Elementary Principles 494 - 11.28(cont'd The plot of CS1 vs...

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11- 23 The plot of C S1 vs. t begins at (t=0, C S1 =3). When t=0, the slope (=dC S1 /dt) is −× = 008 3 024 .. . As t increases, C S1 decreases dC S1 /dt=-0.08C S1 becomes less negative, approaches zero as t . The curve is therefore concave up. The plot of C S2 vs. t begins at (t=0, C S2 =0). When t=0, the slope (=dC S2 /dt) is 008 3 0 .( ) . −= . As t increases, C S2 increases, C S1 decreases (C S2 < C S1 ) dC S2 /dt =0.08(C S1 -C S2 ) becomes less positive until dC S2 /dt changes to negative (C S2 > C S1 ). Then C S2 decreases with increasing t as well as C S1 . Finally dC S2 /dt approaches zero as t →∞ . Therefore, C S2 increases until it reaches a maximum value, then it decreases. The plot of C S3 vs. t begins at (t=0, C S3 =0). When t=0, the slope (=dC S3 /dt) is 004 0 0 0 .( ) . As t increases, C S2 increases (C S3 < C S2 ) dC S3 /dt =0.04(C S2 -C S3 ) becomes positive C S2 increases 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.

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