a) You may wish to plot the data carefully or use an automated, numerical tool f
ID: 474157 • Letter: A
Question
a)
You may wish to plot the data carefully or use an automated, numerical tool for linear regression. Problem 2.43 Batch Reactor A chemical reaction A- B is 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 min t(min) 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 In (CA CAe) (CA0 CAe) J -kt where k is the reaction rate constant.Explanation / Answer
the plot of ln [(CA-CAe)/ (CAO-CAe)] =-Kt
is shown. CAe= Equilibrium concentration which is achierved at infinite time and CAO= concentration at t=0 and K is rate constant
the rate constant for the reaction is K= ( slope)= 0.015/min
at t= 1.5 hr =1.5*60 min
ln[(CA-CAe)/ [CAO-CAe)] = -0.015*1.5*60
ln(CA-0.0495)/ (0.1823-0.0495)= -1.35
(CA-0.0495)/0.1328= 0.2592
CA-0.0495 = 0.2592*0.1328= 0.03442
CA= 0.03442+0.0495= 0.08392
A consumed = B formed
A consumed = A initially present- A remaining = 0.1823-0.08392 =0.09838 g/L
From the stoichiometry, B formed = 0.09838 g/L
volume of tank = 175L . Mass of B formed = 0.09838*175= 17.21 gm
2.
for CA= 1.1 CAe, CA= 1.1*0.0495 =0.05445 g/L,
ln[0.05445-0.0495)/ (0.1823-0.0495)= -0.015*t, t= 219 min , CA =0.05445 g/L, CA reacted= 0.1823-0.05445= 0.12785 g/L. CB= 0.12785 g/L. mass of B= 175*0.12785 =22.37 gm
b) for CA=1.05CAe, CA= 1.05*0.0495 = 0.051975 g/L, A reacted = 0.1823-0.051957=0.130325 g/L, B formed = 0.130325 g/L, mass of B= 0.130325*175=22.81 gm
ln[(0.051975-0.0495)/(0.1823-0.0495) =-0.015*t, t= 265 min
c) for CA=1.01 CAe, CA=1.01*0.0495 = 0.04995 g/L, A reacted = 0.1823-0.04995 = 0.13235 g/L, B formed = 0.13235 g/L, mass of B= 0.13235*175= 23.16 gm
ln [ (0.04995-0.0495)/ (0.1823-0.0495)] = -0.015*t, t= 329 min
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