Let's break this down step by step: ## 1. **Defining the Problem** - **Monomers:** 1,6-hexanediamine (A₂ type) and adipic acid (B₂ type), both at **1.2 mol**. - **Conversion:** Calculate at **p = 1 (100%)** and **p = 0.968 (96.8%)**. - **Impurity:** Adipic acid contains **0.05% inactive impurity** (acts as a mono-functional chain stopper). - **Given Data:** - At **p = 1.0: \(\overline{n} = 39, \overline{M}_n = 8814\)** - At **p = 0.968: \(\overline{n} = 17.6, \overline{M}_n = 3997\)** - **Tasks:** - **A.** How to calculate average degree of polymerization (\(\overline{n}\)) and number average molecular weight (\(\overline{M}_n\)) **without impurity**. - **B.** Recalculate with **0.05% impurity**. --- ## 2. **Degree of Polymerization (DP) Without Impurity** ### **Step 1: For Ideal Case (No Impurity)** For an **A₂ + B₂ polycondensation** with equal mols, the **Carothers equation** applies: \[ \overline{n} = \frac{1}{1 - p} \] where \( p \) is the **fractional conversion** of functional groups. ### **At p = 1.0:** \[ \overline{n} = \frac{1}{1-1} = \infty \] But in practice, you get a finite value due to small deviations or unreacted ends (the data says \( n = 39 \)), so it's probably p = 0.975 or so in practice. (But let's use the data.) ### **At p = 0.968:** \[ \overline{n} = \frac{1}{1-0.968} = \frac{1}{0.032} \approx 31.25 \] But your data gives \( n = 17.6 \), so there may be some stoichiometry imbalance or capping. ### **General formula for unequal functional group conversion:** \[ \overline{n} = \frac{1 + r}{1 + r - 2rp} \] where \( r = \frac{\text{mols of limiting functional group}}{\text{mols of excess functional group}} \). But here, \( r = 1 \). So, again, \[ \overline{n} = \frac{1}{1-p} \] --- ## 3. **Number Average Molecular Weight (\( \overline{M}_n \))** \[ \overline{M}_n = \overline{n} \times M_{\text{repeat unit}} \] ### **Repeat Unit Calculation:** - **Hexanediamine** (H₂N-(CH₂)₆-NH₂): MW = \( 6 \times 14.03 + 2 \times 1.008 + 2 \times (14.01 + 1.008) \) = 116.2 g/mol - **Adipic Acid** (HOOC-(CH₂)₄-COOH): MW = \( 6 \times 14.03 + 2 \times 1.008 + 4 \times 12.01 + 4 \times 1.008 + 4 \times 16.00 \) = 146.1 g/mol But the **repeat unit** is formed by condensation (loss of H₂O): \[ \text{Repeat unit} = \text{Hexanediamine} + \text{Adipic acid} - 2H_2O \] - **Hexanediamine:** 116.2 - **Adipic acid:** 146.1 - **2 H₂O:** 36.0 \[ \text{Repeat unit} = 116.2 + 146.1 - 36.0 = 226.3 \text{ g/mol} \] --- ## 4. **With Impurity (Chain Stopper)** ### **How does impurity affect DP?** - **Mono-functional impurity** (in diacid) acts as a chain stopper. - **Number of moles of impurity:** \( 0.05\% \) of 1.2 mol = \( 0.0005 \times 1.2 = 0.0006 \) mol ### **Adjusted DP formula:** \[ \overline{n} = \frac{\text{Total moles of functional groups} - \text{moles of monofunctional impurity}}{\text{moles of monofunctional impurity}} \] But more generally, for **equimolar A₂ and B₂, with c mols of chain stopper:** \[ \overline{n} = \frac{2N_0}{2c + (1-p)2N_0} \] where \( N_0 \) is initial moles of each monomer. ### **Plugging values:** - \( N_0 = 1.2 \) mol - \( c = 0.0006 \) mol - For 100% conversion (\( p = 1 \)): \[ \overline{n} = \frac{2 \times 1.2}{2 \times 0.0006} = \frac{2.4}{0.0012} = 2000 \] - For \( p = 0.968 \): \[ \overline{n} = \frac{2 \times 1.2}{2 \times 0.0006 + (1-0.968) \times 2 \times 1.2} \] \[ = \frac{2.4}{0.0012 + 0.0768} = \frac{2.4}{0.0780} \approx 30.77 \] --- ## 5. **Summary Table** | Conversion (p) | DP (\(\overline{n}\)), no impurity | DP (\(\overline{n}\)), with impurity | \(\overline{M}_n\), no impurity | \(\overline{M}_n\), with impurity | |----------------|--------------------|---------------------|----------------|-----------------| | 1.0 | \(\infty\) (39)* | 2000 | \(\infty\) (8814)* | 2000 × 226.3 = 452,600 | | 0.968 | 31.25 (17.6)* | 30.77 | 31.25 × 226.3 = 7072 (3997)* | 30.77 × 226.3 = 6960 | _*Values in parentheses are from your data, which may reflect practical yields._ --- ## 6. **Final Answers** ### **A. Without impurity:** - At **p = 1.0:** \(\overline{n} = \frac{1}{1-1} \rightarrow\) very large, in practice your data says 39. - At **p = 0.968:** \(\overline{n} = 31.25\), your data says 17.6. ### **B. With impurity (0.05%):** - At **p = 1.0:** \(\overline{n} = 2000\) - At **p = 0.968:** \(\overline{n} = 30.77\) - **\(\overline{M}_n\) with impurity:** - At **p = 1.0:** \(2000 \times 226.3 = 452,600\) g/mol - At **p = 0.968:** \(30.77 \times 226.3 = 6960\) g/mol --- ### **References** - Carothers, W. H. "Polymers and Polycondensation," *J. Am. Chem. Soc.*, **1932**. - Flory, P. J., *Principles of Polymer Chemistry*, Cornell University Press. --- ## **Formulas Used** **Carothers equation for DP (no impurity, equimolar):** \[ \overline{n} = \frac{1}{1-p} \] **With chain stopper (impurity):** \[ \overline{n} = \frac{2N_0}{2c + (1-p)2N_0} \] **Number average molecular weight:** \[ \overline{M}_n = \overline{n} \times M_{\text{repeat unit}} \] --- ### **If you need further clarification or calculation steps, just ask!**