# Step-by-Step Solution: Estimating Activity Coefficients using UNIFAC Let's estimate the activity coefficients (\(\gamma\)) for an **equimolar mixture of benzene and water** at **300 K** and **350 K** using the **UNIFAC group contribution method**. --- ## 1. **Identifying Functional Groups** ### Benzene (C\(_6\)H\(_6\)) - **Groups:** 6 × Aromatic CH (Ph-CH) ### Water (H\(_2\)O) - **Groups:** 1 × H\(_2\)O --- ## 2. **UNIFAC Group Parameters** | Group | Symbol | \(R_k\) | \(Q_k\) | |---------------|--------|---------|---------| | Aromatic CH | ACH | 0.5313 | 0.4000 | | H\(_2\)O | H2O | 0.9200 | 1.4000 | **Note:** Values from UNIFAC tables. Real UNIFAC uses more detailed groups (e.g., aromatic C, aromatic CH, etc.). --- ## 3. **Compositional Information** - **Mole fraction:** \(x_1 = x_2 = 0.5\) (equimolar) - **Molecule 1:** Benzene (\(n_{\text{ACH}} = 6\)) - **Molecule 2:** Water (\(n_{\text{H2O}} = 1\)) --- ## 4. **Calculate \(R\) and \(Q\) for Each Component** ### Benzene - \(R_1 = 6 \times 0.5313 = 3.1878\) - \(Q_1 = 6 \times 0.4000 = 2.4000\) ### Water - \(R_2 = 1 \times 0.9200 = 0.9200\) - \(Q_2 = 1 \times 1.4000 = 1.4000\) --- ## 5. **Calculate Mixture Parameters** ### **Total \(R\) and \(Q\)** - \(R = x_1 R_1 + x_2 R_2 = 0.5 \times 3.1878 + 0.5 \times 0.9200 = 2.0539\) - \(Q = x_1 Q_1 + x_2 Q_2 = 0.5 \times 2.4000 + 0.5 \times 1.4000 = 1.9000\) ### **Volume and Surface Fractions** For component \(i\): - \(\phi_i = \frac{x_i R_i}{\sum x_j R_j}\) - \(\theta_i = \frac{x_i Q_i}{\sum x_j Q_j}\) #### Benzene (\(i=1\)): - \(\phi_1 = \frac{0.5 \times 3.1878}{2.0539} = 0.775\) - \(\theta_1 = \frac{0.5 \times 2.4000}{1.9000} = 0.632\) #### Water (\(i=2\)): - \(\phi_2 = \frac{0.5 \times 0.9200}{2.0539} = 0.224\) - \(\theta_2 = \frac{0.5 \times 1.4000}{1.9000} = 0.368\) --- ## 6. **Combinatorial Part (\(\ln \gamma_i^C\))** \[ \ln \gamma_i^C = 1 - V_i + \ln V_i - 5 Q_i \left[ 1 - \frac{V_i}{F_i} + \ln \left( \frac{V_i}{F_i} \right) \right] \] where: - \(V_i = \frac{R_i}{\sum x_j R_j}\) - \(F_i = \frac{Q_i}{\sum x_j Q_j}\) ### For Benzene (\(i=1\)): - \(V_1 = 3.1878/2.0539 = 1.552\) - \(F_1 = 2.4000/1.9000 = 1.263\) \[ \ln \gamma_1^C = 1 - 1.552 + \ln(1.552) - 5 \times 2.4 \left[ 1 - \frac{1.552}{1.263} + \ln \left( \frac{1.552}{1.263} \right) \right] \] Calculate step by step: - \(1 - 1.552 = -0.552\) - \(\ln(1.552) = 0.440\) - \(\frac{1.552}{1.263} = 1.229\) - \(\ln(1.229) = 0.206\) - \(1 - 1.229 + 0.206 = -0.023\) - \(-5 \times 2.4 \times -0.023 = 0.276\) Total: - \(-0.552 + 0.440 + 0.276 = 0.164\) **So:** \(\ln \gamma_1^C \approx 0.16\) ### For Water (\(i=2\)): - \(V_2 = 0.9200/2.0539 = 0.448\) - \(F_2 = 1.4000/1.9000 = 0.737\) - \(1 - 0.448 = 0.552\) - \(\ln(0.448) = -0.804\) - \(\frac{0.448}{0.737} = 0.608\) - \(\ln(0.608) = -0.497\) - \(1 - 0.608 - 0.497 = -0.105\) - \(-5 \times 1.4 \times -0.105 = 0.735\) Total: - \(0.552 - 0.804 + 0.735 = 0.483\) **So:** \(\ln \gamma_2^C \approx 0.48\) --- ## 7. **Residual Part (\(\ln \gamma_i^R\))** \[ \ln \gamma_i^R = \sum_k \nu_k^{(i)} \left[ \ln \Gamma_k^{(\text{mix})} - \ln \Gamma_k^{(i)} \right] \] Where: - \(\nu_k^{(i)}\): number of group \(k\) in molecule \(i\) - \(\Gamma_k^{(\text{mix})}\): group activity in mixture - \(\Gamma_k^{(i)}\): group activity in pure \(i\) ### **UNIFAC Interaction Parameters** (\(a_{mn}\)) (in K): | Group Pair | \(a_{mn}\) (K) | \(a_{nm}\) (K) | |--------------|---------------|---------------| | ACH-H2O | 986 | 986 | | H2O-ACH | 986 | 986 | | ACH-ACH | 0 | 0 | | H2O-H2O | 0 | 0 | **Assume only ACH and H2O interact.** #### **Calculate Group Fractions in Mixture (\(X_k\)):** - Total groups: \(0.5 \times 6 + 0.5 \times 1 = 3 + 0.5 = 3.5\) - ACH: \(0.5 \times 6 = 3\), fraction: \(3/3.5 = 0.857\) - H2O: \(0.5 \times 1 = 0.5\), fraction: \(0.5/3.5 = 0.143\) #### **Calculate \(\Theta_k\) (surface fractions):** \[ \Theta_k = \frac{X_k Q_k}{\sum X_j Q_j} \] - For ACH: \(0.857 \times 0.4 = 0.343\) - For H2O: \(0.143 \times 1.4 = 0.200\) - Sum: \(0.543\) - \(\Theta_{\text{ACH}} = 0.343/0.543 = 0.632\) - \(\Theta_{\text{H2O}} = 0.200/0.543 = 0.368\) #### **Calculate \(\Psi_{mn}\):** \[ \Psi_{mn} = \exp \left( -\frac{a_{mn}}{T} \right) \] - At **300 K**: \(\Psi_{\text{ACH,H2O}} = \exp(-986/300) = \exp(-3.287) = 0.0375\) - At **350 K**: \(\Psi_{\text{ACH,H2O}} = \exp(-986/350) = \exp(-2.817) = 0.0599\) - \(\Psi_{\text{ACH,ACH}} = 1\) - \(\Psi_{\text{H2O,H2O}} = 1\) #### **Calculate Residual Activity Coefficients:** \[ \ln \Gamma_k = Q_k \left[ 1 - \ln \left( \sum_{m} \Theta_m \Psi_{mk} \right) - \sum_{m} \frac{\Theta_m \Psi_{km}}{\sum_{n} \Theta_n \Psi_{nm}} \right] \] **For brevity, let's do ACH at 300 K:** - \(\sum_m \Theta_m \Psi_{m,\text{ACH}} = 0.632 \cdot 1 + 0.368 \cdot 0.0375 = 0.632 + 0.0138 = 0.6458\) - \(\sum_n \Theta_n \Psi_{n,\text{ACH}} = 0.6458\) (for denominator) \[ \sum_m \frac{\Theta_m \Psi_{\text{ACH},m}}{\sum_n \Theta_n \Psi_{n,m}} \] - For \(m = \text{ACH}\): \(\Psi_{\text{ACH},\text{ACH}} = 1\), \(\sum_n \Theta_n \Psi_{n,\text{ACH}} = 0.6458\), so: \(0.632/0.6458 = 0.979\) - For \(m = \text{H2O}\): \(\Psi_{\text{ACH},\text{H2O}} = 0.0375\), \(\sum_n \Theta_n \Psi_{n,\text{H2O}} = 0.632 \cdot 0.0375 + 0.368 \cdot 1 = 0.0237 + 0.368 = 0.3917\), so: \(0.368 \cdot 0.0375 / 0.3917 = 0.0138 / 0.3917 = 0.035\) Sum: \(0.979 + 0.035 = 1.014\) \[ \ln \Gamma_{\text{ACH}} = 0.4 \left[ 1 - \ln(0.6458) - 1.014 \right] \] - \(\ln(0.6458) = -0.437\) - \(1 - (-0.437) - 1.014 = 1 + 0.437 - 1.014 = 0.423\) - \(0.4 \cdot 0.423 = 0.169\) So, \(\ln \Gamma_{\text{ACH}} = 0.169\) **Repeat for H2O:** - \(\sum_m \Theta_m \Psi_{m,\text{H2O}} = 0.632 \cdot 0.0375 + 0.368 \cdot 1 = 0.0237 + 0.368 = 0.3917\) - \(\ln(0.3917) = -0.937\) - For \(m = \text{ACH}\): \(\Psi_{\text{H2O},ACH} = 0.0375\), denominator = 0.6458, so \(0.632 \cdot 0.0375 / 0.6458 = 0.0237 / 0.6458 = 0.037\) - For \(m = \text{H2O}\): \(\Psi_{\text{H2O},\text{H2O}} = 1\), denominator = 0.3917, so \(0.368 / 0.3917 = 0.94\) - Sum: \(0.037 + 0.94 = 0.977\) - \(1 - (-0.937) - 0.977 = 1 + 0.937 - 0.977 = 0.96\) - \(1.4 \cdot 0.96 = 1.34\) So, \(\ln \Gamma_{\text{H2O}} = 1.34\) #### **For Pure Components:** - For pure benzene: only ACH present, so \(\Theta_{\text{ACH}}

In the context of UNIFAC calculations, \(\Theta_k\) (surface fraction of group \(k\)) for pure benzene refers to the fractional contribution of the group \(k\) within the pure substance. **Since pure benzene consists solely of aromatic CH groups (ACH):** - The total number of groups in benzene is 6 ACH groups. - The surface fraction \(\Theta_{\text{ACH}}\) in pure benzene is calculated as: \[ \Theta_{\text{ACH}} = \frac{n_{\text{ACH}} \times Q_{\text{ACH}}}{\sum_{k} n_k Q_k} \] where: - \(n_{\text{ACH}} = 6\) (number of ACH groups in benzene) - \(Q_{\text{ACH}}\) is the group surface parameter from UNIFAC tables **Since only ACH groups are present:** \[ \sum_{k} n_k Q_k = 6 \times Q_{\text{ACH}} \] Thus, \[ \Theta_{\text{ACH}} = \frac{6 \times Q_{\text{ACH}}}{6 \times Q_{\text{ACH}}} = 1 \] **Therefore, in pure benzene:** \[ \boxed{ \Theta_{\text{ACH}}^{\text{pure benzene}} = 1 } \] and all other group fractions are zero. **In summary:** - For pure benzene, the surface fraction \(\Theta_{\text{ACH}}\) is **1** because the entire surface is contributed by ACH groups. - Similarly, for pure water, \(\Theta_{\text{H2O}} = 1\). This simplifies the residual activity coefficient calculations for pure components.