QUESTION 15 Consider this data for this and the following problem: For the gas-p

ID: 509867 • Letter: Q

Question

QUESTION 15

Consider this data for this and the following problem:

For the gas-phase dissociation reaction A --> B + C, The following data is collected at 298 K:

Determine the partial pressure of A at each time.

Make a graph of ln(PA) and 1/PA vs time in order to determine the order of the reaction and the rate constant.

The rate law for this reaction is given by

Answer:

2 points

QUESTION 16 (Need HELP!!!) *please show work.

Based on the data from the previous problem, calculate the total pressure at 6,745 seconds.

time (sec) total pressure (atm) 0 1.000 1000 1.359 3000 1.627 5000 1.737 9000 1.834

PA= PAO- (a/deltan)*(PI-Po)

PA= partial pressure of A

PAO= partial pressure of A t=0

PI= total pressure at any time

Po= total pressure at t=0

Deltan= change in no of moles during the reaction= 1+1-1=1

At t= 1000 sec,

PA= 1-(1.359-1)=0.641 atm

At t=3000 sec

PA= 1-(1.627-1)= 0.373 atm

At t=5000 sec, PA= 1-(1.737-1)= 0.263 atm and at t= 9000 sec, PA= 1-(1.834-1)= 0.166 atm

the plot of ln PA or (1/PA) vs time is shown plotted below

the plot is a straightl line for 1/PA vs t. This is the characterisitic of second order equation

for second order reaction -dPA/dt= KPA2 or 1/PA= 1/PAO+Kt

where K = Rate constant , PAO= pressure at zero time and PA= pressrue at any time

from the plot slope =K= 0.0006/atm.sec

the rate equation is -dPA/dt= 0.0006*PA2,

at 6745 sec, PA= PAO-1*(PI-Po)

PAO= 1atm, PA at 6745 sec can be calculated from 1/PA= 1/1+0.0006*6745 =5.047

PA= 0.198 atm

hence 0.198= 1- Pi+1

PI= 2-0.198= 1.802 atm

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