A 200-dm3 constant-volume batch reactor is pressurized to 20 atm with a mixture

Question

A 200-dm3 constant-volume batch reactor is pressurized to 20 atm with a mixture of 75% A and 25% inert. The gas-phase reaction is carried out isothermally at 227 C. Assuming that the ideal gas law is valid, how many moles of A are in the reactor initially? What is the initial concentration of A? If the reaction is first order: -r^zkC, with k = 0.1- Calculate the time necessary to consume 35% of A. v = 200-dm3 P = 20 atm T = 227 C min dm3 mol min If the reaction is second orde -r,-kCA2 with k 0.7 Calculate the time to consume 80% of A. Also calculate the pressure in the reactor at this time if the temperature is 127 C

Explanation / Answer

According to ideal gas law

PV=nRT

CA0=P/RT

P=20 atm

R=0.082 lit.atm/mol K

T=227 C=500.15 K

So CA0=P/RT=20/(0.082*500.15)=0.4876 mol/lit

Now first order reaction

-rA=kCA

dCA/dt=-kCA

Now we know CA=CA0(1-X)

Where X is conversion

So dCA=-CA0dX

So

CA0dX/dt=CA0(1-X)

So integrating

-ln(1-X)=kt

So X=0.35

k=1/min

(1 dm3=1 lit)

So

t=43.07 min

Now if second order

-dCA/dt=kCA2

So integrating equation

(1/CA0)-(1/CA)=-kt

X=0.8

CA=0.4876*(1-0.8)=0.09752 mol/lit

k=0.7 lit/mol min

So

t=11.72 min

CA=P/RT

P=0.09752*0.082*(127+273.15)=3.2 atm

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