11- 23The plot of CS1vs. t begins at (t=0, CS1=3). When t=0, the slope (=dCS1/dt) is −×=−008 3024... As t increases, CS1decreases ⇒dCS1/dt=-0.08CS1becomes less negative, approaches zero as t→∞. The curve is therefore concave up.The plot of CS2vs. t begins at (t=0, CS2=0). When t=0, the slope (=dCS2/dt) is 008 3 0.( ) .−=. As t increases, CS2increases, CS1decreases (CS2< CS1)⇒dCS2/dt =0.08(CS1-CS2) becomes less positive until dCS2/dt changes to negative (CS2> CS1). Then CS2decreases with increasing t as well as CS1. Finally dCS2/dt approaches zero as t→∞. Therefore, CS2increases until it reaches a maximum value, then it decreases. The plot of CS3vs. t begins at (t=0, CS3=0). When t=0, the slope (=dCS3/dt) is 004 0 00.(). As t increases, CS2increases (CS3< CS2)⇒dCS3/dt =0.04(CS2-CS3) becomes positive ⇒CS2increases with increasing t until dC
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.