Let's break down the solution step by step. --- ## **Step 1: Write Down the Given Data** - Diameter of A (\(d_A\)): **442 pm** \(= 442 \times 10^{-12}\) m - Diameter of B (\(d_B\)): **885 pm** \(= 885 \times 10^{-12}\) m - Viscosity (\(\eta\)): **1.27 cP** \(= 1.27 \times 10^{-3}\) kg m\(^{-1}\) s\(^{-1}\) - Temperature (\(T\)): **20°C** \(= 293\) K - Initial [A]: **.200 mol dm\(^{-3}\)** - Initial [B]: **.150 mol dm\(^{-3}\)** - \(1 \text{ cP} = 1 \times 10^{-3}\) kg m\(^{-1}\) s\(^{-1}\) --- ## **Step 2: Find the Diffusion-Controlled Rate Constant (\(k_{\text{diff}}\))** The Smoluchowski equation for the diffusion-controlled rate is: \[ k_{\text{diff}} = 4\pi N_A (D_A + D_B) R_{AB} \] Where: - \(N_A\): Avogadro's number (\(6.022 \times 10^{23}\) mol\(^{-1}\)) - \(D_A, D_B\): Diffusion coefficients of A and B - \(R_{AB}\): Encounter distance (\(= \frac{d_A + d_B}{2}\)) --- ### **Step 2.1: Calculate \(R_{AB}\)** \[ R_{AB} = \frac{d_A + d_B}{2} = \frac{442 + 885}{2} \times 10^{-12} \text{ m} = \frac{1327}{2} \times 10^{-12} \text{ m} = 663.5 \times 10^{-12} \text{ m} = 6.635 \times 10^{-10} \text{ m} \] --- ### **Step 2.2: Calculate Diffusion Coefficients (\(D_A, D_B\))** The Stokes-Einstein equation: \[ D = \frac{k_B T}{6\pi \eta r} \] where \(r = \frac{d}{2}\). #### **For A:** \[ r_A = \frac{442}{2} \times 10^{-12} \text{ m} = 221 \times 10^{-12} \text{ m} = 2.21 \times 10^{-10} \text{ m} \] #### **For B:** \[ r_B = \frac{885}{2} \times 10^{-12} \text{ m} = 442.5 \times 10^{-12} \text{ m} = 4.425 \times 10^{-10} \text{ m} \] #### **Constants:** - \(k_B = 1.38 \times 10^{-23}\) J K\(^{-1}\) - \(T = 293\) K - \(\eta = 1.27 \times 10^{-3}\) kg m\(^{-1}\) s\(^{-1}\) #### **Calculate \(D_A\):** \[ D_A = \frac{1.38 \times 10^{-23} \times 293}{6\pi \times 1.27 \times 10^{-3} \times 2.21 \times 10^{-10}} \] Calculate the denominator: \[ 6\pi \times 1.27 \times 10^{-3} \times 2.21 \times 10^{-10} \approx 6 \times 3.1416 \times 1.27 \times 10^{-3} \times 2.21 \times 10^{-10} \] \[ = 18.8496 \times 1.27 \times 10^{-3} \times 2.21 \times 10^{-10} \] \[ = 23.935 \times 10^{-3} \times 2.21 \times 10^{-10} = 52.894 \times 10^{-13} = 5.2894 \times 10^{-12} \] Numerator: \[ 1.38 \times 10^{-23} \times 293 = 404.34 \times 10^{-23} = 4.0434 \times 10^{-21} \] \[ D_A = \frac{4.0434 \times 10^{-21}}{5.2894 \times 10^{-12}} = .7646 \times 10^{-9} = 7.65 \times 10^{-10} \text{ m}^2 \text{ s}^{-1} \] #### **Calculate \(D_B\):** \[ D_B = \frac{1.38 \times 10^{-23} \times 293}{6\pi \times 1.27 \times 10^{-3} \times 4.425 \times 10^{-10}} \] Denominator: \[ 6\pi \times 1.27 \times 10^{-3} \times 4.425 \times 10^{-10} = 18.8496 \times 1.27 \times 10^{-3} \times 4.425 \times 10^{-10} = 23.935 \times 10^{-3} \times 4.425 \times 10^{-10} = 105.902 \times 10^{-13} = 1.059 \times 10^{-11} \] Numerator: Same as before (\(4.0434 \times 10^{-21}\)) \[ D_B = \frac{4.0434 \times 10^{-21}}{1.059 \times 10^{-11}} = .382 \times 10^{-9} = 3.82 \times 10^{-10} \text{ m}^2 \text{ s}^{-1} \] --- ### **Step 2.3: Calculate \(k_{\text{diff}}\)** \[ k_{\text{diff}} = 4\pi N_A (D_A + D_B) R_{AB} \] \[ = 4\pi \times 6.022 \times 10^{23} \times (7.65 \times 10^{-10} + 3.82 \times 10^{-10}) \times 6.635 \times 10^{-10} \] \[ = 4\pi \times 6.022 \times 10^{23} \times 1.147 \times 10^{-9} \times 6.635 \times 10^{-10} \] Calculate \(1.147 \times 10^{-9} \times 6.635 \times 10^{-10} = 7.612 \times 10^{-19}\) Now, \[ 4\pi \times 6.022 \times 10^{23} \times 7.612 \times 10^{-19} \] Calculate \(4\pi = 12.566\) \[ 12.566 \times 6.022 \times 10^{23} = 75.685 \times 10^{23} \] \[ 75.685 \times 10^{23} \times 7.612 \times 10^{-19} = 576.6 \times 10^{4} = 5.77 \times 10^{6} \] **Unit check:** This is in m\(^3\) mol\(^{-1}\) s\(^{-1}\). Convert to dm\(^3\) mol\(^{-1}\) s\(^{-1}\) (\(1 \text{ m}^3 = 100 \text{ dm}^3\)): \[ k_{\text{diff}} = 5.77 \times 10^{6} \times 10^{3} = 5.77 \times 10^{9} \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1} \] --- ## **Step 3: Calculate the Initial Rate** For a bimolecular reaction: \[ \text{Rate} = k_{\text{diff}} [A][B] \] Plugging in the values: \[ \text{Rate} = (5.77 \times 10^{9}) \times (.200) \times (.150) \] \[ = 5.77 \times 10^{9} \times .03 = 1.731 \times 10^{8} \text{ mol dm}^{-3} \text{ s}^{-1} \] --- ## **Final Answer** \[ \boxed{ \text{The initial rate } \frac{d[P]}{dt} = 1.73 \times 10^{8} \text{ mol dm}^{-3} \text{ s}^{-1} } \] --- **Summary of Steps:** 1. Calculated encounter distance \(R_{AB}\). 2. Used Stokes-Einstein to find \(D_A\) and \(D_B\). 3. Used Smoluchowski equation for \(k_{\text{diff}}\). 4. Plugged values into the rate law for a bimolecular reaction. --- **Let me know if you need clarifications or more details on any step!**