - x$,$P_B = 3x$. Total pressure $P_t = (P_0 - x) + 3x = P_0 + 2x$.At $t = \infty$: $P_A = 0$,$P_B = 3P_0$. Total pressure $P_\infty = 3P_0$,so $P_0 = \frac{P_\infty}{3}$.From $P_t = P_0 + 2x$,we get $x = \frac{P_t - P_0}{2}$.The partial pressure of $A$ at time $t$ is $P_A = P_0 - x = P_0 - \frac{P_t - P_0}{2} = \frac{3P_0 - P_t}{2}$.Substituting $P_0 = \frac{P_\infty}{3}$,we get $P_A = \frac{3(P_\infty / 3) - P_t}{2} = \frac{P_\infty - P_t}{2}$.For a first-order reaction,$K = \frac{1}{t} \ln \left( \frac{P_0}{P_A} ight)$.Substituting the values: $K = \frac{1}{t} \ln \left( \frac{P_\infty / 3}{(P_\infty - P_t) / 2} ight) = \frac{1}{t} \ln \left( \frac{2P_\infty}{3(P_\infty - P_t)} ight)$.

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

For a homogeneous gaseous reaction $A_{(g)} \to 3B_{(g)}$,if the pressure after time $t$ is $P_t$ and after completion of the reaction the pressure is $P_\infty$,select the correct relation for the rate constant $K$.

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