Let’s break down the problem step by step: --- ## **Given Data** - **Bulk polymerization of styrene at 70°C** - Initial monomer concentration, \([M]_0 = 8.3 \text{ mol/L}\) - Initiator: AIBN (MW = 164) - \(k_d = 3.5 \times 10^{-5}\) sec\(^{-1}\) - **Measured Data:** | [I] (g/100ml) | Contraction (%/h) | \( \bar{X}_n \) | |:-------------:|:-----------------:|:---------------:| | 0.061 | 0.84 | 860 | | 0.100 | 1.20 | 625 | | 0.185 | 1.48 | 500 | | 0.352 | 2.09 | 363 | - 18% volume contraction for 100% conversion. - Termination by **disproportionation only**. - **Chain transfer** occurs only to monomer. --- ## **Step 1: Relationship Between Contraction and Rate of Polymerization** Contraction per hour is proportional to the rate of polymerization (\(R_p\)): \[ \text{Fraction polymerized per hour} = \frac{\text{Contraction (\%/h)}}{18\%} \] \[ \text{Moles styrene polymerized per L per hour} = [M]_0 \times \text{Fraction polymerized per hour} \] \[ R_p (\text{mol/L/s}) = \frac{[M]_0 \times \text{Fraction polymerized per hour}}{3600} \] --- ## **Step 2: Expressing \(\bar{X}_n\) in Terms of \(R_p\) and Constants** For free radical polymerization: \[ \bar{X}_n = \frac{\text{rate of monomer consumption}}{\text{rate of chain termination}} \] \[ \bar{X}_n = \frac{R_p}{R_t} \] For termination by disproportionation: \[ R_t = 2 k_t [R^\cdot]^2 \] The rate of polymerization: \[ R_p = k_p [M][R^\cdot] \] So, \[ [R^\cdot] = \frac{R_p}{k_p [M]} \] Substitute into \(R_t\): \[ R_t = 2 k_t \left(\frac{R_p}{k_p [M]}\right)^2 \] So, \[ \bar{X}_n = \frac{R_p}{2 k_t \left(\frac{R_p}{k_p [M]}\right)^2} = \frac{k_p^2 [M]^2}{2 k_t R_p} \] --- ## **Step 3: Calculating Each Quantity** ### (a) **Value of \( k_t / k_p \)** From above: \[ \bar{X}_n = \frac{k_p^2 [M]^2}{2 k_t R_p} \implies \frac{k_t}{k_p^2} = \frac{[M]^2}{2 \bar{X}_n R_p} \] But the question asks for \( \frac{k_t}{k_p} \): \[ \frac{k_t}{k_p} = \frac{k_t}{k_p^2} \cdot k_p \] Let’s first compute \( R_p \) for one row: #### For [I] = 0.061 g/100 ml, Contraction = 0.84 %/h, \( \bar{X}_n = 860 \): \[ \text{Fraction polymerized per hour} = \frac{0.84}{18} = 0.0467 \] \[ \text{Moles polymerized per hour} = 8.3 \times 0.0467 = 0.387 \text{ mol/L/h} \] \[ R_p = \frac{0.387}{3600} = 1.075 \times 10^{-4} \text{ mol/L/s} \] Now, \[ \frac{k_t}{k_p^2} = \frac{(8.3)^2}{2 \times 860 \times 1.075 \times 10^{-4}} \] \[ = \frac{68.89}{2 \times 860 \times 1.075 \times 10^{-4}} \] \[ = \frac{68.89}{0.1849} \] \[ = 372.5 \text{ L mol}^{-1} \text{ s} \] So, \[ \frac{k_t}{k_p^2} \approx 372.5 \text{ L mol}^{-1} \text{ s} \] But for \( \frac{k_t}{k_p} \), you need to know or estimate \( k_p \) (usually literature value), or you can report this as \( \frac{k_t}{k_p^2} \) unless \( k_p \) is given. --- ### (b) **Efficiency of AIBN (\(f\))** The efficiency (\(f\)) is the fraction of initiator radicals that actually start polymer chains. Theoretical rate of initiator radical production: \[ R_i = 2 f k_d [I] \] At steady state, \[ [R^\cdot] = \sqrt{\frac{R_i}{2 k_t}} = \sqrt{\frac{2 f k_d [I]}{2 k_t}} = \sqrt{\frac{f k_d [I]}{k_t}} \] But from rate of polymerization: \[ R_p = k_p [M] [R^\cdot] \] \[ [R^\cdot] = \frac{R_p}{k_p [M]} \] Set equal to above: \[ \frac{R_p}{k_p [M]} = \sqrt{\frac{f k_d [I]}{k_t}} \] Square both sides: \[ \left( \frac{R_p}{k_p [M]} \right)^2 = \frac{f k_d [I]}{k_t} \] \[ f = \frac{k_t}{k_d [I]} \left( \frac{R_p}{k_p [M]} \right)^2 \] You can use values from above and the data table to calculate \(f\). --- ### (c) **Value of \(C = \frac{k_t}{k_p} \)** Already derived above (if you have \( k_p \)), otherwise report as \( \frac{k_t}{k_p^2} \). --- ## **Summary of Assumptions** - **Termination by disproportionation only** - **Chain transfer only to monomer** - **Steady-state radical concentration** - **Bulk polymerization with negligible autoacceleration at low conversion** - **Dilatometrically measured contraction is proportional to polymerization** - **Neglecting gel effect (Trommsdorff effect)** --- ## **Final Expressions** ### **\( \bar{X}_n \) in terms of \( R_p \) and constants:** \[ \boxed{ \bar{X}_n = \frac{k_p^2 [M]^2}{2 k_t R_p} } \] --- ### **\( f \) (efficiency) in terms of \( R_p \), \( [I] \), \( k_p \), \( k_d \), and \( k_t \):** \[ \boxed{ f = \frac{k_t}{k_d [I]} \left( \frac{R_p}{k_p [M]} \right)^2 } \] --- ### **\( C = \frac{k_t}{k_p} \) or \( \frac{k_t}{k_p^2} \) if \( k_p \) is unknown** \[ \boxed{ C = \frac{k_t}{k_p^2} } \] --- Would you like to see the calculations for all the data points, or help with a specific part?