As part of an overall plant re-design, you are given the schematic below and ask
ID: 3741718 • Letter: A
Question
As part of an overall plant re-design, you are given the schematic below and asked to determine the steady-state concentration of a certain chemical species in each of a series of mixing tanks. 0 15 C5 54 025 C 1 C 2 24 0 23 C3 34 003 C 03 Write out the A, x, and b vectors based on a mass balance around each tank. Therefore, Your A matrix should be of dimensions 5x5 because we have 5 tanks The base operating conditions were as follows Concentration C0110 CO3 20 Flowratee Q01 5 Solve for unknown concentrations in the mixing tanks: Q12-3 Q25 1 Q231 C1- C2- Q44 11Explanation / Answer
Answer:
If the reaction 2N2O5?4NO2+O22N2O5?4NO2+O2 follows the following three-step mechanism:
let kfkf and kbkb be forward and backward rate constants N2O5?kfkbNO2+NO3N2O5?kfkbNO2+NO3 NO3+NO2??k2NO+NO2+O2NO3+NO2?k2NO+NO2+O2 NO3+NO???k32NO2NO3+NO?k32NO2
Use the steady-state approximation to derive the rate law. SOLUTION In these steps, NONO and NO3NO3 are intermediates.
You have productionrateofNO=k2[NO3][NO2]productionrateofNO=k2[NO3][NO2] consumptionrateofNO=k3[NO3][NO]consumptionrateofNO=k3[NO3][NO] A steady-state approach makes use of the assumption that the rate of production of an intermediate is equal to the rate of its consumption.
Thus, we have k2[NO3][NO2]=k3[NO3][NO]k2[NO3][NO2]=k3[NO3][NO] and solving for [NO][NO] gives the result, [NO]=k2[NO3][NO2]k3[NO3](1)(1)[NO]=k2[NO3][NO2]k3[NO3] For the other intermediate NO3NO3, productionrateofNO3=kf[N2O5]productionrateofNO3=kf[N2O5] consumptionrateofNO3=k2[NO3][NO2]+k3[NO3][NO]+kb[NO3][NO2]consumptionrateofNO3=k2[NO3][NO2]+k3[NO3][NO]+kb[NO3][NO2]
Applying the steady-state assumption gives: kf[N2O5]=k2[NO3][NO2]+k3[NO3][NO]+kb[NO3][NO2]kf[N2O5]=k2[NO3][NO2]+k3[NO3][NO]+kb[NO3][NO2] Thus,
[NO3]=kf[N2O5]k2[NO2]+k3[NO]+kb[NO2](2)(2)[NO3]=kf[N2O5]k2[NO2]+k3[NO]+kb[NO2] Let's review the three equations (steps) in the mechanism: Step i. is at equilibrium and thus can not give a rate expression.
Step ii. leads to the production of some products, and the active species NONO causes further reaction in step
iii.
This consideration led to a rate expression from step ii. As:
d[O2]dt=k2[NO3][NO2](3)(3)d[O2]dt=k2[NO3][NO2] Substituting (1) in (2) and then in (3) gives d[O2]dt=kfk2[N2O5]kb+2k2=k[N2O5]d[O2]dt=kfk2[N2O5]kb+2k2=k[N2O5] where
k=kfk2kb+2k2k=kfk2kb+2k2.
This is the differential rate law, and it agrees with the experimental results. Carry out the above manipulation yourself on a piece of paper. Simply reading the above will not lead to solid learning yet.
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