A chemical reaction A- Bis carried out in a closed vessel. The following data ar

Question

A chemical reaction A- Bis carried out in a closed vessel. The following data are taken for the concentration of A, CA(g/L), as a function of time, t(min), from the start of the reaction C t(min) t(min) 0 36.0 65.0 100.0 160.0 CA (g/liter) 0.1823 0.12650 0.09880 0.07840 0.06090 0.0495 A proposed reaction mechanism predicts that CA and t should be related by the expression (CA CAe) (CAo CAe) where k is the reaction rate constant. Estimating k EX You may wish to plot the data carefully or use an automated, numerical tool for linear regression. Plot the data appropriately to determine the value of k. 0.00727 min the tolerance is +/-4%

Explanation / Answer

solution:

1)here we are provided with data for concetration Ca and time and relation is

(Ca-Cae)/(Cao-Cae)=e^(-K*t)

here at initial time T0 and at time Tinfinity,value of K is not correct so we just calculate it for in between four data points and take average of them and it is found that they are in tolerance less than 4%,so for

K=ln((Ca-Cae)/(Cao-Cae))/(-t)

for T1=36 min,we get

K1=ln((Ca-Cae)/(Cao-Cae))/(-t)=K=ln((.01265-.0495)/(.1328))/(-36)=.015139

for T2=65 min,K2=.015244

for T3=100 min,K3=.01525

for T4=160 min,K4=.0153451

here average value of K is given as

Kavg=K1+K2+K3+k4/4=.0152445 per min

2)here value of mass of B is given by

Ca gives Cb and no biproduct so we get that

Cb=Cao-Ca

where mass of B is given by

Mb=175Cb

here for T=1.75 hrs or 105 min

Kavg=ln((Ca-Cae)/(Cao-Cae))/(-t)

Ca=.0762952

Cb=Cao-Ca=.10600 g/L

Mb=175*.106=18.55 gm

3)here time required to achieved value of

Ca=1.1 Cae

we have that

t1=ln((Ca-Cae)/(Cao-Cae))/(-Kavg)

t1=215.786 min

corresponding mass of B is

mB1=175*(.1*Cae)=22.3737 gm

for Ca=1.05 Ca

t2=ln((Ca-Cae)/(Cao-Cae))/(-Kavg)=261.257 min

mB2=175*.05*Cae=31.4693 gm

for Ca=1.01 Ca

t2=ln((Ca-Cae)/(Cao-Cae))/(-Kavg)=366.835 min

mB3=175*.01*cae=31.8158 gm

4) in this way above all value can be computed considering equillibrium at each instant.

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