In a first-order reaction of the type: A_{(g) | vedclass

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Question

(A) For the reaction $A_{(g)} 	o 2B_{(g)}$,let the initial pressure of $A$ be $P_i$ at $t = 0$. At time $t$,let the pressure of $A$ decrease by $x$.

Vedclass · Accepted Answer

$k = \frac{1}{t} \ln \frac{P_i}{2P_i - P_t}$

Vedclass · Answer

(A) For the reaction $A_{(g)} 	o 2B_{(g)}$,let the initial pressure of $A$ be $P_i$ at $t = 0$. At time $t$,let the pressure of $A$ decrease by $x$. Then,the pressure of $A$ becomes $(P_i - x)$ and the pressure of $B$ becomes $2x$. The total pressure $P_t = (P_i - x) + 2x = P_i + x$. Therefore,$x = P_t - P_i$. The 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,$k = \frac{1}{t} \ln \frac{P_i}{P_A}$. Substituting $P_A$,we get $k = \frac{1}{t} \ln \frac{P_i}{2P_i - P_t}$.

$S.No.$	$Time/s$	$Total Pressure/(atm)$
$1.$	$0$	$0.1$
$2.$	$115$	$0.28$

In a first-order reaction of the type: $A_{(g)} \to 2B_{(g)}$,the initial pressure is $P_i$ and the total pressure at time $t$ is $P_t$. The rate constant $k$ can be expressed as:

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