# Step-by-Step Solution Let's break down and solve the problem step by step. --- ## **Given** - **:** \( M \rightarrow \) (Dimer, irreversible, second order) - **Reactor type:** Adiabatic CSTR - **Reactor volume, \( V \):** \( .4 \) m³ - **Volumetric flow rate, \( F \):** \( 1.3 \) m³/ksec = \( 1.3 \times 10^{-3} \) m³/sec - **Inlet temperature, \( T_{in} \):** \( 312 \) K - **Monomer feed concentration, \( C_{M} \):** \( 16 \) kmol/m³ - **Standard enthalpy change \( (\Delta H) \) at 300K:** \( -42 \) kJ/mole D formed - **Heat capacity \( (C_p) \):** \( 2. \) J/cm³·K \( = 2. \times 10^6 \) J/m³·K - **Rate constant \( k \):** \( k = 2.70 \times 10^9 e^{-12,185/T} \) m³/mole-ksec (for \( T \) in K) --- ## **Step 1: Write the Mole Balance for a CSTR** For a CSTR at steady-state: \[ F(C_{M} - C_M) = -r_M V \] where - \( C_{M} \): Inlet monomer concentration - \( C_M \): Outlet monomer concentration - \( r_M \): Rate of consumption of M (second order: \( r_M = -2kC_M^2 \)) So, \[ F(C_{M} - C_M) = 2kC_M^2 V \] --- ## **Step 2: Energy Balance for Adiabatic Operation** For adiabatic operation (no heat exchange): \[ \text{Heat generated by reaction} = \text{Increase in temperature of exit stream} \] For 1 mole of D formed, \( \Delta H = -42 \) kJ = \(-42,000 \) J (exothermic). Let \( X \) be the **fractional conversion** of monomer M: \[ X = \frac{C_{M} - C_M}{C_{M}} \] **Moles of D formed per m³ feed:** \[ \text{If } 2 \text{ moles M} \rightarrow 1 \text{ mole D} \implies \text{Fraction of M converted to D is } X/2 \] **Temperature rise:** \[ \text{Heat generated} = \frac{C_{M} X}{2} \cdot |\Delta H| \] \[ \text{Heat absorbed} = C_{M} \cdot C_p \cdot (T - T_{in}) \] But the entire stream's heat capacity absorbs the heat produced: \[ C_{M} \cdot C_p \cdot (T - T_{in}) = \frac{C_{M} X}{2} \cdot |\Delta H| \] \[ C_p \cdot (T - T_{in}) = \frac{X}{2} \cdot |\Delta H| \] \[ T = T_{in} + \frac{X \cdot |\Delta H|}{2C_p} \] --- ## **Step 3: Plug in the Numbers** Plug in values: - \( T_{in} = 312 \) K - \( |\Delta H| = 42,000 \) J/mole D - \( C_p = 2. \times 10^6 \) J/m³·K \[ T = 312 + \frac{X \times 42,000}{2 \times 2. \times 10^6} \] \[ T = 312 + \frac{X \times 42,000}{4,000,000} \] \[ T = 312 + \frac{X \times 42}{400} \] \[ T = 312 + \frac{X \times .0105}{1} \] \[ T = 312 + .0105X \] --- ## **Step 4: Rate Constant \( k \) at Outlet \( T \) (as function of \( X \))** \[ k = 2.70 \times 10^9 e^{-12,185/T} \] --- ## **Step 5: CSTR Mole Balance in Terms of \( X \)** Recall: \[ F(C_{M} - C_M) = 2kC_M^2 V \] But \( C_M = C_{M}(1 - X) \): \[ F(C_{M} - C_{M}(1 - X)) = 2k [C_{M}(1 - X)]^2 V \] \[ F(C_{M} X) = 2k C_{M}^2 (1 - X)^2 V \] \[ F X = 2k C_{M} (1 - X)^2 V \] \[ \frac{F}{2kC_{M}V} X = (1 - X)^2 \] Let \( \alpha = \frac{F}{2kC_{M}V} \): \[ \alpha X = (1 - X)^2 \] --- ## **Step 6: Substitute All Values (except \( k \), which is a function of \( X \))** - \( F = 1.3 \) m³/ksec - \( C_{M} = 16 \) kmol/m³ - \( V = .4 \) m³ - \( k = 2.70 \times 10^9 e^{-12,185/T} \) (where \( T = 312 + .0105X \)) \[ \alpha = \frac{1.3}{2 \cdot k \cdot 16 \cdot .4} \] \[ \alpha = \frac{1.3}{12.8k} \] So, \[ \frac{1.3 X}{12.8 k} = (1 - X)^2 \] \[ X = \frac{12.8 k (1 - X)^2}{1.3} \] --- ## **Step 7: Solve Iteratively for \( X \) (since \( k \) depends on \( X \))** 1. **Guess \( X \) (start with e.g. \( X = .5 \))** 2. Compute \( T = 312 + .0105X \) 3. Compute \( k = 2.70 \times 10^9 e^{-12,185/T} \) 4. Plug into \( X = \frac{12.8k(1-X)^2}{1.3} \) 5. Iterate until \( X \) converges --- ## **Step 8: Initial Guess and Iteration** ### **Iteration 1: Guess \( X = .5 \)** - \( T = 312 + .0105 \times .5 = 312.00525 \approx 312 \) K \[ k = 2.70 \times 10^9 e^{-12,185/312} \] \[ = 2.70 \times 10^9 e^{-39.05} \] \[ e^{-39.05} \approx 1.01 \times 10^{-17} \] \[ k \approx 2.70 \times 10^9 \times 1.01 \times 10^{-17} \] \[ k \approx 2.73 \times 10^{-8}\ \text{m}^3/\text{mol}\cdot \text{ksec} \] Now, \[ X = \frac{12.8 \times 2.73 \times 10^{-8} \times (1 - .5)^2}{1.3} \] \[ = \frac{12.8 \times 2.73 \times 10^{-8} \times .25}{1.3} \] \[ = \frac{12.8 \times .25}{1.3} \times 2.73 \times 10^{-8} \] \[ = \frac{3.2}{1.3} \times 2.73 \times 10^{-8} \] \[ = 2.46 \times 2.73 \times 10^{-8} \] \[ = 6.72 \times 10^{-8} \] So, the conversion is extremely low at this guess—nearly zero. Let's try a much lower \( X \). --- ### **Iteration 2: Try \( X = .01 \)** - \( T = 312 + .0105 \times .01 = 312.000105 \approx 312 \) K So \( k \) is same as above (\( \sim 2.73 \times 10^{-8} \)), so conversion is still extremely low. --- ### **Iteration 3: Try \( X = .9 \)** - \( T = 312 + .0105 \times .9 = 312.00945 \approx 312 \) K Still same \( k \). --- ### **Observation** - The temperature rise is very small due to the large heat capacity. - The rate constant is extremely small at \( 312 \) K. - Thus, the reaction is extremely slow, and **very little monomer is converted**. --- ## **Step 9: Final Answer** ### **Fraction of Monomer Converted (\( X \))** Since all values of \( X \) tried yield extremely low self-consistent values (on the order of \( 10^{-8} \)), the **fraction of monomer that can be converted is extremely small**. \[ \boxed{X \approx 7 \times 10^{-8}} \] Or, in percentage: \[ \boxed{X \approx .000007\%} \] --- ## **Conclusion** ### **Final Answer** > **The fraction of monomer that can be converted to dimer in the adiabatic CSTR is extremely small, on the order of \( 7 \times 10^{-8} \).** > > **This is because the reaction rate at the given (adiabatic) temperature is extremely slow.** --- ### **Summary of Steps** 1. Set up mole and energy balances for the CSTR. 2. Combine balances to relate conversion (\( X \)) to temperature (\( T \)) and rate constant (\( k \)). 3. Substitute values to get a self-consistent equation for \( X \). 4. Iteratively solve for \( X \) and find that conversion is extremely low due to the very small rate constant at adiabatic temperature. --- **If you need to see the calculations for other assumed values or a more detailed iteration, let me know!**