CHAPTER 3 material balance part iI Sem 1, 2015/2016 ERT 214 Material and Energy Balance / Imbangan Bahan dan Tenaga

Content Stoichiometry Limiting and Excess Reactant, Fractional Conversion and Extent of Reaction Chemical Equilibrium Multiple Reaction, Yield and Selectivity Balance on Reactive System

Stoichiometry Stoichiometry – theory of proportions in which chemical species combine with one another. Stoichiometric equation of chemical reaction – statement of the relative number of molecules or moles of reactants and products that participate in the reaction. 2 SO2 + O2 ---> 2 SO3 Stoichiometric ratio ratio of species stoichiometry coefficients in the balanced reaction equation can be used as a conversion factor to calculate the amount of particular reactant (or product) that was consumed (produced). Eg: 2 mol SO3 generated 2 mol SO2 consumed 2 mol SO2 consumed or 1 mol O2 consumed or etc

C4H8 + 6 O2 --------> 4 CO2 + 4 H2O Example: C4H8 + 6 O2 --------> 4 CO2 + 4 H2O Is the stochiometric equation balance? Yes What is stochiometric coefficient for CO2 4 What is stochiometric ratio of H2O to O2 including it unit 4 mol H2O generated/ 6 mol O2 consumed How many lb-moles of O2 reacted to form 400lb-moles CO2 400lb-moles CO2 6 moles O2 reacted = 600 lb-moles O2 reacted 4 moles CO2 produced 100 mol/min C4H8 fed into reactor and 50% is reacted. At what rate water is formed? 100 mol /min fed 4 mol water generated = 200 mol/min water generated 2 1 mol C4H8 reacted

Limiting Reactant & Excess Reactant Limiting reactant : The reactant that would run out if a reaction proceeded to completion, and the other reactants are termed excess reactants. A reactant is limiting if it is present in less than its stoichiometric proportion relative to every other reactant. If all reactants are present in stoichiometric proportion, then no reactant is limiting. Is the amount by which the A in the feed exceeds the amount needed to react completely if the reaction goes to completion. (nA) feed = is the number of moles of an excess reactant A, present in the feed to a reactor. (nA) stoich = stoichiometric requirement of A or the amount needed to react completely with the limiting reactant

Example C2H2 + 2H2 ------> C2H6 Inlet condition: 20 kmol/h C2H2 and 50 kmol/h H2 What is limiting reactant and fractional excess? (H2:C2H2) feed = 2.5 : 1 (50:20) (H2:C2H2) stoich = 2 : 1 H2 is excess reactant and C2H2 is limiting reactant Fractional excess of H2 = (50-40)/40 = 0.25

Fractional Conversion Fractional Conversion (f) The fraction unreacted ; 1- f Example: If 100 moles of a reactant are fed and 90 moles react, the fractional conversion is 0.90 (the percentage conversion is 90%) and the fraction unreacted is 0.10. if20 mol/min of a reactant is fed and the percentage conversion is 80%, then (20)(0.80) = 16 mol/min has reacted and (20)(1 - 0.80) = 4 mol/min remains unreacted.

Extent of Reaction Extent of Reaction, ξ ξ = extent of reaction ni = moles of species i present in the system after the reaction occurred nio = moles of species i in the system when the reaction starts vi = stoichiometry coefficient for species i in the particular chemical reaction equation

N2 + 3H2 ------------> 2NH3 Example N2 + 3H2 ------------> 2NH3 Reactor inlet: 100 mol N2/s; 300 mol H2/s; 1 mol Argon/s (inert gas) If fractional conversion of H2 0.6, calculate extent of reaction and the outlet composition. Unreacted H2 or H2 outlet= (1-0.6) 300 = 120 mol H2/s Solve for extent of reaction : 60 mol/s 120 = 300 – 3 ξ 3 ξ = 300-120 3 ξ = 180 ξ = 60

Test Yourself C2H4 is limiting because [ ]f < [ ]stoi The oxidation of ethylene to produce ethylene oxide proceeds according to the equation 2 C2H4 + O2 ------->2 C2H4O The feed to a reactors contains 100kmol C2H4 and 100kmol O2. a) which is limiting reactant? [C2H4]f = 100 [O2]f = 100 [C2H4 / O2 ]f = 100 / 100 = 1 [C2H4 / O2 ]stoi = 2 / 1 = 2 [O2 / C2H4 ]f = 100 / 100 = 1 [O2 / C2H4 ]stoi = 1/ 2 = 0.5 C2H4 is limiting because [ ]f < [ ]stoi O2 is excess because [ ]f > [ ]stoi

The feed to a reactors contains 100kmol C2H4 and 100kmol O2. 2 C2H4 + O2 ------->2 C2H4O The feed to a reactors contains 100kmol C2H4 and 100kmol O2. b) Percentage of excess of the other reactant? [O2 ]stoi = 100 kmol C2H4 1 mol O2 react = 50 kmol O2 react 2 mol C2H4 react % excess = [O2 ]f - [O2]stoi = 100 - 50 x 100% = 100% (because an AIR) [O2]stoi 50

The feed to a reactors contains 100kmol C2H4 and 100kmol O2. 2 C2H4 + O2 ------->2 C2H4O The feed to a reactors contains 100kmol C2H4 and 100kmol O2. c) If the reaction proceeds to completion, how much of the excess reactant will be left; how much C2H40 will be formed; and what is the extent of reaction? 2 mol C2H4 react with 1 mol O2 If 100kmol C2H4 (½) x 100 kmol = 50 kmol ‘02 react 100kmol O2 feed - 50 kmol ‘02 react = 50 kmol ‘02 left 2 mol C2H4 react produced 2 mol C2H4O If 100kmol C2H4 (2/2) x 100 kmol = 100 kmol ‘C2H4O formed Extent of reaction (y masih tinggal) [O2 ] = 100 kmol O2 feed – 50 kmol O2 left 1 = 50 kmol O2 ξ= extent of reaction ni = moles of species i present in the system after the reaction occurred nio = moles of species i in the system when the reaction starts vi = stoichiometry coefficient for species i in the particular chemical reaction equation

The feed to a reactors contains 100kmol C2H4 and 100kmol O2. 2 C2H4 + O2 ------->2 C2H4O The feed to a reactors contains 100kmol C2H4 and 100kmol O2. d) if fractional conversion for limiting reactant is 50%, what is outlet composition and extent of reaction? so, 0.5 = X/100 = 50kmol C2H4 react 2 C2H4 + O2 ------->2 C2H4O 50kmol C2H4 react so, 2mol C2H4 react with 1 mol O2 50kmol C2H4 react with (1//2)x 50 = 25 kmol O2 react O2 produced (not react) = 100 – 25 = 75 kmol O2 50kmol C2H4 react so, 2mol C2H4 produced 2 mol C2H4 50kmol C2H4 produced (2/2)x 50 = 50 kmol C2H4 produced Extent of reaction [O2 ] = 100 kmol O2 feed – 75 kmol O2 left = 25 kmol O2 1

The feed to a reactors contains 100kmol C2H4 and 100kmol O2. 2 C2H4 + O2 ------->2 C2H4O The feed to a reactors contains 100kmol C2H4 and 100kmol O2. e) if reaction proceed to a point where 60kmol O2 left, what is fractional conversion for C2H4? Fractional conversion of O2 and extent of reaction? -100 kmol O2 feed ….but 60kmol O2 left….so O2 react = 100-60=40kmol O2 react Based on stoi, 1 mol O2 react with 2 mol C2H4 if 40 kmol O2 --------------------40 x 2 = 80 kmol C2H4 react fC2H4= 80/100 =0.8 fO2= 40/100= 0.4 Extent of reaction [O2 ] = 100 kmol O2 feed – 60kmol O2 left = 40 kmol O2 1

Chemical Equilibrium For a given set reactive species and reaction condition, two fundamental question might be ask: 1. What will be the final (equilibrium) composition of the reaction mixture? – chemical engineering thermodynamics deals with this 2. How long will the system take to reach a specified state short of equilibrium? – chemical kinetics deals with this Irreversible reaction reaction proceeds only in a single direction (from reactants to products) the concentration of the limiting reactant eventually approaches zero. Reversible reaction reactants form products for forward reaction and products undergo the reverse reactions to reform the reactants. Equilibrium point is a rate of forward reaction and reverse reaction are equal

Multiple Reaction Some of the chemical reaction has a side reaction which is formed undesired product- multiple reaction occurred. Effects of this side reaction might be: Economic loss Less of desired product is obtained for a given quantity of raw materials Greater quantity of raw materials must be fed to the reactor to obtain a specified product yield.

Multiple Reaction

Selectivity selectivity = moles of desired product moles of undesired product

Yield 3 definition of yield with different working definition Yield = Moles of desired product formed Moles that would have been formed if there were no side reaction and the limiting reactant had reacted completely Yield = Moles of desired product formed Moles of reactant fed Yield = Moles of desired product formed Moles of reactant consumed

Extent of Reaction for Multiple Reaction Concept of extent of reaction can also be applied for multiple reaction only now each independent reaction has its own extent.

Example: Yield and Selectivity in a Dehydrogenation Reactor The reactions take place in a continuous reactor at steady state. The feed contains 85.0 mole% ethane (C2H6) and the balance inerts (I). The fractional conversion of ethane is 0.501, and the fractional yield of ethylene is 0.471. Calculate the molar composition of the product gas and the selectivity of ethylene to methane production. SOLUTION: Basis: 100 mol Feed fC2H6= 0.501 Yield, Y C2H4 = 0.471

the outlet component amounts in terms of extents of reaction are as follows: nl (mol C2H6) = 85.0 mol C2H6 – ξ1- ξ2 n2(mol C2H4) = ξ1 n3(mol H2) = ξ1- ξ2 n4(mol CH4 ) = 2ξ2 n5(mol I) = 15.0 mol I(Inert) -----------------ξ1 -----------------ξ2

Ethane Conversion If the fractional conversion of ethane is 0 Ethane Conversion If the fractional conversion of ethane is 0.501, the fraction unconverted (and hence leaving the reactor)must be (1 - 0.501). Ethylene Yield Given : fC2H6= 0.501 Given : Yield, Y C2H4 = 0.471

Product:

balance on reactive process Sem 1, 2015/2016 ERT 214 Material and Energy Balance / Imbangan Bahan dan Tenaga

Balance of Reactive Processes Balance on reactive process can be solved based on three method: Atomic Species Balance Extent of Reaction Molecular Species Balance Each approach leads to the same results, but anyone of them may be more convenient for a given calculation so it is a good idea to become comfortable with all three.

1. Atomic Species Balance No. of unknowns variables - No. of independent atomic species balance - No. of molecular balance on indep. nonreactive species - No. of other equation relating the variable = No. of degree of freedom

2. Extent of Reaction No. of unknowns variables + No. of independent chemical reaction - No. of independent reactive species - No. of independent nonreactive species - No. of other equation relating the variable = No. of degree of freedom

3. Molecular Species Balance No. of unknowns variables + No. of independent chemical reaction - No. of independent molecular species balance - No. of other equation relating the variable = No. of degree of freedom

Degree-of-Freedom Analysis To carry out degree-of-freedom analyses of reactive systems you must first understand the concepts of : independent equations, independent species, and independent chemical reactions. ?

1. Independent Equation Algebraic equation are independent if we cannot obtain any one of them by adding and subtracting multiples of any of the others Eg: x + 2y = 4 [1] 3x + 6y = 12 [2] Only one independent equation because [2]= 3 x [1] ------------------------------------------------------------------------------------ 2x – z= 2 [2] 4y + z= 6 [3] Although 3 equation, but only two independent equation exist because [3]=2x[1] –[2]

2. Independent Species a) If two MOLECULAR species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get) Similarly b) If two ATOMIC species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get)

a) Independent Molecular Species If two MOLECULAR species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get)

Let make a molecular balance on both species to prove it Example: Since N2 and O2 have a same ratio wherever they appear on the flowchart (3.76 mol N2/ mol O2), only ONE independent balance can obtain. Let make a molecular balance on both species to prove it Balance on O2: n1=n3 [1] Balance on N2: 3.76 n1=3.76n3 n1=n3 [2] Eq. [1] and [2] are SAME. Only ONE INDEPENDENT EQUATION OBTAIN although two species involved.

b) Independent Atomic Species If two ATOMIC species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get)

Atomic N and O are always in same proportion to each other in the process (3.76:1), similar for atom C and Cl which always same ratio too (1:4). Although FOUR atomic species exist, only TWO independent equation can obtain for this cases. Prove: Balance on atomic O: 2n1=2n3 n1=n3 [1] Balance on atomic N: 2(3.76)n1=2(3.76)n3 n1=n3 [2] Balance on atomic C: n2=n4+n5 [3] Balance on atomic Cl: 4n2=4n4 +4n5 n2=n4+n5 [4] Eq. [1]=[2] and [3]=[4], only TWO independent equation (Eq n1=n3 and n2=n4+n5 )obtained

3. Independent Reaction Used when we using either molecular species balance or extent of reaction method to analyze a balance on reactive process Chemical reaction are independent if the stoichiometric equation of any one of them cannot be obtained by adding and subtracting multiples of the stoichiometric equations of the others Only TWO independent eq. can obtained although three equation exist sice [3]=[1] + 2[2].

Example: Atomic Species Balance Dehydrogenation of ethane in a steady-state continuous reactor. The reaction is One hundred kmol/min of ethane is fed to the reactor. The molar flow rate of H2 in the product stream is 40 kmol/min All atomic balance is INPUT=OUTPUT Degree-of-freedom analysis 40 kmol H2/min n1 kmol C2H6/min n2 kmol C2H4/min Reactor 100 kmol C2H6/min

Example: Molecular Species Balance Dehydrogenation of ethane in a steady-state continuous reactor. The reaction is One hundred kmol/min of ethane is fed to the reactor. The molar flow rate of H2 in the product stream is 40 kmol/min 40 kmol H2/min n1 kmol C2H6/min n2 kmol C2H4/min Reactor 100 kmol C2H6/min Degree-of-freedom analysis 2 unknowns variables (n1, n2) +1 independent chemical reaction - 3 independent molecular species balance (C2H6, C2H4, H2) - 0 other equation relating the variable ============================= 0 No. of degree of freedom

Reactor 100 kmol C2H6/min 40 kmol H2/min n1 kmol C2H6/min Dehydrogenation of ethane in a steady-state continuous reactor. The reaction is One hundred kmol/min of ethane is fed to the reactor. The molar flow rate of H2 in the product stream is 40 kmol/min 40 kmol H2/min n1 kmol C2H6/min n2 kmol C2H4/min Reactor 100 kmol C2H6/min

Example: Extent of Reaction Dehydrogenation of ethane in a steady-state continuous reactor. The reaction is One hundred kmol/min of ethane is fed to the reactor. The molar flow rate of H2 in the product stream is 40 kmol/min 40 kmol H2/min n1 kmol C2H6/min n2 kmol C2H4/min Reactor 100 kmol C2H6/min Degree-of-freedom analysis 2 unknowns variables (n1,n2) + 1 independent chemical reaction - 3 independent reactive species (C2H6, C2H4, H2) - 0 independent nonreactive species - 0 other equation relating the variable ============================= 0 No. of degree of freedom

Write extent of reaction for each species C2H6 :n1 = 100-ξ C2H4 :n2= ξ Dehydrogenation of ethane in a steady-state continuous reactor. The reaction is One hundred kmol/min of ethane is fed to the reactor. The molar flow rate of H2 in the product stream is 40 kmol/min 40 kmol H2/min n1 kmol C2H6/min n2 kmol C2H4/min Reactor 100 kmol C2H6/min Write extent of reaction for each species C2H6 :n1 = 100-ξ C2H4 :n2= ξ H2 :40= ξ Solve for n1 and n2 (ξ =40) n1= 60 kmol C2H6/min; n2= 40 kmol C2H4/min

Product Separation & Recycle Overall Conversion Reactant input to Process – reactant output from Process Reactant input to Process Single Pass Conversion Reactant input to Reactor – reactant output from Reactor Reactant input to Reactor

Example: Product Separation & Recycle For example, consider the following labeled flowchart for a simple chemical process based on the reaction A B: 75 mol B/min 100 mol A/min 75 mol A/min Reactor Product Separation Unit 25 mol A/min

Product Separation Unit Purging To prevent any inert or insoluble substance build up and accumulate in the system Purge stream and recycle stream before and after the purge have a same composition. Product Fresh Feed Reactor Product Separation Unit Recycle Purge

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