Question

Question 2. (20 marks) The following liquid-phase consecutive reaction is taking place in a constant volume batch reactor A_ky »B_k >C Derive expressions for the concentrations of A, B and C as functions of time given that the first reaction is first order and the second reaction is zeroth order.

Answer 1

Given set of reaction is ki k2 А B C First reaction is first order and second reaction is zero order Reaction is occurring in batch reactor so input and output to the reactor becomes zero. Taking material balance for reactant A over the batch reactor Input - output + generation – consuption = Accumulation For batch reactor Input = output = 0 For reactant generation = 0 consuption = - AV Vis volume of reactor The reaction is first order so -1A = k1CA dCA Accumulation = V - dt dcA V dt = -k CAV dCA kidt CA By taking integration over limit as CA = concentration of A at any time CAO = Initial concentration

at t=0 Ca = C20 at t=t CA = CA -la RCA DLA СА CAO Skat ICAO In = CA Now take antilog on both sides we get CAO CA ekit By rearranging CA = Cape-kyt For Component B Consumption of B is zero order reaction Consumption of B = -k2V Genaration of B = rg V = k1CAV From the reaction Consumption of 1 mole of A results in to production of 1 mole of B. Consumed Conc. A = initial conc conc. at any time CAO – CA CB = CAO - CA CA = CA0 - CB Genaration of B = k (CAO - C3) dCB Accumulation = V- dt

generation - consuption = Accumulation dCB k (CAO - Cp) - (-k2V)= V dt dCg k,Cao - CB)+kg = dt dCg (kCao + kg) – VCB dt dt = dCo (k1CA0 + k) - k1CB By considering feed is only component A On integrating at t= 0 Cg = 0 at t= t CB = CB t =- 1 -In ((K1CA0 + k) - k1CB) ki hain CO -kt= ln (k Cao + kg) – VCB) By taking antilog on both side (k1CA0 + k_) – kq Cg = e-kat Св (k C20 + k) - e-ket ki

For component C Consumption = 0 Generation = VK2 dcc Accumulation =V dt VK2 = V - dcc dt dcc = K2dt By integrating Lo doc - ( Rude Cc = Kyt