Answer:
Explanation: see attachment below for the equation,
V = volume = 200 dm3
FA0 = molar flow rate in kmol/min = CA0 X volumetric flow
FA0 (kmol/min) = CA0 kmol/m3 X 10 m3/min
Rate law equation for A = B + 2C
-rA = kCA = 0.08 [CA0(1-X)]
Part C
Material balance equation will remain same
Rate law equation for 3A = B
-rA = kCA3 = 0.08 [CA03(1-X)3]