Answer is 371 You are running a series of experiments for the generic, gas-phase
ID: 526162 • Letter: A
Question
Answer is 371
You are running a series of experiments for the generic, gas-phase reaction, 2A rightarrow B in a batch reactor run isothermally. You have carried out two experiments today at 400 degree C and 500 degree C. The time required to reach a conversion of 50% (X 0.5) was 4 hr at 400Â degree C and 1 hr at 500 oC for an initial concentration of species A of C_AO = 0.5 mol/dm^3. You plan on running a reaction overnight at 200 degree C, but before you leave you want to make an estimate of what time you should come in the next morning. Assuming that the reaction is elementary, how long will it take to reach 50% conversion at 200 degree C?Explanation / Answer
For the elementary reaction –dCA/dt= KCA2
when the equation is integrated between t=0 CA =CAO and t= t , CA= CA
KCAOt= XA/(1-XA)
KCAOt is known as Damkohler number.
Where K is rate constant, CAO= initial concentration and XA= conversion
K is function of temperature. Let K1= rate constant at 400 deg.c and K2= rate constant at 500 deg.c
XA= conversion. At 400 deg.c Hence K1*0.5*4= 0.5/(1-0.5), K1= 0.5/M.sec
At 500 deg.c, K2*0.5*1= 0.5/(1-0.5)=1, K2=2/M.sec
From the data given, the activation energy of the reaction
Ln (K2/K1)= (E/R)*(1/T1-1/T2), E= activation energy, T1= 400 deg.c= 400+273= 673K, T2= 500 deg.c= 500+273= 773K, R= gas constant= 8.314 J/mole.K
Ln (2/0.5)= (E/8.314)*(1/673-1/773) , E= 59960 J/mole
Let T3= 200 deg.c,=200+273K= 473 K where rate constant need to be determined which is K3
Ln (K2/K3)=( 59960/8.314) *(1/473-1/773)
K2/K3= 371.45, K3= K2/371.45= 2/371.45 /M.sec
From KCAO*t= XA/(1-XA)= 1, at 200 deg.c, (2/371.45)* 0.5* t= 1
Hence time taken= 371.45 hr