0.5 0.5 = 1 0.5 = HW3 The following reaction is carried out in a PFR under the pressure of 4.05 MPa at 936K. T+H_KqD+G r

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0 5 0 5 1 0 5 Hw3 The Following Reaction Is Carried Out In A Pfr Under The Pressure Of 4 05 Mpa At 936k T H Kqd G R 1 (45.14 KiB) Viewed 8 times

0.5 0.5 = 1 0.5 = HW3 The following reaction is carried out in a PFR under the pressure of 4.05 MPa at 936K. T+H_KqD+G r = kcrc D+H_kz »M+G r = k cc95 M+H+→ N+Grz = ky(cyc* - cy@g/c*K) kz CMCH 95 K = 5, k, = 5.66x10ml/molº..s k2 = 5.866x10m'. /molºss, ką = 2.052x104m-/molº's Yto=25%, Y'ho=75%, flow rate 0.1m/s, XTr-80% What is the length of reactor? You only need to provide the mathematical expression.