Question

5. A 200-dm constant-volume batch reactor is at the pressure 20 atm with pure A. The gas-phase reaction A B is carried out isothermally at 400K. The first order rate expression is -, -kC,, where k=0.1 min. Calculate the reaction time for 80% conversion of A. (20%)

Answer 1

Data provided in the problem statement

First order reaction with rate expression as

-rA = kCA

rate constant ( k) = 0.1 / min

Conversion of A = 80 %

Now taking mole balance around the reactor we have

( Flow rate In ) - ( Flowrate Out ) + ( Rate of generation ) = Rate of accumulation

Here

Flow rate in =0

Flow rate out = 0

We get

Rate of generation = rate of accumulation

ana ra av = dt is constant as Volume &A. V = dua at Here, moilms min seete sa= = Reactia voleme m3 mol a hores of min t= sime . concentration of a can be given as (CA) CA= CA= MA = moles of a v voame. combining above two equations, A = d (NA/0) dt at expression 04 - Sca -0 Now we have reaction rate A = - kca - combining equation @ 80 al

- keq a dia Integrating between time t=0 & 2 = t and concentocation CACAO & CA = CA - Here. A CAO = CA = énitial concentration of final concer treation of 4 As conversicg g A is 80% GA = our CA = CI-0.8) CAO CA= 0.2 CAO - dea C40 -kt a enca -encao = m ( cas ) putring valore sex ca = 0.2 CAO and le- o. mint we get -0.1€ = en ( CAO.0.2) = en 0.2 to enco.2) - 16.og min. v -0.)

Time required for 80% conversion = 16.09 mins