Repeat the Example 9.2 calculation for the case in which the reaction is twice as fast. ( hint: what is the new value of K_r?)

EXAMPLE 9.2 A-C where SOLUTION SAC our = ? C Species A in liquid solution (concentration = 0.74 M) enters a CSTR at 18.3 L/s, where it is con- sumed by the irreversible reaction Freaction = kyky = 0.015/s and is in units of gmol/L) What reactor volume is needed so that the concentration of species A leaving the reactor equals 0.09 M? The density can be assumed to be constant (Note that the steps correspond to the instructions in Chapter 5.) Drawing a diagram for this problem. P=18.3 LS 4 = 0.74 M reaction. 1 = (0.015/s) db Volume=1 =0.09 M We want to construct a mole balance on A. For this case (for a single input and single output stream). the mole balance becomes Am + formation. From pedon. Species A is being consumed, but no species A is being formed, so formurim A = 0. This, along with substituting more convenient forms for the molar flow rates, gives V = Vou + Prompton, (a) The value of the outgoing volumetric flow rate is not specifically given, so we need a total mass balance, which for a single input and single output stream is =tw which, in more convenient terms, is Pull - Au Vou Since the density is constant, this reduces to Vow =V (b) We can now calculate compaion. A using Equations (a) and (b), Equation (a) becomes consumption. A = Vn - Vive = -) gol - (0.74 Komel - 0.09 Set) (1834) - =11.9 mol Up to now, everything we've done is a repeat of the material balances we learned in Chapter 5. The new step is to equate the comprom.A term to the given rate expression times the reactor volume, where in the reactor) = compe="AV = (k) V or V Fromww. kica 11.9 gmods 0.015 gol 0.09 L 8,800L