SUMMER-18 EXAMINATION Model Answer

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1 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page of 7 Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may vary but the examiner may try to assess the understanding level of the candidate. 3) The language errors such as grammatical, spelling errors should not be given more Importance (Not applicable for subject English and Communication Skills. 4) While assessing figures, examiner may give credit for principal components indicated in the figure. The figures drawn by candidate and model answer may vary. The examiner may give credit for any equivalent figure drawn. 5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and there may be some difference in the candidate s answers and model answer. 6) In case of some questions credit may be given by judgement on part of examiner of relevant answer based on candidate s understanding. 7) For programming language papers, credit may be given to any other program based on equivalent concept.

2 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page of 7 Q No. Answer marks A Any 4 8 A-a Dalton s law: It states that the total pressure exerted by a gas mixture is equal to the sum of partial pressures Mathematical Statement: P =P +P +P 3 where P is the total pressure of gas mixture, P,P, P 3 are partial pressures A-b Ideal gas equation is PV=nRT Where P= pressure V=Volume n= number of moles R= Universal gas constant T= absolute temperature Value and unit of R is 8.34 KPa m 3 / kmol K A-c Standard heat of formation : It is the amount of heat liberated or absorbed when one mol of a compound is formed from its elements at standard conditions. Standard heat of combustion: It is the amount of heat liberated when one mol of a compound is combusted or burned in oxygen at standard conditions.(5 0 C and atm pressure) A-d Block diagram for distillation:

3 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 3 of 7 Distillate X kg/h Feed F kg/h distillation residue Y kg/h Overall balance is F = X+ Y A-e 0 kg Cl Molecular weight of Cl =7 Moles of Cl = weight / molecular weight = 0 / 7 = 0.8 kmoles PV=nRT P = 00 KPa T= 98 K V= nrt/p = 0.8 * 8.34 * 98 / 00 = 6.98 m 3 A-f 05.6 KPa.g Absolute pr = Gauge pr + atmospheric pr = = KPa -B Any -B a Basis: 5 kg liquid propane Kmoles of propane = 5/ 44 = 0.34 PV = nrt V = nrt/p

4 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 4 of 7 P = 0.35 KPa R = 8.34 m 3 kpa/ kmol.k T = 73 K V= 0.34 * 8.34 * 73 / 0.35 = m 3 -B b Basis: 00 kmol gas sample Avg. mol.wt of air = M X + M X Density = P* Mav / RT = 3 * * 0.79 = 8.84 = * 8.84/ 8.34 * 503 = 0.48 Kg/m 3 -B c To prove mol% of A = Mol fr of A * 00 Mol % of A = (Moles of A / Total moles of the system) * 00..() Mol fraction of A = (Moles of A / Total moles of the system) () Comparing () and() Mol % of A = Mole fraction of A * 00 Any 4 6 -a Steps involved in solving material balance without chemical reactions:.assume suitable basis of calculation as given in problem.. Adopt weight units in case of problem of process without chemical reaction. 3. Draw block diagram of process 4. Show input and output streams 5. Write overall material balance 6. Write individual material balance 7. Solve above two algebraic equations 8. Get values of two unknown quantities. 9.Write balances as follows: 4

(ISO/IEC - 700-005 Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 5 of 7 Unchanging component Outgoing component -b Basis: 00 kg of groundnut seeds.

5 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 5 of 7 Unchanging component Outgoing component -b Basis: 00 kg of groundnut seeds. feed product Component removed kg of solid=45kg kg of oil=45kg unchanging component is solid let weight of cake=x kg solid balance 0.8x=45 Therefore x=45/0.8=56.5kg Oil in cake=56.5*0.05 =.8kg Therefore oil recovered=45-.8 =4.9

6 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 6 of 7 % recovery of oil =(4.9/45)*00 =93.75 % -c SO + ½ O SO 3 SO fed = 00 kg. moles SO 3 formed = 80 kg. moles kg. mole SO reacted = kg. mole SO 3 formed? = 80 kg. mole SO 3 formed kg. mole SO reacted = 80 % conversion of SO =( SO reacted /SO fed)* 00 = 80*00/00 = 80% -d i)stoichiometric Equation : The stoichiometric equation of a chemical reaction is the statement indicating relative moles of reactant and products that take part in the reaction. For example, the stoichiometric equation CO + H CH 3 OH Indicates that one molecule of CO react with two molecules of hydrogen to produce one molecule of methanol ii)stoichiometric ratio : It is the ratio of stoichiometric coefficient of two molecular species or Components in the balanced reaction For example, in the stoichiometric equation CO + H CH 3 OH Stoichiometric ratio of H to CO = : (Students may write other suitable example)

7 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 7 of 7 -e Basis 50kmol SO 50 kmol air Reaction SO + O = SO 3 Air fed = 50kmol O in air = 50 (0.) = 3.5 kmol Theorctical requirement of O Kmol SO 0.5 kmol O = 50 = 5 kmol % excess of O used = = 00 = 6 % excess air used = 6% Theo.air reqd = 5 = 9.05 Kmol % excess air used = = 6% -f C 0 pm = kJ / (K mol. K) Moles of air (n) = 3 OR

8 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 8 of 7 T = 473K T 0 = 98 Heat added Q = n* C 0 pm ( T T 0 ) = 3*9.3955(473-98) = kj 3 Any 6 3-a Basis : 00 Kmol of feed Feed contains 60 kmol A, 30 kmol B and 0 kmol inerts Let X be the kmol of A reacted by reaction : A + B C A reacted = 0.8*60 = 48 kmol C From reaction B reacted = (/)* 48 = 4 kmol C formed = (/)* 48 = 4 kmol Product stream contains unreacted A, unreacted B, product C and inert Unreacted A = = kmol B unreacted = (30-4) kmol =6 kmol Total moles of product stream = =5 kmol Composition of product stream: component kmoles Mol% A 3.08 B 6.54 C inert b Basis :000 Kg of desired acid

9 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 9 of 7 Let X and Y be the kg of dilute acid and Commercial Grade required to prepare desired acid. Overall Material balance Kg dilute acid + kg Commercial Grade = Kg of desired acid X + Y = () Material Balance of H SO 4 H SO 4 in dilute acid + H SO 4 in Commercial Grade = H SO 4 in desired acid 0.5 X Y = 0.65 * X Y = X Y = Y= X put in equation () X + ( X ) = 000 X= 45 kg We have, X +Y = Y =000

10 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 0 of 7 Y = 548 Kg Dilute acid required = 45 Kg Commercial Grade Acid required =548 kg 3-c Basis : 000 kmol Benzen- Toluene mixture Top product X kmol Feed 00 kmol Distillation Residue Y kmol Let X and Y be the mass flow rates of distillate and bottom product respectively Overall Material Balance: X + Y = (i) Material Balance of benzene: ( 5/00)*X + (5/00)*Y = (8/00)* *X *Y = 80 By solving X =489.36Kg/hr Y= kg/hr Mass flow rates of distillate = Kg/hr ---- ans. (a) Mass flow rates of bottom Product = kg/hr ---- ans.(a) Benzene in distillate = 0.5 * = Kg/hr Benzene in feed = 0.8 *000 = 80 Kg/hr

11 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page of 7 benzene in distillate % recovery of benzene = * 00 Benzene in feed % recovery of benzene = * % recovery of benzene = % ans. (b) 4 Any 6 4-a Basis mol of n propanol liquid ) C (s) O (g) = CO (g) H = ) H (g) + O (g) = H O (e) H = ) C 3 H 7 OH (ʵ) + 4.5O = 3CO + 4H O Hc = H + C 3 H 7 OH = Heat of formation of n propanol = 3 H + 4 H - H 0 C = 3 (-393.5) + 4 (-85.83) (-08.9) = kj/mol. 4-b Basis : 00 Kg of Soyabean seeds It contains 8.6 kg oil,69 kg solids and.4 kg moisture Let X be the Kg of cake obtained Material balance of Solids : Solids in seeds = Solids in cake 0.69 * 00 = 0.877* X

12 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page of 7 X = Kg Material balance of Oil : Oil in seeds = Oil in cake + Oil recovered 0.86* 00 = 0.008* Oil recovered 8.6 = Oil recovered Oil recovered = =7.97 Kg Oil recovered % recovery of Oil = * 00 Oil in Seeds 7.97 % recovery of Oil = * % recovery of Oil = 96.6 % ans. 4-c To prove Pressure % = Mol % = Volume % Ideal gas law is PV = nrt.() Ideal gas law for component gas A(partial pressure) is P A V = n A RT () () / () P A V / PV = n A RT / nrt P A / P = n A / n Multiplying both sides by 00, we get (P A / P)* 00 = (n A / n) * 00 Pressure % = Mol %..(3) Ideal gas law for component gas A(pure component volume) is P V A = n A RT (4)

13 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 3 of 7 (4) / () P V A / PV = n A RT / nrt V A / V = n A / n Multiplying both sides by 00, we get (V A / V)* 00 = (n A / n) * 00 Volume % = Mol %..(5) Comparing (3) and(5) We can write Pressure % = Mol % = Volume % 5 Any 6 5-a Basis: 000 kg wet ONA 000 Kg feed 90% solid dryer Water Xkg Product Y kg 0.5% moisture Overall balance is 000 = X+ Y Balance for solid 0.90 * 000 = * Y Y = kg X = 95.48kg Water removed = 95.48kg Product obtained = kg % of original water removed = (Water removed/original water 0*00

14 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 4 of 7 5-b Basis: 00 Kmol/min of CO Q= Heat added = (95.48/00)*00 =95.48% T = n ʃ Cp dt T T 0 = n ʃ [ x 0-3 T x 0-6 T + T x 0-9 T 3 ] dt = n [.3655 (T T ) x 0-3 / (T T ) x 0-6 /3 (T 3 T 3 ) x 0-9 /4 (T 4 T 4 ) ] Where n= 00 kmol/min, T = 383 K, T =98 K = 00 [.3655 (383 98) x 0-3 / ( ) x 0-6 /3 ( ) x 0-9 /4 ( ) ] = KJ/min = KJ/s Q = KW Ans c Basis: mol of gaseous ethanol reacted C H 5 OH(g) CH 3 CHO(g) + H (g) ΔH o R = Standard heat of reaction =[ Σ ΔH o c] reactant - [ Σ ΔH o c] product =[ xδh o c C H 5 OH(g) ] - [ xδh o c CH 3 CHO(g) + x ΔH o c H ] 3

15 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 5 of 7 = [ x )] [ x (-9.65) + x (-85.83) ] = = KJ Ans. Note: Instead of -9.65, if the student calculated with the value of -.965,the answer will be -.33kJ. Full marks should be given. 6 Any a Vander Waal s equation of state: (P+a/V )(V-b)= nrt Where a & b are constants. A= 7 R T c / 64 P c lit. Mpa / mol B = RT c / 8 P c Tc & Pc = Critical Temperature and Pressure 6-b Water evaporated X kg/hr Feed F Kg/hr X % solid Evaporator 3 Thick liquor Y kg/hr X % solid Overall Material balance is F = X + Y 6-c Heat capacity: It is the amount of heat required to increase the temperature of one kg of substance by K. It is expressed on a unit mass or unit mole basis. Heat of combustion: It is the amount of heat liberated when one mole of a compound is combusted or burned in oxygen. The combustion reaction proceeds with a reduction in enthalpy of a system, hence heat of combustion is

16 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 6 of 7 negative. 6-d Hess s law of constant heat summation : It states that the enthalpy change i.e. heat evolved or absorbed in a particular reaction is the same whether the reaction takes place in one or several steps. For Example : Carbon can be converted into CO by two ways Path : C (s) + O ( g) > CO (g) Δ H Path : (i) C (s) + ½ O ( g) > CO (g) Δ H (ii)co (g) +½ O ( g) > CO (g) Δ H (i) + (ii) C (s) + O ( g) > CO (g) Thus Δ H = Δ H + Δ H 6-e Differentiate Conversion and Yield : Conversion Yield.Conversion is the ratio of the. Yield of a desired product is the amount of reactant reacted to the ratio of the quantity of the desired initial amount of the reactant product actually obtained to its quantity maximally obtainable.. Conversion gives us idea. The Yield of a desired product regarding how efficient a given tell us how efficient is a given chemical process is from the point chemical process is in terms of the of view of utilization of the reaction product. starting materials. 3. Higher values of Conversion is 3. Higher values of Yield is the the indication of minimum indication of minimum occurrence amount of the limiting reactant of side reactions. left unreacted. mark each for any 4 points

17 (ISO/IEC Certified) SUMMER-8 EXAMINATION Subject Title: Stoichiometry Subject code : 735 Page 7 of 7 4. Conversion is applicable to single reactions as well as to Complex reaction. 5. Conversion may have low to high values maximum 00% 4. Yield is applicable to Complex reaction 5. Yield should have always higher values,maximum : less than 00% 6-f Recycling: It is returning back a portion of stream leaving a process unit to the entrance of the process unit for further processing. Reasons for performing recycling: (any four). Maximum utilization of the valuable reactant. Improvement of the performance of the equipment/ operation 3. Utilization of the heat being lost in the exit stream. 4. Better operating conditions of the system 5. Improvement in the selectivity of a product 6. Enrichment of a product ¾ marks each

WINTER-15 EXAMINATION Model Answer

WINTER-15 EXAMINATION Model Answer Subject code :(735) Page of 9 Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the