Taking A as the basis of calculation, A + 9/2 B → C + 2D + 2E

We are going to rework the class problem for the case of 3.5% naphthalene and 96.7% air and for a pressure of 10 atm and a temperature of 500K. A feed under these conditions naphthalene is the limiting reactant.

**Stoichiomentric Table - Batch System**

__Species____Symbol____In____Change____Out__Naphthalene A N _{A0}-N _{A0}XN _{A}= N_{A0}(1-X)Oxygen B N _{B0}= N_{A0}-9/2N _{A0}XN _{B}= N_{A0}( - 9/2X)Phthalic Anhydride C N _{C}= 0+N _{A0}XN _{C}= N_{A0}XCarbon Dioxide D N _{D}= 0+2N _{A0}XN _{D}= 2N_{A0}XWater E N _{E}= 0+2N _{A0}XN _{E}= 2N_{A0}XNitrogen I N _{I}= N_{A0}--- N _{I}= N_{A0}Total N _{T0}N _{T}= N_{T0}+ N_{A0}X= [1 + 2 + 2 - 9/2 - 1] = -1/2 **Stoichiometric Table - Flow System**__Species____Symbol____In____Change____Out__Naphthalene A F _{A0}-F _{A0}XF _{A}= F_{A0}(1-X)Oxygen B F _{B0}= F_{A0}-9/2F _{A0}XF _{B}= F_{A0}( - 9/2X)Phthalic Anhydride C F _{C}= 0+F _{A0}XF _{C}= F_{A0}XCarbon Dioxide D F _{D}= 0+2F _{A0}XF _{D}= 2F_{A0}XWater E F _{E}= 0+2F _{A0}XF _{E}= 2F_{A0}XNitrogen I F _{I}= F_{A0}--- F _{I}= F_{A0}Total F _{T0}F _{T}= F_{T0}+ F_{A0}X- Determine each of the following soley as a function of the conversion of naphthalene, X for a constant-pressureisothermal flow reactor.
- Find the concentration of O
_{2}**Gas Phase Flow System:**

**Constant pressure and isothermal: P = P**_{o}; T = T_{o}

- Find the volumentric flowrate

**Constant pressure and isothermal: P = P**_{o}; T = T_{o}

- Find the reaction rate

**Rate Law:**-r_{A}= k_{A}C^{2}_{A}C_{B}

- Find the concentration of O

- The concentration of O
_{2}**For Batch:**V = V_{0}, C_{B}= N_{B}/V = N_{B}/V_{0}

- The total pressure, P.

- The rate law

**Rate Law:**-r_{A}= k_{A}C^{2}_{A}C_{B}