The irreversible, elementary gas phase reaction shown below is carried out in an

Question

The irreversible, elementary gas phase reaction shown below is carried out in an isothermal, isobaric CSTR reactor. Calculate the volume of the reactor that reaches conversion equal to 80%. A rightarrow B + C, - r_ A = k C_A Data: y_ A compositefunction = 1/2., k = 0.1 s^-1, F_A compositefunction = 2 mols, C_A compositefunction = 1 mol/liter A plot of in C_A (M) vs t (s) gives a slope = - 0.03 and an intercept = 0 with R^2 = 0.99. A plot of l/C_A (M) vs t (s) gives a slope = 0.02 and an intercept = 1 with R^2 = 0.6. A plot of C_A (M) vs t (s) gives a slope = 0.01 and an intercept = 1 with R^2 = 0.70. Calculate the order of reaction and the k value. Justify using the integral method.

Explanation / Answer

XA= conversion and K= rate constant , CAO=initial concentration = 1 mol/Liter, FAO= 2 mol/sec

For a CSTR, T= space time= V/VO=VCAO/FAO= CAOXA/KCAO*(1-XA)

V/FAO= XA/(KCAO*(1-XA)= 0.8/(0.1*1*0.2)= 40

V = 40*2= 80 Llters

for 1st order reaction, n=1, -dCA/dt= KCA when integrated

lnCA= lnCAO-Kt, where CAO= initial concentration of reactant A, k is the rate constant and t is the time. So a plot of lnK vs time gives a straight line whose slope is –K.

for second order reaction –dCA/dt= KCA2, when integrated , 1/CA= 1/CAO+Kt

so a plot of 1/CA vs t gives straight line.

For zero order reaction – dCA/dt= K

When integrated, CA= CAO-Kt,

So a plot of CA vs t gives a straight line.

For the case given , which ever gives higher value of R2 will be the best fit. Accordingly, 1st order reaction represents the data having value of R2=0.99. The rate constant is 0.03/s

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