Question please calculation 3. 3A + 2B 3C + 5D elemantary gas phase reaction is carried out isothermally at 427°C and 28.7 atm in a flow reactor. By creating the stoichiometric table, write the rate expression; a) (10 P) in terms of the concentration of each component, b) (S P) in terms of the conversion ratio, c) (5 P) in terms of the total molar flow rate. d) (10 P) Find the required CSTR volume to achieve 75% conversion rate. The inlet volumetric flow rate is 10 L/s and the reaction rate constant is 200 L/mol's. The feed contains equal amounts of A and B.

  Soluction:given Data:the elemntry gas phase Reaction:\begin{array}{l}3 A+2 B \stackrel{K}{\rightarrow} 3 C+50 \\\text { at } T=427^{\circ} \mathrm{C} \text { \& } P=28.7 \mathrm{~atm} .\end{array}Reaction is fint adir with respect ANow,(a) Expreds the Rate low; & each peeive ad a function of time,the Rata low Exprestion is ad fallows.-r_{A}=k_{1} C_{A}^{x} c_{B}^{y}where x in axe arder of Raction w.r. to A d B.Here x=y=1So,\left[r_{A}=K_{1} C A \cdot C_{B}\right]and Conantration of eash species ad a function of Convirdion.\begin{aligned}C_{A} & =C_{0}\left(1 x_{A}\right)\left(m_{0}\left(_{B}^{3}\right)\right. \\C_{B} & =\left(C_{A 0}-\frac{2}{3} C_{A_{0}} X_{A}\right)\left(m a\left(m^{3}\right)\right. \\C_{C} & =\left(C_{0}+\left(\frac{3}{3}\right) C_{A_{0}} X_{A}\right)\left(m a m_{4}\left(m^{3}\right)\right. \\C_{D} & =\left(C_{D}+\frac{5}{3} C_{A_{0}} X_{A}\right)\left(m_{0}\left(m^{3}\right)\right.\end{aligned}\Rightarrow Mow, pant:(b)\nabla_{0}=\left.1000 \mathrm{dm}^{3}\right|_{\text {ce }} \text { (volumstic flow Rate) }Rate Constantk_{1}=20 \mathrm{dm}^{3} \text { imelise. }fred is squimear in A &B.for x A=.90Ritermine the valume of ffR:\begin{array}{l}\text { Hare } \rightarrow \text { aqume } \operatorname{Sar}(A ß B \text { ) } \\C A_{0}=\frac{1}{2}\left(\frac{P}{R T}\right)=\frac{1}{2} \times\left(\frac{28.7 \times 101.325 \times 10^{3}}{700 \times 8.314)}\right. \\\left.C_{A_{0}}=249.83 \mathrm{mal} / \mathrm{m}^{3}\right]=0.249 \frac{\mathrm{mal}}{\mathrm{dm}^{3}} \\\end{array}ad we thow, ferformance eof. for ffR:\begin{array}{l}\left(\frac{v}{f_{A_{0}}}\right)=\frac{1}{k \times a_{A_{0}} \times c_{B_{0}}} \int_{0}^{x_{A}} \frac{d x_{A}}{\left(1-x_{A}\right) \frac{\left(3-2 x_{A}\right)}{3}} \\\frac{V \times K C C_{B} C_{0}}{\text { HAO }_{0}}=3 \int_{0}^{x_{A}} \frac{d x_{A}}{\left(1-x_{A}\right)\left(3-2 x_{A}\right)} \\\frac{V \times 20 \times C A_{0} C_{0}}{C_{0} \nu_{0}}=3 \int_{0}^{x_{A}} \frac{d x_{A}}{\left(1 x_{A}\right)\left(32 x_{A}\right)} \\\left(\frac{V \times 20 \times \cdot 249}{1000}\right)=3 \times \int_{0}^{x_{A}} \frac{d x_{A}}{\left(1-x_{A}\right)\left(3-2 x_{A}\right)} \\V \times .00166=\int_{0}^{x_{A}} \frac{d x_{A}}{(1-x A)(3-2 x A)} \\V x .00166=\int_{0}^{x_{A}}\left(\frac{1}{1-x_{A}}-\frac{2}{\left(3-2 x_{A}\right)}\right) d x_{A} \\\end{array}\begin{array}{c}V x \cdot 00166=\left(-\ln (1-x)+\frac{2}{2} \ln (3-2 x y)\right)_{0}^{+9} \\V \times \cdot 00166=-\ln (1-.9)+\ln (3-2 x+9) \\V x \cdot 00166=2.4849 \\V=1496.9277 \mathrm{dm}^{3}\end{array}\Rightarrow Molo, part(c) If V=50 \mathrm{dm}^{3}, How many taberequired to achieve the proceds.Heve, calceletel totel rolemeRequived for 90 \% Convarion.V_{\text {Tate }}=1496.927 \mathrm{dm}^{3}if one PfR volume =50 \mathrm{dm}^{3}\begin{aligned}\text { Mombereff PfR(Stube) usel } & =\left(\frac{1496.927}{50}\right) \\n & =29.938 \\n & \cong 30 \text { tubed Ans }\end{aligned} ...