Chapter 9: Reaction Mechanisms, Pathways, Bioreactions and Bioreactors


Example 9.5: Calculating Xn for Stoichiometric Inbalance

Case A: \( r = 0.99 \), \( p = 1 \)

\(\bar{X_n} = \frac{r+1}{r-1} = \frac{0.99+1}{1-0.99} = \frac{1.99}{0.01} = 199\)

Case B: \( r = 1 \), \( p = 0.99 \)

\(\bar{X_n} = \frac{r+1}{r+1-2rp} = \frac{1+1}{1+1-2(0.99)} = \frac{1}{1-p} = \frac{1}{0.01} = 100\)

Case C: \( p = 0.99 \), \( r = 0.99 \)

\(\bar{X_n} = \frac{r+1}{r+1-2rp} = \frac{1.99}{1.99-2(0.99)(0.99)} = 66.78\)

Case D: \( r = 1 \), \( p = 0.999 \)

\(\bar{X_n} = \frac{1}{1-p} = \frac{1}{0.001} = 1000\)

Case E: \( r = 0.99 \), \( p = 0.999 \)

\(\bar{X_n} = \frac{r+1}{r+1-2rp} = \frac{1.99}{1.99-2(0.99)(0.999)} = \frac{1.99}{0.01198} = 166.1\)

Summary

r p Xn
1 0.99 100
1 0.999 1000
0.99 1 199
0.99 0.99 66.8
0.99 0.999 166.1

Only a 1% imbalance reduces the number average chain length by almost an order of magnitude

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