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Multiple choice question for engineering

Set 1

1. Which type of graph plot is used to analyze the effect of temperature on the rates of chemical reactions?
a) Dot plot
b) Normal probability plot
c) Arrhenius plot
d) Line weaver-Burk plot

Answer: c [Reason:] An Arrhenius plot displays the logarithm of kinetic constants (ln (k), ordinate axis) plotted against inverse temperature (1/T, abscissa). The Arrhenius equation can be written as:

ln (k) =ln (A) – (Ea/R) (1/T) Where : k= Rate constant A= Pre-exponential factor Ea= Activation energy R= Gas constant T= Absolute temperature, K In the right, which represents the y-coordinate axis.

2. From the below representation of growth phase, which is the missing growth phase?

a) Exponential Growth Phase
b) Final Lag Phase
c) Initial Log Phase
d) Final Log phase

Answer: a [Reason:] The Log/ Exponential Growth phase is a period characterized by cell doubling. If growth is not limited, doubling will continue at a constant rate so both the number of cells and the rate of population increase doubles with each consecutive time period. For this type of exponential growth, plotting the natural logarithm of cell number against time produces a straight line.

3. If the substrate is given in molar concentration how would you define it’s Molarity and Molality based on the temperature condition?
a) Molarity and Molality both will change on effect of temperature
b) Molarity will change but not Molality
c) Molality will change but not Molarity
d) Normality of the reaction will change

Answer: b [Reason:] Molarity is the number of moles of solute per kilogram of the Solvent. That is because molarity does not depend on temperature, (neither on number of moles of solute nor mass of solvent will be affected by changes of temperature). While molarity changes as temperature changes because molarity is affected by temperature. This is because it is based on the volume of the solution, and the volume of a substance will be affected by changes in temperature.

4. What is the generation time of a bacterial population that increases from 10,000 cells to 10,000,000 cells in four hours of growth?

a) 24 minutes
b) 30 minutes
c) 34 minutes
d) 60 minutes

Answer: a [Reason:] The generation time is the time interval required for the cells (or population) to divide. G (generation time) = (time, in minutes or hours)/n(number of generations) G = t/n t = time interval in hours or minutes B = number of bacteria at the beginning of a time interval b = number of bacteria at the end of the time interval n = number of generations (number of times the cell population doubles during the time interval) b = B x 2n (This equation is an expression of growth by binary fission) Solve for n: Logb = logB + nlog2 n = logb – logB log2 n = logb – logB .301 n = 3.3 logb/B G = t/n

5. On which type of plot the best fit line is represented?
a) Dot plot
b) Line weaver- Burk plot
c) Scatter plot
d) Bland- Altman plot

Answer: c [Reason:] A line of best fit is a straight line drawn through the center of a group of data points plotted on a scatter plot. Scatter plots depict the results of gathering data on two variables. The line of best fit shows whether these two variables appear to be correlated and can be used to help identify trends occurring within the dataset. Analysts may use the line of best fit when determining a relationship between two variables where one variable is independent and one variable is being examined for dependency.

6. In the case of Rota meter, which is used for measuring the water flow rate?
a) Only one variable is involved, i.e. independent variable ( the Rota meter reading)
b) Only two variables are involved: one independent variable( the Rota meter reading) and one dependent variable (the water flow rate)
c) Only two variables are involved: one independent variable(the water flow rate) and one dependent variable (the Rota meter reading)
d) Only one variable is involved, i.e. independent variable ( the water flow rate)

Answer: b [Reason:] Rota meter is a device which measure flow rate. This is done by allowing the cross-sectional area the fluid travels through to vary, causing a measurable effect, i.e. water flow rate is dependent on Rota meter.

7. The diauxic growth curve mainly found in E.coli in case of two types of nutrients is found in Batch culture is based on which equation?
a) Monod equation
b) Michaelis – Menten equation
c) Schrodinger equation
d) Lyapunov equation

Answer: a [Reason:] Jacques Monod discovered diauxic growth in 1941 during his experiments with Escherichia coli and Bacillus subtilis. While growing these bacteria on various combinations of sugars during his doctoral thesis research, Monod observed that often two distinct growth phases are clearly visible in batch culture.

8. Essentially in an experiment involving bacteria, we measure Optical density of the media where the bacteria grow at different time point from the same flask. If we have 2 or more types of bacteria and want to investigate whether the growth curve of those bacteria are significantly different, what do you think the best analysis can be applied for this purpose?
a) Linear regression analysis or general regression analysis
b) Non- linear regression analysis or general regression analysis
c) Logistic regression analysis
d) Stepwise regression analysis

Answer: b [Reason:] If the growth rates fit a polynomial curve, you can use General Regression and enter the bacteria type as categorical variable. Otherwise, you may need to use nonlinear regression separately for each type.

9. What is the basic Optical Density (O.D) of the bacterial cells?
a) 680nm
b) 600nm
c) 200nm
d) 280nm

Answer: b [Reason:] 600nm is a wavelength of light that is minimally absorbed by material within cells. Therefore, they can be more reliably used than other wavelengths (in most common situations). To simply, measure the scattering of light due to turbidity (just general presence of particles in a solution).

10. For what MANOVA is used for?
a) Detection of biological samples
b) Detection of the best fit curve
c) Comparison between monovariate samples
d) Comparison between multivariate samples

Answer: d [Reason:] In statistics, multivariate analysis of variance (MANOVA) is a procedure for comparing multivariate sample means. As a multivariate procedure, it is used when there are two or more dependent variables, and is typically followed by significance tests involving individual dependent variables separately.

Set 2

1. “Heat transferred to raise or lower the temperature of a material is called Specific heat”.
a) True
b) False

Answer: b [Reason:] Heat transferred to raise or lower the temperature of a material is called sensible heat; change in the enthalpy of a system due to variation in temperature is called sensible heat change, whereas the term specific heat refers to heat capacity expressed on a per-unit-mass basis.

2. The symbol “Û” refers to?
a) Amount
b) Rate of enthalpy
c) Molar flow rate
d) Specific internal energy

Answer: d [Reason:] The symbol “Û” refers to Specific internal energy in a close system in Enthalpy change in non-reactive processes whereas in an open- system symbol “Ĥ” refers to specific enthalpy.

3. Acetone (denoted as Ac) is partially condensed out of a gas stream containing 66.9 mole % acetone vapor and the balance nitrogen. Process specifications and material balance calculations lead to the flowchart shown below.
The process operates at steady state. Calculate the required cooling rate.
Reference states for acetone and nitrogen are-
N2 (g, 25°C, 1 atm), Ac (l, 20°C, 5 atm)

a) – 2390kW
b) – 2320kW
c) – 3560kW
d) – 3570kW

Answer: b [Reason:] Basis: 100 mole/s of mixed gas (Ac & N2) Write and simplify the energy balance. Note the following points about the table – (i) Nitrogen has only one state for inlet and outlet. (ii) Acetone has one inlet state but two outlet state (iii) Since the liquid acetone leaving the system is at the reference state, so its specific enthalpy is zero. Calculate all unknown specific enthalpies –

4. Calculate how much energy (kJ) is required to heat a bottle containing 300 mL of liquid water from room temperature (20 °C) to its normal boiling point (100 °C), while still remaining a liquid.
a) 104.4 kJ
b) 100.4 kJ
c) 140.4 kJ
d) 104.5 kJ

Answer: b [Reason:] Enthalpy (“Water”, 20, 000,”l”)

5. Calculate ΔH for a process in which 2.0 mole of NaOH is dissolved in 400 mol H2O at 25C.
(ΔH ̂s at r = 200, 25° C is – 42.26 kJ/ mol)
a) –84.52 kJ
b) -80.42 kJ
c) –64.52 kJ
d) –60. 42 kJ

6. What term is used for Temperature measured by Thermometer?
a) Wet- bulb temperature
b) Dry- bulb temperature
c) Dew point temperature
d) Normal temperature

Answer: b [Reason:] The dry-bulb temperature (DBT) is the temperature of air measured by a thermometer freely exposed to the air but shielded from radiation and moisture. DBT is the temperature that is usually thought of as air temperature, and it is the true thermodynamic temperature.

7. Estimate the following properties of humid air at 41°C and 10% relative humidity:
Absolute humidity = 0.0048 kg H2O/ kg DA
Wet-bulb temperature = 19°C
Humid volume = 0.895 M3/ kg DA
Dew point = 3°C
Specific enthalpy = 54.2 – 0.7 = 53.5 kJ/ kg DA
What is the amount of water in 150 m3 of air at these conditions?
a) 0.605 kg H2O
b) 0.705 kg H2O
c) 0.805 kg H2O
d) 0.905 kg H2O

8. What is the enthalpy of 130 g formic acid at 70°C and 1 atm relative to 25°C and 1 atm?
Cp for formic acid in the temperature range of interest is 0.524 cal g-1 °C-1.
a) 4.68 kcal
b) 3.06 kcal
c) 2.06 kcal
d) 2.68 kcal

Answer: b [Reason:] ΔH = ( 130 g) (0.524 cal g-1 °C-1) (70-25)°C

ΔH = 3065.4 cal or ΔH = 3.06 kcal

Relative to H=0 at 25°C the enthalpy of formic acid at 70°C is 3.06 kcal.

9. Processes for phase change of vapour to liquid is called?
a) Vaporization
b) Fusion
c) Sublimation
d) Condensation

Answer: d [Reason:] Condensation is the change of the physical state of matter from gas phase into liquid phase, and is the reverse of evaporation.

10. 50 g benzaldehyde vapour is condensed at 179°C, What is the enthalpy of the liquid relative to the vapour?
Given: The molecular weight of benzaldehyde is 106.12, the normal boiling point is 179.0°C and the standard heat of vaporization is 38.40 kJ gmol-1. For condensation the latent heat is – 38.40 kJ gmol-1.
a) -18.09 kJ
b) -17.09 kJ
c) – 18.06 kJ
d) – 17.06 kJ

Answer: a [Reason:] ΔH = 50 g (- 38.40 kJ gmol-1). |(1 gmol)/(106.12 g)| = -18.09 kJ Therefore, the enthalpy of 50 g benzaldehyde liquid relative to the vapour at 179°C is – 18.09 kJ. As heat is released during condensation, the enthalpy of the liquid is lower than the vapour.

Set 3

1. Aerobic reactions are not batch operations.
a) True
b) False

Answer: a [Reason:] Aerobic reactions are not batch operations in the strictest sense; the low solubility of oxygen in aqueous media means that it must be supplied continuously while carbon dioxide and other off-gases are removed.

2. In a perfectly mixed reactor _________
a) The output composition is different from input composition
b) The output composition is identical from input composition
c) Both output and input composition are constant
d) Both output and input composition are not constant

Answer: b [Reason:] In a perfectly mixed reactor, the output composition is identical to composition of the material inside the reactor, which is a function of residence time and rate of reaction.

3. Convert vmax = 2.5 mmol m-3 s-1 into mM h-1.
a) 2500
b) 900
c) 25
d) 9

4. Convert vmax = 7 mmol m-3 s-1 into mM h-1.
a) 25
b) 20
c) 25.20
d) 20.25

5. In batch reaction time with enzyme deactivation, calculate the first-order deactivation rate constant.
(Given – so = 12 mM; vmax = 9 mM h-1; Km = 8.9 mM; sf = 1.2 mM; th = 4.4 h).
a) 0.150 h-1
b) 0.158 h-1
c) 0.155 h-1
d) 0.154 h-1

6. By taking the parameters of Q5, and the batch reaction time tb.
a) 5.0 h
b) 10.0 h
c) 15.0 h
d) 20.0 h

7. Zymomonas mobilis is used to convert glucose to ethanol in a batch fermenter under anaerobic conditions. The yield of biomass from substrate is 0.06 g g- 1; YPX is 7.7 g g- 1. The maintenance coefficient is 2.2 g g-1 h-1; the specific rate of product formation due to maintenance is 1.1 h-1. The maximum specific growth rate of Z. mobilis is approximately 0.3 h-1.5 g bacteria are inoculated into 50 litres of medium containing 12 g l-1 glucose. Determine batch culture times required to produce 10 g biomass.
a) 3.1 h
b) 3.3 h
c) 3.5 h
d) 3.7 h

Answer: d [Reason:] Yxs = 0.06 gg-l; Ypx= 7. 7 gg-l; μmax = 0. 3 h-l; ms=2.2 gg-l h-l; mp= 1.1 h-l; x0=5g/501= 0.1 gl-l;s0= 12 gl-l.

If 10 g biomass are produced by reaction, the final amount of biomass present is (10 + 5) g = 15 g. Therefore xf= 15 g/ 50 l = 0.3 g l-1.

8. By taking the pararmeters of Q7 and Determine batch culture times required to achieve 90% substrate conversion.
a) 5.1 h
b) 5.3 h
c) 5.5 h
d) 5.7 h

9. By taking the parameters question 8 and question 7, determine batch culture times required to produce 100 g ethanol.
a) 3.2 h
b) 3.4 h
c) 3.6 h
d) 3.8 h

10. The Zymomonas mobilis cells are used for chemostat culture in a 60 m3 fermenter. The feed contains 12 g l-1 glucose; Ks for the organism is 0.2 g l-1. What flow rate is required for a steady-state substrate concentration of 1.5 g l-1?
a) 15.6 m3 h-1
b) 15.8 m3 h-1
c) 15.4 m3 h-1
d) 15.2 m3 h-1

Answer: a [Reason:] Yxs = 0.06 g g-1; Ypx = 7.7 g g-1max = 0. 3 h-1; Ks = 0.2 g-1; ms = 2.2 g g-1 h-l; si = 12 g 1-1; V=60 m3. qp = 3.4 h-1, YPS = Ypx Yxs = 0.46 g g-1.

s = 1.5 g l-1

11. By taking Q10 and at the flow rate of Q10, what is the cell density?
a) 0.42 g l-1
b) 0.44 g l-1
c) 0.46 g l-1
d) 0.48 g l-1

Answer: a [Reason:] When synthesis of product is coupled with energy metabolism as for ethanol, x is evaluated. Therefore:

12. By taking Q10 and Q11 into account and at the flow rate of Q10, what concentration of ethanol is produced?
a) 5.1 g l-1
b) 5.3 g l-1
c) 5.5 g l-1
d) 5.7 g l-1

Answer: c [Reason:] Assuming ethanol is not present in the feed, Pi = 0. Steady-state product concentration is given by:

13. Which of the following type is of the perfusion culture?
a) Batch
b) Conc. Batch
c) Continuous
d) Conc. Fed-Batch

Answer: d [Reason:] Concentrated Fed-Batch- In a process that can be considered a form of perfusion, the culture is perfused to generate ultrahigh cell concentration, greater than 108 cells/mL; and the product is also retained in the vessel.

14. Immobilised lactase is used to hydrolyse lactose in dairy waste to glucose and galactose. Enzyme is immobilised in resin particles and packed into a 0.5 m3 column. The total effectiveness factor for the system is close to unity; Km for the immobilised enzyme is 1.32 kg m-3; Vmax is 45 kg m-3 h-1. The lactose concentration in the feed stream is 9.5 kg m-3; a substrate conversion of 98% is required. The column is operated with plug flow for a total of 310 d per year. At what flow rate should the reactor be operated?
a) 1.56 m3 hh-1
b) 1.58 m3 hh-1
c) 1.50 m3 h-1
d) 1.54 m3 h-1

Answer: a [Reason:] For 98% substrate conversion, sf = (0.02 si) = 0.19 kg m-3

15.By taking the parameters of Q14 into consideration, estimate how many tonnes of glucose are produced per year?
a) 56.3 tonnes yr-1
b) 56.6 tonnes yr-1
c) 56.7 tonnes yr-1
d) 56.5 tonnes yr-1

Answer: c [Reason:] The rate of lactose conversion is equal to the difference between inlet and outlet mass flow rates of lactose:

F(si – sf)= 1.56 m3 h-1 (9.5 – 0.19) kg m-3= 14.5 kg h-1

Converting this to an annual rate based on 310 d per year and a molecular weight for lactose of 342:

The enzyme reaction is: lactose + H20 → glucose + galactose. Therefore, from reaction stoichiometry, 315 kg mol glucose are produced per year. The molecular weight of glucose is 180; therefore:

Set 4

1. Which is the correct order for the equation of mass balance of the system?

{A} – {B} + {C} – {D} = {Mass accumulated within system}

a) A. Mass consumed within system, B. Mass in through system boundaries, C. Mass out through system boundaries, D. Mass generated within system
b) A. Mass generated within system, B. Mass in through system boundaries, C. Mass out through system boundaries, D. Mass consumed within system
c) A. Mass in through system boundaries, B. Mass out through system boundaries, C. Mass generated within system, D. Mass consumed within system
d) A. Mass consumed within system, B. Mass generated within system, C. Mass out through system boundaries, D. Mass in through system boundaries

Answer: c [Reason:] A mass balance for the system can be written in a general way to account for these possibilities: { Mass in through system boundaries} – { Mass out through system boundaries} + { Mass generated within system} – { Mass consumed within system} = { Mass accumulated within system}.

2. If the Reactant iron is combined with Reactant Sulfur to form the product Iron sulfide, then what will be the atomic mass of the product?
a) 88
b) 44
c) 22
d) 80

Answer: a [Reason:] Fe ( iron) + S ( Sulfur) = FeS ( Iron sulfide), Atomic masses: Fe = 56, S = 32 One atom of each element on each side of the equations Law of conservation of mass balance: 56 + 32 = 88.

3. If the reactant Magnesium reacts with Reactant Hydrochloric acid to form the product Magnesium chloride + Hydrogen, then what will be the atomic mass of both the product?
a) 90
b) 97
c) 80
d) 87

Answer: b [Reason:] Mg ( Magnesium) + 2 HCl ( Hydrochloric acid) = MgCl2 ( Magnesium chloride) + H2 ( Hydrogen), Atomic masses: Mg = 24, H= 1, Cl= 35.5 one atom of Mg, 2 atoms of H and 2 atoms of Cl on both sides of the equation, Law of conservation of mass balance: 24 + 2 × ( 1+ 35.5) = 24+ (2 × 35.5) + (2×1) = 97 ( Both equal 97) Note the subscript 2 after the Cl in magnesium chloride or the 2 after the H in the hydrogen molecule, means two atoms of that element. The 2 before the HCl doubles the number of hydrochloric acid molecules.

4. A continuous process is set up for treatment of wastewater. Each day, 103 kg cellulose and 105 kg bacteria enter in the feed stream, while 102 kg cellulose and 1.5 x 102 kg bacteria leave in the effluent. The rate of cellulose digestion by the bacteria is 8 x 102 kg d-1. The rate of bacterial growth is 4x 102 kg d-l; the rate of cell death by lysis is 6 x 102 kg d -1. Write balances for cellulose and bacteria in the system.
a) 1× 103 kg, 9× 103 kg
b) 1× 102 kg, 9.965 × 104 kg
c) 1× 103 kg, 9.964 × 104 kg
d) 1× 102 kg, 9× 103 kg

Answer: b [Reason:] Cellulose is not generated by the process, only consumed. Using a basis of 1 day, the cellulose balance in kg is : (103 – 102 + 0 – 8 x 102) = accumulation

Therefore, 1× 102 kg cellulose accumulates in the system each day.

Performing the same balance for bacteria:

(105 – 1.5 x 102 + 4x 102 – 6 x 102) = accumulation

Therefore, 9.965 × 104 kg bacterial cells accumulate in the system each day.

5. Which of the following system follows the differential equation?
a) Semi- Batch process
b) Batch process
c) Fed- Batch process
d) Continuous process

Answer: d [Reason:] Amounts of mass entering and leaving the system are specified using flow rates, A mass balance based on rates is called a differential balance. Whereas, each term of the mass-balance equation in this case is a quantity of mass, not a rate. This type of balance is called an integral balance. Differential balances for continuous systems operating at steady state, and integral balances for batch and semi-batch systems between initial and final states are used.

6. 2SO2+O2→ 2SO3. What is the stoichiometric ratio of SO2 to SO3?
a) 3
b) 1
c) 2
d) 0

Answer: b [Reason:] stoichiometric ratio of SO2 to SO3 = (2 mole of SO23 produced) = 1.

7. Two methanol-water mixtures are contained in separate tanks. The first mixture contains 40.0 wt% methanol and the second contains 70.0 wt% methanol. If 200 kg of the first mixture is combined with 150 kg of the second, what are the mass and composition of the product (Total balance, Methanol balance, Physical constraint) ?
a) 350 kg, 0.529, 0.471
b) 320 kg, 0.529, 0.521
c) 360 kg, 0.620, 0.471
d) 350 kg, 0.690, 0.760

Total Balance: m1 + m2 = m3 Methanol-Balance: m1xM1 + m2xM2 = m3xM3 Water-Balance: m1xw1 + m2xw2 = m3xw3

(choose only 2 equations since one of them is no longer independent)

Physical Constraint (applied to mixture 3):

xM3 + xW3 = 1.00

Always start with the equation with the least number of unknowns if possible and minimize solving equations simultaneously.

Total Balance (m3) ↓ Methanol Balance (xM3) ↓ Physical Constraint (xW3) Total balance: m3 = (200 kg) + (150 kg) m3 = 350 kg

CH3OH balance: (200 kg)(0.40) + (150 kg)(0.70) = (350 g)xM3 xM3 = 0.529

Physical constraint: xW3 = 1.00 – xM3 = 1 – 0.529 xW3 = 0.471.

8. A solution of common salt in water is prepared by adding 20 kg of salt to 100 kg of water, to make a liquid of density 1323 kg/m3. Calculate the concentration of salt in this solution as a (a) weight fraction, (b) weight/volume fraction, (c) mole fraction, (d) molal concentration.
a) 15.7%, 23.1%, 0.058, 3.88 moles in m3
b) 14.5%, 22.4%, 0.059, 2.88 moles in m3
c) 16.7%, 22.1%, 0.058, 3.77 moles in m3
d) 14.7%, 23.1%, 0.059, 2.77 moles in m3

Answer: c [Reason:] (i) Weight fraction:

20/(100+20) = 0.167; %weight/weight = 16.7%

(ii) Weight/volume:

A density of 1323kg/m3 means that lm3 of solution weighs 1323kg, but 1323kg of salt solution contains

( 20×1323kg of salt) / (100+20) = 220.5 kg salt/m3

1 m3 solution contains 220.5 kg salt. Weight/volume fraction = 220.5 / 1000 = 0.2205 And so weight / volume = 22.1%

(iii) Moles of water = 100 / 18 = 5.56 Moles of salt = 20 / 58.5 = 0.34 Mole fraction of salt = 0.34 / (5.56 + 0.34) = 0.058

(iv) The molar concentration (M) is 220.5/58.5 = 3.77 moles in m3.

9. Calculate the degree of reduction for ethanol, where the degree of reduction of C=4, H=1,
O= -2.
a) ϒ = 9
b) ϒ = 6
c) ϒ = 3
d) ϒ = 2

Explanation: Ethanol (C2H5OH) = 2(4) + 6(1) + 1(-2) = 12, ϒ = 12/2 = 6

As, The number of equivalents of available electrons per gram atom C is the measure of degree of reduction.

10. Propane (C3H8) burns in this reaction:

C3H8 + 5O2 = 4H2O + 3CO2

If 200 g of propane is burned, how many g of H2O is produced?
a) 327.27 g
b) 345.5 g
c) 323.2 g
d) 232.3 g

Since you cannot calculate from grams of reactant to grams of products you must convert from grams of C3H8 to moles of C3H8 then from moles of C3H8 to moles of H2O. Then convert from moles of H2O to grams of H2O.

Step 1: 200 g C3H8 is equal to 4.54 mol C3H8. Step 2: Since there is a ratio of 4:1 H2O to C3H8, for every 4.54 mol C3H8 there are 18.18 mol H2O. Step 3: Convert 18.18 mol H2O to g H2O. 18.18 mol H2O is equal to 327.27 g H2O.

Set 5

1. If air consists of 77% by weight of nitrogen and 23% by weight of oxygen calculate:
(i) the mean molecular weight of air,
(ii) the mole fraction of oxygen,
(iii) the concentration of oxygen in mole/m3 and kg/m3 if the total pressure is 1.5 atmospheres and the temperature is 25 degreeC.
a) 29.8, 0.21, 0.010 mole/m3
b) 28.8, 0.21, 0.013 mole/m3
c) 27.6, 0.30, 0.010 mole/m3
d) 29.8, 0.30, 0.013 mole/m3

Answer: b [Reason:] (i) Taking the basis of 100 kg of air: it contains 77/28 moles of N2 and 23/32 moles of O2,

Total number of moles = 2.75 + 0.72 = 3.47 moles. So mean molecular weight of air = 100 / 3.47 = 28.8 Mean molecular weight of air = 28.8 (ii) The mole fraction of oxygen = 0.72 / (2.75 + 0.72) = 0.72 / 3.47 = 0.21 Mole fraction of oxygen = 0.21 (iii) In the gas equation, where n is the number of moles present: the value of R is 0.08206 m3 atm/mole K and at a temperature of 25 ⁰C = 25 + 273 = 298 K, and where V= 1 m3 pV = nRT and so, 1.5 x 1 = n x 0.08206 x 298 n = 0.061 mole/m3 weight of air = n x mean molecular weight = 0.061 x 28.8 = 1.76 kg / m3 and of this 23% is oxygen, so weight of oxygen = 0.23 x 1.76 = 0.4 kg in 1 m3 Concentration of oxygen = 0.4kg/m3 or 0.4 / 32 = 0.013 mole / m3

When a gas is dissolved in a liquid, the mole fraction of the gas in the liquid can be determined by first calculating the number of moles of gas using the gas laws, treating the volume as the volume of the liquid, and then calculating the number of moles of liquid directly.

2. In the carbonation of a soft drink, the total quantity of carbon dioxide required is the equivalent of 3 volumes of gas to one volume of water at 0⁰C and atmospheric pressure. Calculate (i) the mass fraction and (ii) the mole fraction of the CO2 in the drink, ignoring all components other than CO2 and water.
Basis 1 m3 of water = 1000 kg
Volume of carbon dioxide added = 3 m3
From the gas equation, pV = nRT
1 x 3 = n x 0.08206 x 273 n = 0.134 mole.
Molecular weight of carbon dioxide = 44
And so weight of carbon dioxide added = 0.134 x 44 = 5.9 kg
a) 6.9 × 10-3, 3.41 × 10-3
b) 8.9 × 10-3, 2.41 × 10-3
c) 5.9 × 10-3, 2.41 × 10-3
d) 9.9 × 10-3, 3.41 × 10-3

Answer: c [Reason:] (i) Mass fraction of carbon dioxide in drink = 5.9 / (1000 + 5.9) = 5.9 × 10-3 (ii) Mole fraction of carbon dioxide in drink = 0.134 / (1000/18 + 0.134) = 2.41 × 10-3.

3. Each year 80,000 people move into a city, 70,000 people move out, 18,000 are born, and 11,000 die. Write a balance on the population of the city.
a) 17,000 P/yr
b) 15,000 P/yr
c) 20,000 P/yr
d) 22,000 P/yr

Answer: a [Reason:] Let P denotes people, Input+ Generation- Output- Consumption = accumulation 80,000 P/yr + 18,000 P/yr – 70,000 P/yr – 11,000 P/yr = A(P/yr) A = 17,000 P/yr Each year the city’s population increases by 17,000 P/yr.

4. According to the Balances on steady-state processes, the accumulation is equal to?
a) 1
b) 0
c) 100
d) 2

Answer: b [Reason:] The process is said to be operating at steady-state when all process variables do not change with time. The accumulation term in a balance must equal to zero to ensure that the amount/mass of material in the process do not change with time. Steady state means accumulation = 0 Input + generation – output – consumption = 0 Input + generation = output + consumption.

5. A process unit involves 3 chemical components. How many mass balances can be written?
a) 3
b) 5
c) 6
d) 4

Answer: d [Reason:] We can write 4 balance equations: A total balance equation and 3 component balance equations. Independent balances: Not all balances are independent since the total balance is the sum of all of the component balances. Thus, the number of independent balances we can write = the number of components.

6. The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor.

C2H4 + HBr = C2H5Br
The product stream is analyzed and found to contain 51.7 mole% C2H5Br and 17.3% HBr. The feed to the reactor contains only ethylene and hydrogen bromide.
Calculate the fractional conversion of the limiting reactant and the percentage by which the other reactant is in excess. If the molar flow rate of the feed stream is 165 mol/s, what is the extent of reaction?
a) 56.2 mol/s
b) 45.6 mol/s
c) 55.6 mol/s
d) 44.6 mol/s

Answer: a [Reason:] Determine the limiting reactant:

7. What is the fractional and percent excess of H2 from the following equation:
H2 + Br2 = 2HBr
H2= 25 mol/hr
Br2= 20 mol/hr

a) 0.45, 45%
b) 0.25, 25%
c) 0.75, 75%
d) 0.65, 65%

Answer: b [Reason:] Fractional excess H2 = (25-20)/ 20 = 0.25 Percent excess = 25%.

8. When pure carbon is burned in air, some of it oxidizes into CO2 and some to CO. The molar ratio of N2 to O2 is 7.18 and the ratio of CO to CO2 is 2.0 in the product gas. What is the percent excess air used? The exit gases contain only N2, O2, CO and CO2.
a) 50 %
b) 60 %
c) 40 %
d) 20 %

Answer: c [Reason:] The process involves two reactions: R1 = C + O2 = CO2 R2 = C + ½ O2 = CO As the product (output) gas composition is specified more clearly, we may use it as starting point. Take N2 in the product as nN2 = Take N2 in the product as, then O2 = 1 mol N2 as an inert gas: input = output = 7.18 mol O2 input with air = 7.18 (21/79) = 1.91 mol Total O2 consumption = 1.91-1 = 0.91 mol If R1 uses n1 mol O2, generating n1 mol CO2, R2 uses 0.91-n1 mol O2, generating 2 (0.91-n1) mol CO. Since CO/CO2 in the product gas = 2, 2 (0.91-n1)/n1 = 2 n1 = 0.91/2 = 0.455 mol Therefore, total moles of C input = 3 n1 = 1.365 mol O2 required for complete combustion of 1.365 mol C = 1.365 mol (for R1 only) O2 excess % = (1.91 -1365)/1.365 x 100%= 39.9% =40% .

9. Convert 800 mmHg into bars.
a) 1.064 bars
b) 1.066 bars
c) 1.054 bars
d) 1.055 bars