Let's walk through the solution step-by-step: --- ### **Question Recap** Estimate the **activity coefficients** (γ) for an **equimolar mixture** of **benzene and water** at **300 K** and **350 K** using **UNIFAC**. --- ## **Step 1: Identify Components and Groups** **Benzene (C₆H₆):** - Structure: Six carbon ring, all carbons bonded to one hydrogen - UNIFAC Groups: 6 × CH (aromatic), or sometimes C₆H₆ as a special group **Water (H₂O):** - UNIFAC Groups: 1 × H₂O --- ## **Step 2: Find UNIFAC Group Parameters** You need: - **Rk** (volume parameter) - **Qk** (surface area parameter) - **Interaction parameters a_ij** between groups **Standard UNIFAC groups:** - **CH (aromatic):** Sometimes denoted as ACH - **H₂O:** Water group Standard values (from UNIFAC tables): | Group | Rk | Qk | |------------|------|------| | CH (ACH) | 0.531| 0.400| | H₂O | 0.920| 1.400| **Benzene:** 6 × CH (aromatic) - \( R_{benzene} = 6 \times 0.531 = 3.186 \) - \( Q_{benzene} = 6 \times 0.400 = 2.400 \) **Water:** 1 × H₂O - \( R_{water} = 0.920 \) - \( Q_{water} = 1.400 \) --- ## **Step 3: Calculate Mole Fractions** **Equimolar mixture:** - \( x_{benzene} = 0.5 \) - \( x_{water} = 0.5 \) --- ## **Step 4: Calculate Combinatorial and Residual Parts** ### **A. Combinatorial Part (γᶜ):** The combinatorial part (for each component \( i \)) is: \[ \ln \gamma_i^C = 1 - V_i + \ln(V_i) - 5 q_i \left[1 - \frac{V_i}{F_i} + \ln \frac{V_i}{F_i}\right] \] Where: - \( r_i = \sum_k v_{ik} R_k \), \( q_i = \sum_k v_{ik} Q_k \) - \( V_i = \frac{r_i}{\sum_j x_j r_j} \) - \( F_i = \frac{q_i}{\sum_j x_j q_j} \) **Calculate:** - \( r_{benzene} = 3.186 \) - \( r_{water} = 0.920 \) - \( q_{benzene} = 2.400 \) - \( q_{water} = 1.400 \) \[ \sum_j x_j r_j = 0.5 \times 3.186 + 0.5 \times 0.920 = 1.593 + 0.460 = 2.053 \] \[ \sum_j x_j q_j = 0.5 \times 2.400 + 0.5 \times 1.400 = 1.200 + 0.700 = 1.900 \] Now, compute for each: - **Benzene:** - \( V_{benzene} = \frac{3.186}{2.053} = 1.552 \) - \( F_{benzene} = \frac{2.400}{1.900} = 1.263 \) - **Water:** - \( V_{water} = \frac{0.920}{2.053} = 0.448 \) - \( F_{water} = \frac{1.400}{1.900} = 0.737 \) Plug these into the combinatorial formula for each. --- ### **B. Residual Part (γᴿ):** The residual part uses group interaction parameters and is more complex: \[ \ln \gamma_i^R = \sum_k v_{ik} \left[ \ln \Gamma_k - \ln \Gamma_k^{(i)} \right] \] - \( v_{ik} \): number of group \( k \) in molecule \( i \) - \( \Gamma_k \): activity coefficient of group \( k \) in the mixture - \( \Gamma_k^{(i)} \): activity coefficient of group \( k \) in pure component \( i \) You need the UNIFAC group-group interaction parameters \( a_{mn} \) for the pairs: - (CH, CH) - (CH, H₂O) - (H₂O, H₂O) You'd use the UNIFAC equations to compute \( \Gamma_k \) for each group at each temperature (using the \( a_{mn} \) parameters, which are temperature-dependent). --- ## **Step 5: Example Calculation (Summary)** Since the **actual calculation requires UNIFAC tables** (for \( a_{mn} \)), I'll outline the steps: 1. **List all functional groups and their \( v_{ik} \).** 2. **Find all relevant \( a_{mn} \) values for 300 K and 350 K** (from UNIFAC tables). 3. **Calculate combinatorial part** (\( \ln \gamma_i^C \)) using the equations above. 4. **Calculate residual part** (\( \ln \gamma_i^R \)) using group compositions and interaction parameters. 5. **Sum both parts** to get the total activity coefficient: \[ \ln \gamma_i = \ln \gamma_i^C + \ln \gamma_i^R \] 6. **Exponentiate** to get \( \gamma_i \). --- ## **Step 6: Reference UNIFAC Calculators** For a real assignment, you can use: - [UNIFAC Online Calculator](https://www.ddbst.com/unifac.html) - Aspen Plus, ChemCAD, or other process simulators **Typical Results:** - The activity coefficients for benzene in water are **very high** (γ ≫ 1) due to poor solubility. - Water in benzene also has a high γ. --- ## **Summary Table (Expected Qualitative Results)** | T (K) | γ_benzene | γ_water | |-------|-----------|---------| | 300 | Large | Large | | 350 | Slightly lower, but still large | Slightly lower, but still large | --- **Note:** - The actual numbers depend on the group interaction parameters from the UNIFAC tables. - If you need the actual values, let me know if you want a sample calculation with assumed group parameters! --- **Would you like to see a sample calculation with made-up group interaction values, or do you want to try an online calculator for the exact numbers?**