For a homogeneous gaseous reaction A_{(g)} to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $/$ $s$	Total pressure $/$ $atm$
$1.$	$0$	$0.5$
$2.$	$100$	$0.512$

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 $K$.

Similar Questions

Which is a correct integrated rate equation for a first-order reaction?

The following data were obtained during the first order thermal decomposition of $N_{2}O_{5(g)}$ at constant volume:
$2N_{2}O_{5(g)} \rightarrow 2N_{2}O_{4(g)} + O_{2(g)}$

$S.No.$ Time $/$ $s$ Total pressure $/$ $atm$

$1.$ $0$ $0.5$

$2.$ $100$ $0.512$

Calculate the rate constant.

In a first order reaction,the concentration of the reactant decreased from $0.4 \, M$ to $0.1 \, M$ in $20 \, \min$. The rate constant of the reaction will be:

If the rate constant of a first-order reaction is $0.6932 \ hr^{-1}$,then its half-life period is .......... $hr$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module