St. Joseph’s College of Engineering Department of Chemical Engineering CHEMICAL PROCESS CALCULATIONS – CH CH 2254 QUESTION BANK – 2013 2013 -14 UNIT I and UNIT II
1. State Dalton’s and Amagat’s Amagat’s law. (May 2013) Dalton’s Law: If Law: If Pi is the partial pressure exerted by a single component in a mixture with the mole fraction yi then total pressure PT = ∑Pi and Pi = PT yi Av. Molecular Weight = ∑ Mi Yi Where Mi = Molecular weight of i th component. component. and Yi = Mole fraction of the i th component. Amagat’s Law: If Law: If Vi is the t he volume of pure component i present in the mixture, total volume of the gas V is given by, V = ∑Vi 2. Define mass fraction and mole fraction of a component in a mixture (May 2013)
Mass fraction of a component in a mixture is the ratio of its mass to the total mass of t he sample. Individual weight of the component in the sample Mass fraction of a component = [------------------------------------[-------------------------------------------------------------------------------------] -------] Total weight of the sample Individual Moles of the component in the sample Mole % of a component =[------------------------------=[--------------------------------------------------------------------------------] --------] x 100 1 00 Total number Moles of the sample Similarly, mole fraction of a component in a mixture is the ratio of its moles to the total moles of the sample. 3. Define compressibility factor (May 2013)
Compressibility factor is the ratio of the product of pressure and volume to the product of its absolute temperature and gas constant.
Z
pV RT
For ideal solutions it is 1. For real real gases values of Z can can be obtained obtained from the Nelson-Obert
generalized compressibility charts. 4. A wet paper contains 12% water on a wet basis. Calculate the % of water on a dry basis. (May 2012)
Given 100 Kg of wet sample contains 12 Kg of water. On dry basis: (100-12) = 88 Kgs of dry sample contains 12 Kg of water Hence % of water on dry basis =
12 100 13.64% 88 o
5. Calculate the volume of 50 Kgs of O 2 at a pressure of 730 mm Hg and 40 C. (May 2012)
Basis: 50 Kg of CO2 1
760 mmHg 1 atm
1 730 0.96 760
730 mm Hg
nCO2=Kg mol of Co2=50/44=1.14 Kg Mol V = nRT/P=1.14*0.082058 nRT/P=1.14*0.082058 (273+40)/0.96 = 30.5 m3 6. Show that for a mixture of ideal gas mixtures pressure% pressure% = mole% = Volume%. (May 2012)
All the components in the gas mixture occupy the total volume and hence Vi is the volume occupied by the component i, if it is present at the pressure p and temperature T of the mixture.
n RT Where ni is the number of moles of component i. V i i p Here Vi is proportional to ni. That is volume % of a component component in a gas mixture equals to mole% of it. Partial pressure of component i in a mixture of gases is the pressure exerted by the component when it occupies the same volume as that of the mixture at the same temperature.
n RT i Here Pi is directly proportional to ni. That is pressure% in a gas mixture equals to mole% of V
Similarly P i it.
Hence for ideal gases pressure% = Volume% = Mole% 7. Express the Vander Waals constant ‘a’ and ‘b’ in terms of critical properties of the constituent substance. (May 2011) The values of values values of „a‟ and „b‟ in terms of critical properties: critical properties: 2
a 27
2
R T c
64 P c
and b
RT c
8 P c
where Pc and T c are critical properties.
8. Define partial pressure pressure and vapor pressure. pressure. (May 2011)
The partial pressure of a component in a gaseous mixture is defined as the pressure that the component exerts when it occupies the same volume as that of the mixture at the same temperature of the mixture. The vapor pressure of liquid is defined absolute pressure at which the liquid and its vapor are in equilibrium at a given temperature. 9. What are primary primary dimensions? dimensions? (Dec 2011)
Seven “primary” dimensions are : mass [M], length [L], time [T], amount of a substance [N], electric cu r rent, rent, temperature, and luminous intensity. 10. Define mole fraction. (Dec 2011)
Mole fraction of a component in a mixture is the ratio of its moles to the total number of moles in the mixture. Mole fraction = Individual number of moles/Total number of Moles. 11. What is an ideal gas? (Dec 2011)
An ideal gas in one in which there are no intermolecular forces of attraction or repulsion and the actual volume occupied by the molecules is negligible compared to the total volume occupied by the gas. 12. What is the weight of 1 litre of methane at NTP? (May 2010)
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At NTP, P = 101.325 kPa; T = 273.15 K
n= PV/RT = (101.325 x 103) (1 x 10-3) /8.314 x 103 (273.15) = 0.446 x 10-3 Kg moles
Mass of chlorine = (Kg moles) x (molecular weight) = (0.446 x 10-3)(35.5) = 1.58 x 10-3 Kgs. 13. A solution of NaCl in water contains 230 g of NaCl per litre of solution. The density of the solution is 1.148 g/cc. Find the molality. (April 2010)
Molality is defined as the moles of solute dissolved in one Kg of the solvent. Density of the solution = 1.148 g/cc 1 cc 1.148 g 1000 cc 1 litre 1148 g. 1.148 Kg of solute contains 230 g of NaCl 230/58.5 = 3.93 moles. Molality moles of solute /Mass of solvent in Kgs = 3.93/(1.148-0.230) = 4.28 M 14. The nitrogen content of an urea sample is 22% by weight. Calculate the purity of the sample. (May 2010)
The molecular weight of urea (NH2CONH2) is 60. 60 g of urea contains 28 g of nitrogen. Given sample contains 22 g in 100 g. 28/60 = 0.467 corresponds to 100% purity 22/100 = 0.22 corresponds to (100/0.467) x 0.22 = 47.1% Purity. 15. Calculate the density of air at 2.5 atm and 45 oC. (May 2010)
Molar volume of air V = RT/P = (8.314x103)(273.15 + 45)/2.5 (1.01325 x 105) = 10.44 m3/Kg mole. Air is assumed to be made up of Nitrogen (0.78 Mole fraction) and Oxygen (0.22 Mole fraction) Average Molecular weight of air = 28(0.79) + 32(0.21) = 28.84 Kg/Kg mole. Density of the gas = Molecular weight / Molal volume = 28.84/10.44 = 2.76 Kg/m3. 16. Define mole fraction and explain how it is related to partial pressure. (May 2010)
Mole fraction of a component is a mixture is the ratio of its moles to the total number of moles of the mixture. Partial pressure on any component in a mixture can be determined by multiplying its mole fraction with the total pressure of the mixture. Mole fraction = Partial pressure/Total pressure. 17. Do the following conversions (Nov 2010)
(i). 7000 Ao to nm 1 Ao = 0.1 nm. Hence 7000 Ao = 700 nm. (ii) 4000 cm to Ao a cm = 1 x 10Ao Hence 4000 cm = 4000 x 10 8 Ao (iii). 0.94 g/cc in to specific volume. Specific volume cc/gm Inverse of density. Ans: 1.064 cc/g. (iv) 4.5 g water to mole of water. Mole of water = (Wt of water)/(Mol. Wt of water) = 4.5/18 = 0.25 g/g mol. 18. Calculate the molarity of the solution when 34.2 gram C12H22O11 is dissolved in one litre of water. (Nov 2010)
Molecular weight of sucrose C12H22O11 = (12*12 +1*22+16*11) = 342 34.2 gm 34.2/342 = 0.1 moles. Molarity = moles of solute/Litres of solvent = 0.1/1 = 0.1 M 19. The volumetric flow rate of a liquid of specific gravity 0.8 is 120 ft 3/min. Find the mass flow rate in Kg/s. (May 2012) Density of the liquid = 0.8 x 1000 = 800 Kg/m3 3
3 ft 3 m 3 1 min 1 3 Volumetric flow rate = 120 0.567 m s min ft 3.28 sec 60
Mass flow rate = (Density ) (Vol Flow rate) = 800 x 0.567 = 45.36 Kg/S. 3
20. Find the value of gas constant R in ft 3 atm/lbmol oR (May 2012)
Normal Temperature = 273.15K; 1K 1.8 oR; 273.15 K = 491.67 oR. Normal pressure = 1 atm 3
m 3 ft 3 3.28 Kgmol 1 3 Volume 22.4136 22.4136 358.89 ft3/lb mol Kgmol m 1 lbmol 2.205 R = PV/T = (1 x 358.89)/491.67 = 0.730 ft3 atm/lb mol oR PART B
1. 1000 litres of a mixture of H2, N 2, and CO2 at 150oC was found to have the following ratio for the partial pressures of the gases: PH2:PN2:PCO2 is 1:4:3. If the total pressure is 2 atm absolute, calculate (i). Mole fraction of each of these gases. (ii). Weight percent of each of these gases. (iii). Average Molecular weight (May 2013) (iv). Weight of CO2 in Kg. 2. A natural gas has the following composition by volume: CH4 = 82%, C2H6 =12%, and N2= 6%. Calculate the (May 2013) density of gas at 288 K (15oC) and 101.325 KPa and composition in weight percentage. 3. Estimate the density of the chlorine gas at temperature 503 K and 15.2 MPa pressure using, (i). Ideal gas law (ii) The van der Walls equation.
(May 2013)
4. A gas mixture entering an absorber contains 20% CO 2 and 80% air by volume. In the absorber, 95% CO2 is removed. The gas enters at 60oC and 750 mm HG and leaves at 40oC and 740 mm Hg. Calculate (i). Volume of gas leaving per 100 m3 entering the absorption apparatus. (ii). % composition by volume of gases leaving. (May 2012) (iii). Weight of CO2 removed per 100 m3 of gas entering the absorber. 5. A gas has the following composition by volume. CO2 = 9.5%, CO = 0.2%, O2 = 9.6% and N2 = 80.7%. Using ideal gas law, calculate, (i). Composition by Wt% (ii). Volume occupied by 1 Kg of the gas at 40oC and 740 mm Hg. (May 2012) (iii). Density of the gas in Kg/m3 at 40oC and 740 mm Hg. 6. A liquid mixture contains three components A (MW = 72), B (MW = 58) and C (MW = 56) in which A and B are present in the ratio 1.5% : 1 and the weight percent of B is 25%. The specific gravities of pure liquids are 0.67, 0.60 and 0.58 respectively for A, B and C and there is no volume change on mixing. Calculate the following, (i). The analysis of the mixture in mole percent. (ii). The MW (Molecular Weight) of the mixture. (iii). The volume percent of C on a B-free basis. (Nov 2011) (iv). The specific gravity of the mixture. 7. Natural gas is piped from the well at 300 K and 400 kPa. The gas is found to contain 93.0% methane, 4.5% ethane and the rest nitrogen. Calculate the following: (i). The partial pressure of nitrogen. (ii). The pure component volume of ethane in 10 m3 of the gas. (iii). Density at standard conditions in Kg/m3 (iv). Average molecular weight of the gas. (Nov 2011) (v). Composition in weight percent. 8. The feed to absorption column consists of 30% H2S and 70% inerts. Only H2S is removed from the gas by absorbing in an alkaline solution. The as enters the absorber at 700 kPa, 350 K and leaves at 600 kPa, 300 K containing 5% H2S. If H2S is removed at a rate of 100 Kg/hr, calculate the following: (i). Number of moles of each component entering and leaving the system. (ii). Cubic meters of gas entering per hour. 4
(iii). Cubic meters of gas leaving per hour. (iv). Percentage recovery of H2S
(Nov 2011)
9. Check the dimensional consistency of the following empirical equation for a heat transfer coefficient:
hi 0.023G 0.8 k 0.67 c p0.33 D 0.2 0.47 Where h is the heat transfer coefficient, G is the mass velocity, k is the thermal conductivity, c p is the specific heat, D (May 2010) is the diameter and µ is the absolute viscosity. 10. Define molarity, normality and molality. A solution of sodium chloride in water contains 25% NaCl (by mass) at 323 K. The density of the solution is 1.135 Kg/litre. Find the molarity, norality and molality of the solution. (Nov 2010) UNIT III MATERAL BALANCS
UNIT IV
HUMIDITY AND SATURATION
PART A 1. What is the use of Humidity chart? (May 2013)
Humidity charts are convenient for determining the properties of moist air and is useful for delineating the various humidification processes. The use of humidity charts eliminates the input of physical properties in heat and mass balance calculations. 2. Write the relationship between the absolute humidity and molar humidity. (May 2013, May 2012)
Absolute humidity H = 0.622 Hm where H m is the molar humidity. 3. What do you mean by limiting and excess reactant? (May 2013)
A limiting reactant is one which decides the conversion in the reactions and is available in less quantity than the other reactants as per the stoichiometric requirement. An excess reactant in one which is in excess amount over the stoichiometric requirement of the reactant as determined by the desired chemical reaction. 4. What are the advantages of using recycle stream? (May 2013)
Recycling operations has the following objectives: (i) To utilize the valuable reactants to their maximum and avoid wastage (ii) to utilize the heat being lost in the outgoing steam (iii) to improve the performance of the equipment (iv) to control the operating variables in a reaction and (v) to improve the selectivity of a product. 5. An evaporator is fed with 10,000 Kgs/hr of a solution containing 1% solute which is concentrated to 1.5% by wt of solute. Determine the amount of weight of vapor. (May 2012)
By component balance : FXf = PX p 10,000(0.01) = P (0.015) ; P = 6666.67 Kgs/hr. V= F- P = 10,000 – 666.67 = 3333.33 Kg/hr. 6. Define limiting reactant and degree of completion. (May 2012)
Limiting reactant is one that is available in less quantity than other reactants, as per the stoichiometric proportions. Degree of completion, the extent to which a reaction progresses. 7. Define purge stream with a simple schematic. (May 2012)
When a process uses a recycle loop, there can often be a buildup of some undesired material within the system. By using a purge, a fraction of the recycle loop material is removed. This purge fraction is generally only a few percent of the recycle flow rate. It is done to bring down the concentration of unwanted substance in a gas mixture by addition of inert gas such as nitrogen or argon in tank and then venting the mixture to atmosphere.
5
8. CO combines with Cl 2 in the presence of a catalyst to form phosgene as per the reaction CO(g) + Cl 2 COCl2(g). After reaction the products contained 12 moles of COCl 2, 3 moles of Cl 2 and 8 moles of CO. Assuming that the original reactant mixture is free of phosgene, find out the percentage excess of the excess reactant used. (May 2012)
As per the Stoichiometry, 1 mole of CO reacts with 1 mole of Cl2 to give 1 mole of COCl2. Hence reaction mixture composition: 20 moles of CO and 15 moles of Cl2. Excess reactant is CO, and the limiting reactant is Cl2. % Excess = (18-15/15) x 100 = 20%. 9. Wood containing 40% moisture is dried to 5% moisture. What mass of water in kilograms is evaporated per kilogram of dry wood? (May 2012)
Basis: 100 Kg of moist wood. Amount of water 40 Kgs; amount of dry wood 60 Kgs. After drying amount of water 5 Kgs. Total amount of water evaporated = 35 Kgs. Mass of water in Kgs evaporated per Kg of dry wood = (35/60) x 100 = 58.33% 10. Explain relative humidity and percentage humidity. (May 2012) Relative Humidity or Percentage relative humidity: It is the ratio of the partial pressure of water vapor in air to the vapor pressure of water at the dry bulb temperature.
RH = {PA / Ps} x 100 Percentage Humidity or Percentage Absolute humidity: It is defined as the ratio of the actual absolute humidity to the saturation humidity.
% Humidity: = {H/Hs} x 100 11. Explain bubble point and dew point with a diagram. (May 2012)
12. Define Reflux ratio and recycle ratio. (May 2012)
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Reflux ratio: The ratio of the amount of condensate being refluxed to the amount being withdrawn as product. Generally, the higher the reflux ratio, the greater is the degree of separation of the components in a distillation system. Recycle ratio: In a reactive process, there is generally some unreacted feed material found in the product. In order to reduce cost and increase efficiency, the unreacted material is often separated and reused in a recycle loop. Recycle ratio is the ratio of quantity of recycle feed to the quantity of fresh feed.
13. What are steady state and unsteady state processes? (Nov 2011)
Steady state processes are those in which the parameters will not change with respect to time. In Put = Output On the other hand, in unsteady state processes variables will change with respect to time. In put = Output + Accumulation. 14. What is a tie element? (Nov 2011)
Tie element is one whose quantity does not change during the unit operation. Example: Concentration of a solution in an evaporator in which the dissolved solids do not change. During drying bone dry material will not change which is a tie element, but the solvent amount changes. 15. Define adiabatic saturation temperature. (May 2010)
Adiabatic Saturation Temperature: Adiabatic process is the one in which no heat flows into or out of the system but during which thermal changes usually occur within the system. When a definite quantity of water is allowed to evaporate in a stream of air adiabatically, the dry bulb temperature of the gas decreases and the Humidity of the air increases. The final temperature of the intimately mixed stream is adiabatic saturation temperature. For air water systems WB and AST are practically same. o 16. Air at a temperature of 20 C and 750 mm Hg has a relative humidity of 80%. Calculate the molar humidity of the air if the vapor pressure of water at 20 oC is 17.5 mm Hg. (May 2010) Relative humidity = (PA/Ps) x 100; 80 = (PA/17.5) x 100; Therefore PA = 14 mm Hg. Molar humidity = PA/P - PA = 14/750-14 = 0.019 moles of water/moles of dry air. 17. A wet material weighing 1000 Kg is dried from 40% moisture to 15% moisture. What will be final weight of the material? (May 2010) Amount of initial moisture content = 1000(0.4) = 400 Kg. Amount of final moisture content = 1000(0.15) = 150 Kg. Amount of water evaporated = 400 – 150 = 250 Kgs. After evaporation final weight of the sample = 1000 – 250 = 750 Kgs. 18. A mixture of acetone vapor and nitrogen contains 14.8% acetone by volume. Calculate the relative o o saturation and percentage saturation at 20 C and 745 mm Hg. The vapor pressure of acetone at 20 C is 184.8 mm Hg. (May 2010)
Partial pressure of Acetone P A = 745 (0.148) = 110.26 Relative saturation = Amount of vapor / Amount of vapor at saturation condition = (110.26/184.8) x 100 = 59.66 %
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110.26 745 110 . 26 100 53.87% % Saturation = 184 . 8 745 184.8
19. Define humid heat and humid volume. Humid Heat: It is defined as the heat capacity of 1 kg dry gas and the moisture contained in it. It is the heat required to raise the temperature of unit mass of gas and its accompanying vapor by one degree at constant pressure.
Cs = CA H + CB where CA and CB are specific heat of water and air. Cs = 1.84 H + 1.006 Humid Volume: It is the volume of unit mass of dry gas and its accompanying vapor at the prevailing temperature and pressure.
VH = RT/ pA MA
and
VHs = RT/ ps MA
20. Clearly distinguish between the terms Conversion, Selectivity and yield.
Conversion is defined as the ratio of the reacting amount of a component to its initial amount. It is the measure of the degree of completion of the chemical reaction. The yield and selectivity are the measure of the degree of the desired reaction predominates over other side reactions. The yield is defined as quantity of limiting reactant reacted to produce the desired product expressed as percentage of the total quantity of limiting reactant reacted. PART B
1. It is required to make 1000 Kg mixed acid containing 60% sulphuric acid, 32% Nitric acid and 8% water by blending, (i). Spent acid containing 11.3% nitric acid, 44.4% sulphuric acid and 44.3% water. (ii) Aqueous 90% nitric acid and (iii) aqueous 98% sulphuric acid. All percentages are by mass. Calculate the (May 2013) quantities of each of three acids required for blending. 2. After crystallization operation, the solution of calcium chloride in water contains 60 g of CaCl2 per 100 g of water. Calculate the amount of this solution necessary to dissolve 200 Kg of CaCl2. 6 H2O crystals at a temperature of 298 K (May 2013) (25oC). The solubility of CaCl2 at 298 K is 819.2 g of CaCl2 per 1000 g of water. 3. Wet solid containing 50% water and 50% solids are to be dried to get solids with 5% water by weight. Fresh air contains 0.010 Kg water per kg of dry air and air leaving the dryer contains 0.05 Kg of water per kg of dry air. If 100 Kg of dry air enters the dryer per kg of dry solids, calculate the fraction of air recirculated and recycle ratio. (May 2013) 4. In synthesis of ammonia the fresh feed containing 24.75% nitrogen 74 .25% hydrogen and 1% inerts (on mole basis) are mixed with recycle feed. Mixed feed entering into reactor resulted in 25% conversion to ammonia. The product mixture is passed through condenser where ammonia gets condensed and the remaining gases are recycled after purging small portion of gas stream. The recycle stream contains 12.5 mol% inerts. Calculate recycle ratio and purge (May 2013) ratio. 5. In order to obtain barium in a form that may be put in to solution, the natural barites, containing only pure barium sulfate and infusible matter is fused with an excess of pure anhydrous soda ash. Upon analysis of the fusion mass it is found to contain 11.3% barium sulfate, 27.7% sodium sulfate and 20.35% sodium carbonate. The remaining is barium carbonate and infusible matter. Calculate, (i). The % conversion of barium sulfate to barium carbonate. (ii). The composition of original barites and (iii). The % excess of sodium carbonate used above the theoretical amount required for reaction with all the barium (May 2013) sulfate. 6. The dry bulb temperature and dew point of ambient air were found to be 303 K and 298 K respectively. Calculate (i). The absolute molar humidity. 8
(ii) The absolute humidity. (iii). The percentage saturation and (iv) The humid heat. Data: Vapour pressure of water at 289 K = 1.818 kPa Vapour pressure of water at 303 K = 4.243 kPa (May 2013) Barometric pressure= 101 kPa.
7. A crystallizer is charged with 6400 Kgs of an aqueous solution containing 29.6% anhydrous Na2SO4. The solution is cooled and 10% of the initial H2O is lost in evaporation. The crystals obtained are Na2SO4 10H 2O. If the mother liquor is found to contain 18.3% Na2SO4 calculate the weight of mother liquor and crystals. (May 2012) 8. Antimony is produced by heating powdered Sb2 S3 and Fe followed by removal of Sb from the reaction vessel. If 1 Kg of Sb2 S3 and 0.6 Kg of Fe are heated to give 0.4 Kgs of Sb, calculate, (i). Limiting reactant (ii). Degrees of completion (May 2012) (iii). Percentage conversion of Fe 9. A 1:2 mixture of CO and H 2 is used to produce methanol with 15% conversion based on CO. The methanol is separated completely and the unconverted CO and H2 are recycled back to the reactor after mixing with the fresh feed. The concentration of impurities in the fresh feed is 0.1% by volume and in the recycle stream is 1% by volume, determine: (i). Rate of purge stream (May 2012) (ii). Ratio between the purge and fresh feed streams. 10. Calculate the (i). Molar humidity (ii). Absolute humidity (iii). Relative humidity (iv). Percentage humidity for air which is at 29oC and 750 mmHg. Dew point of air is 18oC. Given: Vapor pressure of water at 18 oC = 15.40 mm Hg (May 2012) Vapor pressure of water at 29 oC = 30.01 mm Hg. UNIT V – FUELS AND COMBUSTION 1. Write Antoine equation. (May 2013).
ln P A
B
T C
where P is the vapor pressure in kPa, T is the temperature in K and A, B and C are constants.
2. Define theoretical flame temperature. (May 2013)
Theoretical flame temperature is the calculated temperature of the combustion products, with the assumption that the burner is adiabatic and no dissociation of the products occurs. 3. How heat of combustion of white petrol determined? (May 2012)
White petrol is kerosene based aviation fuel. The heating value of a fuel may be determined either by a calculation from a chemical analysis or by burning a sample in a calorimeter. 4. Pure butane gas is burnt in a furnace using 10% excess air. Calculate the mole ratio of butane to oxygen in the feed to the furnace. (May 2012).
C4H10 + 6.5O2 4CO2 + 5H2O Theoretical requirement of O2 per mole of butane = 6.5 Oxygen used 10% excess = 6.5 x 1.1 = 7.15 moles. Hence mole ratio of butane to oxygen in the feed = 1:7.15 5. Define excess air. (Nov 2011) 9
Excess air is the amount of additional air used for combustion than the stoichiometrically required amount. 6. Enumerate between wet gas analysis and dry gas analysis. (May 2011)
Wet gas analysis is the analysis in which moisture content present the gas stream is also accounted. When the analysis if done by eliminating the moisture content it is termed as dry gas analysis. 7. Differentiate between proximate and ultimate analysis. (Nov 2010)
The ultimate analysis of coal gives the composition of the coal as the percentage by weight of the various elements such as C, H, N, O, S, etc and the ash. The proximate analysis of coal analyses only moisture, volatile combustible matter, ash and fixed carbon as percentage by weight of coal. 8. Give the use of Orsat apparatus. (Nov 2010)
Orsat apparatus is used to estimate the composition of the flue gas on a dry basis. It measures the amount of CO2, CO, and O2 in a given sample of flue gas. 9. What is a fuel? What are flue gases?
A fuel is a combustible substance which on burning releases a large amount of heat energy. The combustibles in the fuel are carbon, hydrogen and sulphur. The products of combustion of a fuel are known as flue gases. 10. Define calorific value of a fuel?
The calorific value of a fuel is the standard heat of combustion of the fuel. It is defined as the total heat produced when unit mass of the fuel is completely burn with pure oxygen. The calorific value is also known as heating value. 11. What is net calorific value?
When water vapor is present in the flue gas, the latent heat of vaporization is not available as heat energy of the fuel. The calorific value of the fuel determined, considering that water is present in the form of vapor is known as net calorific value or lower heating value or net heating value. 12. What is gross heating value?
Gross heating value is the total heat recovered after burning the fuel, including the latent heat of vaporization of the water vapor that is vaporized during the combustion operation. 13. What is theoretical oxygen demand?
Theoretical oxygen demand (ThOD) is the minimum amount of oxygen required to burn the fuel completely so that the combustibles in it namely carbon, hydrogen, an sulphur are converted to carbon dioxide, water vapor and sulphur dioxide. 14. Define the percentage of excess air?
Percentage of excess air is the ratio of the difference between the actual air supplied and theoretical air demand to the theoretical air demand. Actual air supplied – Theoretical air demand Percentage of excess air = ---------------------------------------------------------- x 100 Theoretical air demand 15. How will you determine the calorific value of fuels?
The calorific value of a fuel is determined in a calorimeter in which the fuel is burnt with pure oxygen and the heat liberated is absorbed in water. The rise in temperature of water gives the calorific value of the fuel, 10
NCV = GCV – (Wt% Hydrogen) (9) (ƛ ) KJ/Kg 16. Define the thermal efficiency of a boiler or a furnace.
Thermal efficiency of a boiler is the ratio of useful heat gain to the total heat input. Useful heat gain will be always lower than total heat input because heat is lost in the flue gases, in the refuse which remains at the end of combustion, in vaporizing the moisture content of the fuel, blow downs, and radiations. 17. Name any two methods of improving the thermal efficiency?
Thermal efficiency can be improved by (i) recovering the heat of flue gases by employing economizers or pre heaters and (ii) minimizing the use of excess air. 18. How will you determine the rank of the coal?
Rank of the coal is determined by the fuel ratio value. Fuel ratio is the ratio of the amount of fixed carbon to the amount of volatile matter per unit amount of fuel. 19. Estimate the amount of dry air required to burn 1 Kg of carbon.
C + O2 CO2 32 Kgs of Oxygen is required to burn 12 Kg of Carbon. For one Kg of carbon, 32/12 = 2.67 Kgs of Oxygen is required. Assuming the wt fraction of O2 in air is 0.23, Amount of dry air required to burn 1 Kg of Carbon = 2.67/0.23 = 11.60 Kgs. 20. Pure carbon is completely burnt in oxygen. The flue gas analysis is 70%CO 2, 20%CO and 10%O 2. Estimate the amount of excess oxygen used.
Excess oxygen used is zero%. 10% of O2 in the flue gas is due to partial combustion of 20% of carbon to CO. PART B
1. Methanol is being burnt as a fuel; both methanol and air are available at 298 K. If 40% excess air is supplied calculate the adiabatic flame temperature assuming the combustion to be complete.(May 2013) (i). Net calorific value of methanol at 298 K = 638100 J/mol (ii). Heat capacity data C p Component CO2 H2O O2 N2
a bT cT 2 d 3 J / mol . K
a 37.174 25.1574 18.4331 31.1182
b x 103 23.2371 21.2818 21.7174 3.1969
c x 106 -7.3788 -5.34 -8.3048 -0.4052
d x 109 0.8213 0.4825 1.1558 0.0023
2. The ultimate analysis of coal sample is given below. Carbon = 61.5%; Hydrogen = 3.5% Sulphur = 0.4% Ash = 14.2% Nitrogen = 1.8% and the rest is Oxygen. Calculate, (i). Theoretical oxygen requirement per unit weight of coal. (ii). Theoretical dry air requirement per unit weight of coal. (iii). The orsat analysis of flue gases when the coal is burned with 90% excess dry air. (May 2013) 3. Natural gas containing 90% CH4, 6% C2H6 and 4% C3H8 is burnt with 40% excess air. Assuming that 90% of hydrocarbons are converted to CO 2 and the rest to CO, determine, (i). Flue gas analysis. (ii). Orsat analysis of the flue gas. (May 2012) 4. Propane is burnt with excess air to ensure complete combustion. If 55 Kgs of CO2 and 15 Kgs of CO are obtained when propane is completely burned with 500 Kgs air, determine the following, (i). The mass of propane burnt (In Kilograms) (ii). The percent excess air. 11
(iii). The composition of fuel gas. (May 2012) 5. A producer gas contains CO2 = 9.2%, C2H4 = 0.4%, CO = 20.9%; H2 = 15.6%, CH4 = 1.9%, N2 = 52%. When it is burnt the product of combustion are found to contain CO2 = 10.8%, CO = 0.4%, O2 = 9.2% and N2= 79.6%. Calculate (May 2011) m3 of air used per m3 of producer gas, % excess air and % N 2 that has come from producer gas. 6. A sample of fuel has the following composition by mass C = 84%; H2 = 15.2% and the rest comprising noncombustibles. The fuel was completely burnt with excess air in a internal combustion engine. The dry exhaust gas has 8.6% CO2 by volume. Estimate the amount of air supplied per Kg of fuel and find the products of combustion. (May 2011) (Air contains 23.1 weight% of Oxygen). 7. A fuel gas constitutes of CO2 = 3.4%, C2H4 = 3.7%. C6H6 = 1.5%, O2=0.3%, CO = 17.4%, H2 = 36.8%, CH4 = 24.9% and N2 = 12% (on mole basis). It is burnt with air in a furnace. The pyrite analyzer indicated 10.0 mole% CO2 (on dry basis) in the flue gases. Find the percent excess air used and the complete Orsat analysis. (May 2011) 8. In an ultimate analysis the following are estimated C = 65.93%, N=1.3%, available H = 3.5%, combined H2O = 6.31%, free moisture = 4.38%. Calculate the % of ash. If 2 grams of coal is fired how much ash is expected? 9. Ethane is burnt with 60% excess air. The percentage conversion of ethane is 90% of the ethane burnt, 25% reacts to form CO and the balance forms CO 2. Calculate the composition of the flue gas. (May 2010) 10. Blast furnace gas having an analysis by volume on basis of: CO2 – 13.0 %, CO – 25.0 %, H2 – 3.5 %, N2 – 58.5 %, is burnt in a furnace. Calculate: (i). Percentage of excess air when the dry product of combustion contains 3.5 % O2 (ii). Percentage of excess air when the dry flue gases contain 19.5 % CO2 5.8% o2 AND 74.7 % N2 UNIT VI – THERMO PHYSICS: PART A 1. Define Kopp’s rule. (May 2012)
Kopp‟s rule predicts that the heat capacity of solid compound at room temperature is approximately equal to the sum of the heat capacities of the individual components. 2. Define heat capacity. (May 2012)
Heat required to raise the temperature of the one Kg of any substance by 1 K is termed as the heat capacity of that substance. 3. Diffentiate between sensible heat and latent heat? (May 2011)
The heat absorbed by any material that results in its temperature increase is called sensible heat. The heat required to change the state of unit mass of a substance from one phase to another is known as the latent heat. This will not be perceived as increase in temperature. 4. Define specific heat. (Nov 2010)
Specific heat of any material is the heat required to raise the temperature of unit amount of material by one degree centigrade. J/Kg oC 5. What is Trouton’s rule? (May 2010)
Trouton‟s rule states that the ratio of molar heat of vaporization at the normal boiling point to the boiling point in absolute units is constant and is approximately equal to 57.8 J/mol K
H vb T b
K 87.9 J/ Mol K
6. What is the effect of pressure on heat capacity of a gas?
12
Heat capacities for the ideal gas state are independent of pressure. However, they are functions of temperature. 7. Define law of conservation of energy. According to law of conservation of energy, energy can neither be created nor be destroyed, but it can be transferred from one form to another form. 8. List out the properties of an ideal gas. The properties of an ideal gas are: An ideal gas consists of a large number of identical molecules. The volume occupied by the molecules themselves is negligible compared to the volume occupied by the gas. The molecules obey Newton's laws of motion, and they move in random motion. The molecules experience forces only during collisions; any collisions are completely elastic, and take a negligible amount of time. 9. What is kinetic energy?
The energy possessed by the body by virtue of its motion is called kinetic energy.
KE
1 2
mv 2
10. What is potential energy?
The energy possessed by the body by virtue of its position is called potential energy.
PE mgz
11. What is flow energy?
Flow energy of a fluid entering a conduit is the force acting on the fluid responsible for the flow of the fluid and it the product of pressure and the cross sectional area of the conduit. This work which is equal to PV for unit mass of the fluid is done by the flowing fluid and therefore is called flow energy. 12. The potential energy of a body of mass 20.0 Kg is 3.5 KJ. What is the height of the body from the ground? If a body of mass 20 Kg is moving at a velocity of 50 m/s what is its kinetic energy?
PE mgz ; 3.5 x 103 = (20)(9.81)(z); Hence Z = Height of the body from the ground = 17.83 m 1 1 KE mv 2 = (20)(50)2= 25 KJ 2 2 13. What is internal energy?
The energy stored in the system by virtue of the configuration and motion of the molecules constituting the system is called its internal energy. 14. What is enthalpy?
Enthalpy may be defined as total energy as it includes both the intrinsic energy (U) and the energy due to the expansion possibilities of the system or the energy that the system possess because of it occupying a space (PV). 15. How will you quantify mean heat capacity?
As the heat capacity is a strong function of temperature, the mean heat capacity at constant pressure may be denoted by Cpm and can be calculated as:
13
T 2
C pm
C dt p
Q (T 2 T 1 )
T 1
(T 2 T 1 )
T 2
1 (T 2
(a bT cT T ) 1
2
)dT
T 1
b 2 c 3 2 3 ( ) ( ) ( ) a T T T T T T 2 1 2 1 2 1 (T 2 T 1 ) 2 3 1
16. How will you evaluate the heat capacity of mixture of gases?
The heat capacity of mixture of gases can be evaluated as the sum of the heat capacities of the individual constituents each weighted according to the mole fraction in the mixture.
C p ,mixture
n
y C i
p ,i
i 1
17. Write the Clapeyron equation.
The Clapeyron equation predicts the dependence of vapour pressure on the temperature.
dP s dT
H v T V
18. Write and explain the Clausius-Clapeyron equation.
ln
P 2 s P 1 s
H v 1 1 Where P2S and P1Sare the saturation pressures at temperatures T2 and T1 respectively, R T 1 T 2
and H v is the latent heat of vaporization. 19. Write the Watson equation.
Watson proposed the following empirical equation for calculating latent heat of vaporization in terms of latent heat of vaporization at the normal boiling point.
H v 2 1 T r 2 H v1 1 T r 1
0.38
Where H v 2 and H v1 are the heat of vaporization at temperature T1 and T2 respectively.
T r 2 and
T r 1 are the respective reduced temperatures.
20. Write the Kistyakowsky equation.
The Kistyakowsky equation is an exact thermodynamic equation that can be utilised for estimating the enthalpy of vaporization of non polar liquids at their normal boiling points. Expressing the heat of vaporization at the normal boiling point in J/mol and temperature in K, the equation takes the following form.
H vb T b
36.63 8.31ln T b
PART B
1. 1 Kg of tin is to be heated from 30 oC to 300oC. Tin remains in the molten condition at 300oC. Determine the heat required for this process. Given: Molecular weight = 118.7, melting point = 231.8oC, λ f = 1720 Kcal/Kg mole, C p of solid Tin = 5.05+0.0048T Kcal/Kg mole K and T is in K. (May 2012) C pm of molten tin between melting point and 300oC is 7.64 Kcal/Kg mole K 14
2. Calculate the amount of heat given off when 1 m3 of air at standard condition cools from 600oC to 100oC at constant pressure. (May 2012) C p(air) Kcal/Kg mole K = 6.386 + 1.762 x 10-3T – 0.2656 x 10-6T2 3. Explain in detail the determination of heat capacities of solids and liquids.
(May 2011)
4. The analysis of 15, 000 lit of a gas mixture at standard condition is as follows. SO2 = 10%, O2=12% and N2= 78%. How much heat must be added to this gas to change its temperature from 30oC to 425oC The C pm values are in cal/g mole oC Gas C pm at 30oC C pm at 425oC
SO2 10 11
O2 6.96 7.32
N2 6.80 7.12
(May 2011)
5. Chlorinated diphenyl is heated from 313 K to 553 K at the rate of 4000 Kg/h in an indirectly fixed heater. In this particular temperature range the heat capacity of the fluid is given by equation, C1 = 0.7511 + 1.465 x 10-3 T KJ/Kg. K Where T is in K. Heat capacity data for chlorinated diphenyl at 313 K and 553 K are 1.1807 and 1.5198 KJ/Kg K respectively. Calculate the heat to be supplied to the fluid in the heater using the heat capacity equation. (Nov 2010) 6. An evaporator is to be fed with 10,000 Kg/hr of a solution having 1% solids. The feed is at 38oC. It is concentrated to 2% solids. Steam at 108oC is used. Find the weight of steam used if the enthalpies of feed, product and vapor are 38.1 Kcal/Kg, 100.8 Kcal/Kg, and 540 Kcal/Kg and 644 Kcal/kg respectively. (May 2010) 7. Calculate the heat input to raise the temperature of 144 Kg of CO2 from 105oC to 1000oC. Perform the calculation in the following ways: (i). By integrating the expression. (ii). By using mean heat capacity value, C pCO2 = 6.85 + 8.5333 x 10-3 – 2.475 x 10-6T2 cal/gm mole K. T in K. 8. Flue gas leaving the boiler stack at 523 K (250 oC) have following composition on mole basis. CO2 = 11.31%, H2O = 13.04%, O2 = 2.17% and N2 = 73.48% Calculate the heat lost in 1 kmol of gas mixture above 298 K (25 oC) using heat capacity data given below:
Cpo a bT cT2 dT3 Gas CO2 H2O O2 N2
a 21.3655 32.4921 26.0257 29.5909
b x 103 64.2841 0.0796 11.7551 -5.141
c x 106 -41.0506 13.2107 -2.3426 13.1829
d x 109 9.7999 -4.5474 -0.5623 -4.968
9. Air containing 21 mole % O2 and 79 mole % N 2 is to be heated from 303 K (30 oC) to 423 K (150 oC). Calculate the heat required to be added if the air flow rate is 3 m3 (N.T.P) per minute. Data given below
Cpo a bT cT2 dT3 Gas a b x 103 c x 106 d x 109 O2 26.0257 11.7551 -2.3426 -0.5623 N2 29.5909 -5.141 13.1829 -4.968 o 10. Calculate the Enthalpy of Zinc Vapour at 950 C and atmospheric pressure, Relative to the solid at 30oC Data: C p solid = 0.105 cal./gm.oC. Melting Point = 419oC C p liquid = 0.109 cal./gm.oC Boiling Point = 907oC Latent Heat of Vaporization = 26900 cal/gm.atom C p Vapor = 4.97 cal/gm atom oC Latent heat of fusion=1660 Cal/gm.atom Atomic Weight of Zinc = 65.38 UNIT VII THERMOCHEMISTRY PART A
15
1. State Hess law. (May 2013)
Hess‟s law states that, the net heat evolved or absorbed in a chemical reaction is the same whether the reaction takes place in a single step or in a series o steps. This permits us to treat all thermo chemical equations as algebraic equations. 2. How heat of combustion of white petrol determined? (May 2012)
Heat of combustion is determined in a calorimeter in which the fuel is burnt with pure oxygen and the heat liberated is absorbed in water. The rise in temperature of water gives the heat of combustion value of the fuel. 3. Explain heat of solution and heat of mixing. (May 2012)
Heat of solution is the amount of heat evolved or absorbed when one mole of the sample dissolves in a l arge excess of solvent, at constant pressure. Heat of mixing: The heat of mixing is the enthalpy when pure species are mixed at constant pressure and temperature to form one mole of the solution. 4. How is heat of fusion estimated? (Nov 2011)
The Clausius-Clapeyron equation is employed to estimate the heat of fusion. With the data of variation of melting point with pressure following equation can be used to estimate the heat of fusion.
H f T f
Constant.
The value of constant lies between 8.1 and 12.6 for elements, between 21 and 29.4 for inorganic compounds and between 37.8 and 46.2 for organic compounds. 5. Define adiabatic reaction temperature. (Nov 2011)
When no heat is added to the system from the surroundings or no heat is removed from the system during a reaction, the reaction can be termed as adiabatic reaction. When the products get heated up utilizing the heat liberated during the reaction, the temperature attained is termed as the temperature of the reaction. In the case of adiabatic reactions, the temperature is the adiabatic reaction temperature. 6. Define standard heat of reaction. (May 2011).
The standard heat of reaction is the change in enthalpy resulting from the reaction starting and ending with the system at 25oC and 1 atm pressure per formula weight of reactant. 7. Distinguish between exothermic and endothermic reactions. (Nov 2010)
Exothermal reactions are those reactions in which heat is liberated from the system during the reaction, whereas the endothermic reactions absorb heat from the surroundings for its progress. 8. Calculate the enthalpy of sublimation of iodine form the following reactions and data: H2(g) +I2(S) H2(g) +I2(g)
2HI (g) ΔH = 57.9 KJ/mol 2HI (g) ΔH = -9.2 KJ/mol (May 2010)
Given H2(g) +I2(S) 2HI (g) ΔH = 57.9 KJ/mol (1) H2(g) +I2(g) 2HI (g) ΔH = -9.2 KJ/mol (2) Subtracting (2) from (1) : I2 (s) – I2 (g); I2 (s) I2 (g) 57.9 - -(9.2) ; ΔH = 67.1 KJ/mol Enthalpy of sublimation = 67.1 KJ/mol. 16
9. Calculate the standard heat of reaction for the reaction CaC2 + 2H 2O Ca (OH)2 + C2H2. The standard heat of reaction of CaC 2 = -15000, H 2O = -68317.4; Ca (OH) 2 = -235800 and C 2H2 = 54194 cal/mol (May 2010)
Heat of reaction = Sum of heat of formation of products – Sum of heat of reaction of reactants = (54194 -235800) – (-15000 – (2) 68317.4) = - 29971.2 cal/mol
10. Define heat of mixing. (May 2010)
The heat of mixing is the enthalpy change when pure species are mixed at constant pressure and temperature to form one mole of solution. 11. Define standard heat of combustion. (May 2010)
The standard heat of reaction is the enthalpy change accompanying a reaction when both the reactants and products are at their standard states at constant temperature T. When the reaction under consideration is a combustion reaction, the heat of reaction is known as heat of combustion. 12. What is heat of formation? A formation reaction is defined as s reaction, which forms a single compound form the elements contained in it. e.g. C + ½ O 2 +2H2CH3OH The heat of formation is based on 1 mol of the compound formed. The equations should indicate the physical state of each reactant and product whether it is gas, liquid or solid. The heat of formation of a chemical compound is the standard heat of reaction where the reactants are the necessary elements for the formation of compound which is the only product formed. Eg. C(s) + +O2 (g)CO2 Hf =94051 cal/g mol For all the elements, the heat of formation is zero. 13. Define heat of absorption When the solvent and solute form an ideal system, the heat liberated during the absorption system is equal to the latent heat of condensation of the solute. The mixing of non-ideal liquids is accompanied by the evolution of heat and the heat of absorption is equal to the algebraic sum of the heats of condensation and mixing. 14. What is Flame temperature? When a fuel is burnt, lot of heat is generated since the reaction is exothermic. If the combustion chamber is sealed from surroundings, the heat produced is utilized to heat the produce (flue gases). Hence the flue gas comes out at a higher temperature which is called Flame temperature or Theoretical Flame Temperature. 15. State the Hess’s law. Hess‟s law of heat summation states that the net heat evolved or absorbed in the chemical process is the same whether the reaction takes place in one or more steps. (i.e.)H0533 (or) (H0 R +H0298+H0P) Consider the reaction CH4 (g) + 2O2(g)CO2 (g) + 2H2O(g) 16. How will you quantify the heat of reaction from heat of formation data?
The heat of reaction may be calculated as the difference between algebraic sum of the heat of formation or combustion of the reactants and the algebraic sum of the heat of formation or combustion of the products.
H o
o f
Pr oducts
o f Re ac tan ts
17. What is the role of thermo chemistry? Thermo chemistry plays a very significant role in the design and analysis of chemical processes. 18. What is integral heat of solution at infinite dilution?
17
The heat of solution depends on the moles of solvent used per mole of the solute. The higher is the amount of solvent, higher will be the heat of solution. But increase in the heat of solution is not directly proportional to the amount of solvent used. Ultimately the heat of solutions becomes constant and does not change with the further addition of solvent. This limiting value of the heat of solution is known as the integral heat of solution at infinite dilution. 19. What is the effect of pressure on heat of reaction?
Usually the effect of pressure on the heat of reaction is insignificant at low pressures. However a correction is required for those reaction which takes place at very high pressures. 20. List out few important data sources where thermodynamic data are available.
TRC – Thermodynamic Research Centre - USA is one of the oldest data centres in the world. Design institute for physical Property Engineers (DIPPD) sponsored by American institute of chemical Engineers, provides data for over 2000 chemicals. PART B
1. Dry methane and dry air at 298 K and 1 bar pressure are burnt with 100% excess air. The standard heat of reaction is -802 KJ/gm mole of methane. Determine the final temperature attained by the gaseous products if combustion is adiabatic and 20% of heat produced is lost to the surroundings. (May 2012) C pm values j/gm mole K for components are O2 = 31.9, N2 = 32.15, H2O = 40.19 and CO2 = 51.79. 2. The analysis of 15,000 litre of gas mixture at standard condition is as follows: SO2 = 10%, O2 = 12% and N2 = 78%. How much heat must be added to this gas to change its temperature from 30 oC to 425oC? The C pmean values are in cal/gm mole oC.
o
C pm at 30 C C pm at 425oC
SO2 10 11
O2 6.96 7.32
N2 6.80 7.12
(May 2012)
3. A tank contains 10 Kg of a salt solution at a concentration of 2% by weight. Fresh solution enters the tank at a rate of 2 Kg/min at a salt concentration of 3% by weight. The contents are stirred well and the mixture leaves the tank at a rate of 1.5 Kg/min. (i). Express the salt concentration as a function of time and (May 2012) (ii). At what instant of time the salt concentration in the tank will reach 2.5% by weight. 4. A gaseous, mixture at 600 K and 1 bar consisting of methane and steam in the ratio 1:2 by mole is sent to a catalytic reformer where the following reaction occurs, CH4 (g) + H2O CO (g) + 3H2 (g) CO (g) + H2O CO2 (g) + H2 (g)
ΔHo298 = 206.284 KJ ΔHo298 = - 41.20 KJ
Methane is completely converted during the reaction and the product stream contains 17.4% (mol) CO. Heat is supplied in the reactor so that the products reach a temperature of 1300 K. The mean heat capacities in the temperature range of 298 – 600 K are 43.8 KJ/k mol K for methane and 34.8 KJ/Kmol K for steam, and the mean heat capacities in the temperature range of 298 – 1300 K are 31.8 KJ/K mol K for CO, 29.8 KJ/ Kmol K for H2, 50.1 KJ/k mol K for CO2 and 38.9 KJ/k mol K for steam respectively. Calculate the heat requirement for the reactor per (Nov 2011) mole of methane. 5. Estimate the theoretical flame temperature of a gas containing 20% CO and 80% N 2 when burnt with 100% excess air. Both air and gas are initially at 25oC. Assume complete combustion. C p CO2 = (6.339) + (10.14 * 10-3) T – (3.415 * 10-6) T2 C p O2 = (6.117) + (3.167 * 10 -3) T – (1.005 * 10-6) T2 C p N2 = (6.457) + (1.389 * 10 -3) T – (0.069 * 10-6) T2 The values of C p are in (Kcal/Kg mole K) and the temperature is in K. ΔHreaction25oC = (-67,636) K cal. 18
6. Estimate the theoretical flame temperature that can be reached by combustion of methane with 20% excess air. Both air and methane enter the burner at 25oC. Assume complete combustion ΔHo formation 298 K, CO2 = (- 393,509) J/mole ΔHo formation 298 K, H2O = (- 241,818) J/mole ΔHo formation 298 K, CH4 = (- 74,520) J/mole C pideal gas = R[A+(B*T)+(C*T2)+(D*T 3)] R = Gas constant, T in K, Species CO2 H2O O2 N2
A 5.457 3.470 3.639 3.280
B*103 1.045 1.450 0.506 0.593
C*106 0 0 0 0
D*10-5 -1.157 0.121 -0.227 0.040
7. Ethyl alcohol reacts with acetic acid to give ethyl acetate. Heat of combustion of ethyl alcohol, acetic acid and ethyl acetate are -326,700 Cal, -208,340 cal, -553,760 cal respectively. Calculate the standard heat of reaction. (Nov 2010) 8. In the reaction 4FeS (s) +11O 2 (g) 2Fe2O3 (s) + 8 SO2 (g) the conversion from FeS 2 to Fe2O3 is only 80% complete. If the standard heat of reaction is -197.7 Kcal/ gmol what will be the heat liberated per Kg of FeS2 fed? (May 2010) 9. SO2 gas is oxidized in 100% excess air with 80% conversion to SO2. The gases enter the reactor at 410 oC and leave at 460oC. How much heat is to be transferred from the reactor on the basis of 1 mole of entering gas.
C p R
SO2 SO3 O2
a bT
c T 2
; T is in K.
A
b x 103
c x 105
(ΔHof )295, J/mol
5.699 8.06 3.28
0.8 1.06 0.59
-1.01 -2.03 0.04
-296830 -395720 (May 2010)
10. Calculate the heat of the gas phase reaction. C2H2(g) + H2O (g) C2H5OH At 300O C, The standard heat of reaction is -10135 J Heat Capacity Data. C2H2 = 2.830 + 28.601x10-3 T-8.726x10-6T2 H2O = 7.256 + 20.298x10-3 T+0.283x10-6T2 C2H5OH = 6.990 + 39.741x10-3 T-11.926x10-6T2
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