industrial chemistry mcq ans

8/20/2019 Industrial Chemistry MCQ Ans

1/16

Multiple Choice Answers

1 C

2 B Petroleum is a natural raw material.

3 C

4 D Bulk chemicals can be made from other raw materials like air and water.

Hydrogen is a bulk chemical. It can be obtained from electrolysis of water,

steam-methane reforming process, etc.

5 A Wool and cotton are natural materials. Polyethene is seldom used to make

clothes, instead, it is usually used to make bags, wrapping film for food, boxes,

buckets, etc.

6 D Bulk chemicals are chemicals that are sold thousands of tonnes a year and are

used for the production of synthetic products. Water and petroleum are raw materials.

Plastic is a synthetic product.

7 C Nitrogenous fertilizers can be produced by the reaction between ammonia and

sulphuric acid. NH3(aq) + H2SO4(aq) (NH4)2SO4(aq)

8 D Disposable foam food box is made of expanded polystyrene. Low-density

polyethene is used to make wrapping film for food. PVC instead of polypropene is

used to produce shower curtain.

9 A Bulk chemicals are simple and basic chemicals which can be used for production

of synthetic products.

10 D

11 C

12 A

13 C In fact, the majority of petroleum fractions can be used. The light fractions are

used in the automobile industry. The heavy fractions are used to make a lot of

different synthetic products.

14 C Ethene can be obtained by cracking of petroleum. Ethanol is the product obtained

by catalytic hydration of ethene.

15 D

16 B Chlorine is a useful industrial chemical. However, it is not a petrochemical.

Petrochemical industry produces any chemicals from petroleum or natural gas.

17 D

18 C Apart from Reichstein process, two-stage fermentation process can be used for

ascorbic acid production. Glucose is used as feedstock for both processes. Reichstein

process is a two-stage process which combines fermentation and chemical synthesis.

19 C

20 B Night blindness is mainly caused by the deficiency of vitamin A.

21 D The chemical name for vitamin C is ascorbic acid. The intermediate compoundsformed in Reichstein process and two-stage fermentation process are called DAKS


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and KGA respectively. The overall production cost of vitamin C by two-stage

fermentation process is lower compared with Reichstein process. The overall yields

of vitamin C produced from both processes are about 60%.

22 B

23 B

24 C Calcium carbonate is an example of bulk chemical. Calcium carbonate can be

obtained from natural raw materials like rocks and seashells.

25 B

26 C Some plastics are made from other substances such as biodegradable substances

(e.g. starch).

27 B

28 D Rate equations can also be written respectively for the forward and backward

reactions of a reversible reaction. Besides, rate equation shows the quantitative

relationship between concentration and reaction rate, but not rate constant.

29 D The order of reaction with respect to a reactant can only be found through

experiments.

30 A Let the rate equation for the reaction be rate = k [A] x[B]

y. Since [B] is in largely

excess, [B] y is a constant. The rate equation can then be modified to rate = k’[A]

x,

where k’ = k [B] y. When [A] was doubled, the rate of the reaction doubled. Therefore,

the reaction is first order with respect to A. In other words, the rate is directly

proportional to [A].

31A The half-life of A is

k

693.0 only if the reaction is a first order reaction.

32B k =

yr 2010

6930.= 3.45 × 10

4 yr

1

33 D

34 D The value of rate constant is equal to the reaction rate only if the reaction is in

zeroth order.

35 A

36 B

37C k =

1hr 0.169

693.0

= 4.1 hr

38 C

39C 30 min =

k

693.0, k = 0.0231 min

1, that iss60

0231.0 = 3.85 × 10

4 s1.

40 C Rate constant of the reaction depends on the temperature, but not the

concentration of any reactants.

41 D


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42A Number of half-lives that 214Bi has passed after 59.1 min =

min719

min159

.

.= 3. Mass of

214Bi after 59.1 min = 0.001 × (2

1)3 g = 1.25 × 10

4 g.

43

B Number of half-lives that55

Cr has passed after 10.8 hr = hr 81

hr 810

.

.= 6. Let the mass

of 55Cr material before shipping is M .

1.0 = M × (2

1)6

∴ M = 64

44 A

45 A

46 A Let the rate of reaction when [C] = 0.60 mol dm3 be r .

r

13 sdmmol020.0 =

2

3

3

2

3

3

)dmmol60.0(

)dmmol0.1(

r = 0.0093

47 B By putting the numbers into the rate equation, rate = k [X]2[Y]

R1 = 0.729k mol dm3

s1

; R2 = 2.92k mol dm3

s1

; R3 = 2.92k mol dm3

s1

; R4 =

5.83k mol dm3 s1

48 C

By substituting the given information into the rate equation, we have

12.0 = k [2.25] 21

[2.00] z (1)

54.0 = k [4.00] 21

[3.00] z (2)

Dividing (2) by (1),

0.12

0.54 = (

25.2

00.4) 2

1

× (00.2

00.3) z

z = 3

49 A If [A] is largely greater than [B], the rate of reaction is independent of [A].∴

[A] x is a constant. The rate equation for the reaction becomes rate = k 2[B]

y, where k 2

= k [A] x = constant. If y = 0, then rate = k

2. A horizontal straight line is obtained when

the graph of rate against [B] is plotted. Therefore, (2) is correct. (3) is correct only if x

= y.


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50 C

51 C A graph of rate against [B]2 is a straight line passing through the origin.

52 B There is no direct relationship between the order of reaction with respect to a

reactant in the rate equation and the stoichiometric coefficient of that reactant in the

chemical equation. Besides, the reaction may involve other reactants that the orders

of reaction with respect to them are zero. Option C is wrong. Reaction rate should be

quadrupled when [A] was doubled while keeping [B] constant. Option D is also

wrong. The overall effect on reaction rate = (3)2 × (

2

1)3 =

8

9

53 D This is a third order reaction. A plot of [Y] against time should be a curve. The

reaction rate increased by eight times when both [X] and [Y] were doubled. The unit

of rate constant =33

13

)dm(mol

sdmmol

= mol2 dm

6 s

1

54 C

Let the rate equation be rate = k [A] x[B]

y[C]

z, where x, y and z are the orders of

reaction with respect to A, B and C respectively.

By substituting the given information into the rate equation, we have

0.48 × 105

= k (0.10) x(0.05)

y(0.04)

z (1)

1.92 × 105 = k (0.20) x(0.05)

y(0.08)

z (2)

1.44 × 105

= k (0.10) x(0.15)

y(0.04)

z (3)

0.24 × 10

5 = k (0.20) x(0.10) y(0.02) z (4)


5

5

1048.0

1044.1

=

z y x

z y x

k

k

)04.0()05.0()10.0(

)04.0()15.0()10.0(

3 = (3) y

∴ y = 1


5

5

1024.0

1092.1

=

z y x

z y x

k

k

)02.0()10.0()20.0(

)08.0()05.0()20.0(

8 = (0.5) y(4)

z

Since y = 1, so 16 = (4) z

∴ z = 2


5

5

1048.0

1092.1

=

z y x

z y x

k

k

)04.0()05.0()10.0(

)08.0()05.0()20.0(

4 = (2) x(2)

z


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Since z = 2, so, 1 = (2) x

∴ x = 0

55 C From Experiments 1 and 2, under the same initial concentration of Y(g), initial

rate quadrupled when the initial concentration of X(g) was doubled. Therefore, the

reaction is second order with respect to X(g).

From Experiments 1 and 3, under the same initial concentration of X(g), initial rate

doubled when the initial concentration of Y(g) was doubled. Therefore, the reaction is

first order with respect to Y(g).

The rate equation for the reaction is rate = k [X(g)]2[Y(g)].

Use the information given from Experiment 1,

k =)dmmol00.2()dmmol400.0(

sdmmol1020.6 323

133

= 0.0194 mol2 dm6 s1

Initial rate of reaction in Experiment 4

= 0.0194 × (0.500)2 × (1.50) mol dm

3 s1

= 7.28 × 103 mol dm3 s1

56D Overall effect on rate of reaction =

2

1× (2)

2 = 2

57 D The order of reaction with respect to a particular reactant can be any realnumbers. However, the order of reaction with respect to a particular reactant can only

be found through experiments.

58 B

59 B

60 A For (2), the energy of the products is less than that of the reactants. For (3), the

activation energy is independent of the type of reaction.

61 D The activation energy is the difference in energy between the reactants and the

transition state. Therefore, the activation energy for the reaction, B A = 70 20 kJmol

1 = 50 kJ mol

1

62 A For an exothermic reaction, the energy of the products is less than the energy of

the reactants. Besides, there should be an energy barrier to overcome before the

reaction takes place.

63 B Activation energy is the minimum energy required for a chemical reaction to

occur.

64 C The species that exists at the point of maximum energy is called the activated

complex, which contains partially broken and partially formed bonds.65 A Apart from overcoming the energy barrier, the reactant particles must be

correctly oriented to one another. The forward reaction and the backward reaction of


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a reversible reaction generally have different activation energies. Activation energy is

the minimum energy required to break bonds for a reaction to occur.

66 C For multi-step reactions, there are more than one transition state.

67 B For C, a change in temperature affects the equilibrium position of the reaction but

not the activation energy. For D, since the E a of the backward reaction is greater than

that of the forward reaction, the rate constant of backward reaction is smaller than

that of the forward reaction.

68 B For A, x-axis is the reaction coordinate, y-axis is the energy. For C, the activated

complex contains partially broken and partially formed bonds. For D,Δ H is the

difference in energy between the reactants and the products.

69 B For an endothermic reaction, the enthalpy change, Δ H is positive. Therefore, the

energy of the products is greater than that of the reactants.

70 D

71 D

72 A

73 D For A, the area under the curve remains the same because the number of particles

in the system does not change at any temperatures. D is incorrect because there is no

maximum energy for a particle.

74 A

Referring to the diagram,

the activation energy for the backward reaction= 57.2 kJ mol

1 (+54.0 kJ mol

1) = 3.2 kJ mol

1

75 B (1) is incorrect because the number of particles in the system does not change

when temperature rises. (3) is incorrect because the maximum point of the curve is

shifted to the right and gets lower when temperature rises. The number of particles

with the most probable kinetic energy decreases.

76 D The most probable kinetic energy is on the left of the average kinetic energy. The

curve is asymmetric. The area under the curve represents the total number of particlesin the system.

77 C The change in temperature does not affect activation energy. The area under the

2NO (g)2

E n e r g y

E aΔ H

N O (g2 4

Reaction coordinate


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Maxwell-Boltzmann distribution curve is a constant as the number of particles is the

same. An increase in reaction rate with a rise in temperature is mainly due to the

increased number of particles possessing energy equal to or greater than the

activation energy.

78 B

79 B The change in temperature does not affect the activation energy.

80 A The area under the curve remains the same because the number of particles in the

system does not change when the temperature rises.

81 C The change in temperature does not affect the activation energy of a reaction.

82 B The average kinetic energy is on the right of the maximum point.

83 B

By substituting values of k 1, k 2, T 1 and T 2 into the Arrhenius equation and then

solving for E

a,

k 1 = 0.5 × 1010 T 1 = 30 + 273 K = 303 K

k 2 = 22 × 1010 T 2 = 100 + 273 K = 373 K

)11

(3.2

loglog12

a21

T T R

E k k

)303

1

373

1(

314.83.21022log105.0log a1010

E

Solving for E a = 50.7 kJ mol1

84 A

By applying RT E

Aek a

0.5 × 1010 = )303(314.8

80050

Ae

Solving for A = 0.0286

85 D


solving for E

a,

k 1 = 5.49 × 106 T 1 = 5000 K

k 2 = 9.86 × 108 T 2 = 10 000 K

)11

(3.2

loglog12

a21

T T R

E k k

)5000

1

10000

1(

314.83.21086.9log1049.5log a86

E



8/16

86 D


solving for E

a,

k 1 = k T 1 = 25 + 273 K= 298 K

k 2 = 2

k

T 2 = 35 + 273 K = 308 K

)11

(3.2

loglog12

a21

T T R

E k k

)298

1

308

1(

314.83.22loglog a

E k k


87 C At a low temperature, the reaction rate decreases because the number of particles

with energy equal to or greater than E a decreases. In addition, for a reaction with high

E a, the number of particles with energy equal to or greater than E a decreases.

88 C

By substituting values of k 1, k 2, T 1 and T 2 into the Arrhenius equation.

k 1 = k 37C T 1 = 37 + 273 K= 310 K

k 2 = k 15C T 2 = 15 + 273 K = 288 K

)310

1

288

1(

3.2loglog aC15C37

R

E k k

)310

1

288

1(314.83.2

00087log

C15

C37

k

k

C15

C37

k

k 13.2

89 D


solving for E

a,

k 1 = 1.3 × 1011 T 1 = 270 + 273 K= 543 K

k 2 = 4.5 × 1010 T 2 = 350 + 273 K = 623 K

)11

(3.2

loglog12

a21

T T R

E k k

)543

1

623

1(

314.83.2105.4log103.1log a1011

E


90 D (1) is incorrect because the activation energy of a reaction remains unchanged at

any temperatures. However, in the presence of a catalyst, the reaction will proceed in

an alternative pathway with lower activation energy.


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91D Taking logarithm of the Arrhenius equation gives log k = log A

R

E

3.2a

By plotting a graph of log k againstT

1, its slope is equal to

R

E

3.2a

and the

y-intercept is equal to log A.

92 D By substituting values of k 1, k 2, T 1 and T 2 into the Arrhenius equation and

then solving for E a,

k 1 = 0.096 T 1 = 350 + 273 K = 623 K

k 2 = 0.400 T 2 = 400 + 273 K = 673 K

)11

(3.2

loglog12

a21

T T R

E k k

)623

1

673

1(

314.83.2400.0log096.0log a

E


93 D


solving for E

a,

k 1 = k T 1 = 80 + 273 K = 353 K

k 2 = 5k T 2 = 120 + 273 K = 393 K

)11

(3.2loglog 12

a21

T T R

E k k

)353

1

393

1(

314.83.25loglog a

E k k


94 D


solving for E

a,

k 1 = k T 1 = 500 K

k 2 = 4k T 2 = 520 K

)11

(3.2

loglog12

a21

T T R

E k k

)500

1

520

1(

314.83.24loglog a

E k k


95 D



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solving for E a,

k 1 = 0.0503 T 1 = 25 + 273 K = 298 K

k 2 = 6.71 T 2 = 60 + 273 K = 333 K

)11

(3.2

loglog

12

a21

T T R

E k k

)298

1

333

1(

314.83.271.6log0503.0log a

E


96 D

97 A

98 A

99 B

100 C Catalysts do not affect the yield of products because they do not affect the

equilibrium position. Catalysts can be classified into positive catalysts and negative

catalysts. Positive catalysts speed up reactions while negative catalysts slow down

reactions.

101 B For (1), the catalyst usually reacts with one of the reactants to form an

intermediate. The intermediate then reacts with another reactant to give the product

and the catalyst is regenerated at the same time. For (2), as the catalyst is regenerated

at the end of the reaction, its amount is not changed. For (3), there is a limit where the

amount of catalyst used increases, the reaction rate will not increase.

102 B Referring to the energy profile of the endothermic reaction, the activated complex

has the highest energy while the reactants have the lowest energy.

103 D I and II represent the activation energy and the enthalpy change of the reaction in

the absence of a catalyst respectively. III is equal to I + II. The catalyst does not affect

the activation energy and the enthalpy change of the original reaction. In fact, it provides an alternative reaction pathway with lower activation energy.

activated complex

E n e r g y

products E a

Δ H = +vereactants

Reaction coordinate


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104 C Catalysts do not lower the E a of the original reaction. They allow reaction to

proceed in a different way with lower E a.

105 C

106 D A catalyst increases the rate of reaction by providing an alternative pathway with

lower activation energy for the reaction to proceed. It does not change the activation

energy and the yield of the reaction.

107 B The catalysts do not affect the yield of a reaction because they do not affect the

equilibrium position of a reaction. The catalysts can be in solid, liquid or gaseous

phase.

108 B Positive catalyst increases the rate of reaction without changing the equilibrium

position. The yield of product is not changed. Therefore, A and D are incorrect. C

shows the effect of negative catalyst on the reaction.

109 A The positive catalyst can speed up the reaction while the negative catalyst can

slow down the reaction. Since the catalyst increases the rates of both forward and

backward reactions to the same extent, it does not affect the equilibrium position and

then the yield of products.

110 D The catalyst increases the rates of both forward and backward reactions to the

same extent in an equilibrium. It does not affect the equilibrium position, so the

equilibrium concentrations of the products remain unchanged.

111 C The catalyst increase the rates of both forward and backward reactions to the

same extent in an equilibrium. It does not affect the equilibrium position. Besides,

they have no effect on the enthalpy change of the reaction.

112 D The catalyst does not affect the activation energy of the original reaction. Instead,

it provides an alternative reaction pathway with lower activation energy.

113 D They change chemically during a reaction. During a reaction, a catalyst forms an

intermediate with one of the reactants. Then the intermediate reacts with another

reactant to give the product and the catalyst is regenerated. In fact, the catalyst

remains chemically unchanged at the end of the reaction.

114 C (1) is incorrect because the homogeneous catalysts are in the same phase with the

reactants and the products.115 A Platinum acts as a heterogeneous catalyst as it is in solid phase while the

reactants and the products are in gaseous phase. The catalyst provides a surface for

the adsorption of the reactants, not absorption.

116 D The heated catalyst is an aluminium oxide which is a heterogeneous catalyst

because it is in solid phase while the reactants and the products are in gaseous phase.

117 D Fe2+(aq) is a homogeneous catalyst. It is oxidized by S 2O82(aq) to Fe3+(aq).

118 D

119 C The reaction between peroxodisulphate ions and iodide ions can be catalysed byeither iron(II) ions or iron(III) ions.


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120 D X is a catalyst because it is regenerated at the end of the reaction. Z is an

intermediate because it is formed in the first step and then is consumed in the next

step.

121 C

122 A

If Cu+(aq) is used as the catalyst, the reaction is:

Cu+(aq) + Fe3+(aq) Cu2+(aq) + Fe2+(aq)

Cu2+(aq) + V3+(aq) Cu+(aq) + V4+(aq)

If Cu2+(aq) is used as the catalyst, the reaction is:

Cu2+(aq) + V3+(aq) Cu+(aq) + V4+(aq)

Cu+(aq) + Fe3+(aq) Cu2+(aq) + Fe2+(aq)

123 A

124 A During the hydrogenation of ethene in the presence of nickel, the C=C bonds

within the ethene molecules are weakened only after adsorbing on the surface of

nickel. Besides, CH bonds are formed between a hydrogen atom and one of the

carbon atoms of ethene.

125 C

126 C A is not a cracking. B is an addition polymerization.

127 A Finely divided iron is used as a heterogeneous catalyst.

128 C Nickel is used as a catalyst in the hydrogenation of unsaturated oils to make

margarine.

129 C The catalyst used is finely divided iron. The kinetic energy of molecules is not

affected by a catalyst. A catalyst has no effect on the equilibrium position.

130 A Rhodium or platinum catalyses the reaction between carbon monoxide and

nitrogen monoxide to form carbon dioxide and nitrogen.

131 D

132 B

133 A

134 D Finely divided iron is used as a heterogeneous catalyst in the Haber process. It is

in solid phase while the reactants and the product are in gaseous phases.

135 A

136 C In fact, the catalysts change chemically during a reaction.

137 C The hydrogen and alkene molecules are adsorbed on the surface of the catalyst.

138 A A high pressure can shift the equilibrium position to the side with less number of

molecules and increase the yield. (1) is correct. A high pressure means same number

of particles in a smaller volume. It can increase the chance of effective collisions. (2)

is correct. Liquefying the ammonia in the condenser facilitates storage of ammonia.139 B Molar mass of CON2H4 = 12.0 + 16.0 + 14.0 × 2 + 1.0 × 4 g mol

1 = 60.0 g


13/16

mol1

∴ mass of urea used = 25.0 ×20.14

kg = 53.6 kg0.60

140 a ofA Haber process is exothermic, so a lower temperature favours higher yield

product. B and C are incorrect. For the reaction, N2(g) + 3H2(g) 2NH3(g),

increasing pressure will shift the equilibrium position to the side with less moles of

gaseous materials (i.e. NH3). Therefore, A is correct.

141

mmonia which is useful in fertilizers. Meanwhile, the cost of

C The significance of the Haber process is the conversion of nitrogen (sufficient

supply from air) into a

nitrogen is very low.

142

ly affects the precision of the data but it will not lead to

A Aqueous barium nitrate should be added in excess so that all the sulphate ions in

the fertilizer would be precipitated. It is almost an immediate reaction. Hence, B and

D are incorrect. Option C on

large error in measurement.

143 n, both reactants and

ation for the reaction is:

D As the steam-methane reforming is a reversible reactio

products exist in equilibrium. The equ

CH4(g) + H2O(l) 3H2(g) + CO(g)

144 lts that contain one or more of the elementsC Fertilizers are usually inorganic sa

nitrogen, phosphorus and potassium.

145 C A mixture of hydrogen and carbon monoxide is called synthesis gas or syngas.

146 monia are 200 atm, 450C and withD The optimum conditions for producing am

the use of finely divided iron as the catalyst.

147 of

e reaction proceeds slowly. So a

B According to Le Châtelier’s Principle, a low temperature favours a high yield

NH3 at equilibrium. But at low temperatures, th

moderately high temperature (500C) is used.

148 ber, the NH3 is liquefied and both N2 and H2 remain,

ich are being recycled.

A From the catalytic cham

wh

149

1 mole of CH4 produces

D

Let the volume of NH3 synthesized be V L.

38 moles of NH3.

1 L of CH4 produces3

8 L of NH3.

7V = 5.6 × 10 ×

3 × 75% = 1.1 × 10

8 8

150

lem, but this is not a

A ‘Volatile’ means the ammonia cannot be handled easily. Ammonia is so alkaline

that it will change the soil pH. Unpleasant odour may be a prob

consideration of using ammonium nitrate instead of ammonia.151 NO3 plant. ThenC NH3 is generated from the reaction between N2 and H2 in the H

HNO3 produced further reacts with aqueous NH3 to form NH4 NO3.


14/16

152 uitable forA The product of the nitrogen fixation is ammonia NH3, which is not s

use as fertilizer. It must go through further treatment to produce nitrate.

153 metal and thenA In a flowing mercury cell, sodium ions are reduced to sodium

sodium metal dissolves to form sodium amalgam at the cathode.

154 age

ed by exhaust gas

B It has a potential hazard of mercury leakage. Algal bloom is caused by sew

with excess amount of nutrient. Acid rain and smog are caus

(includes particulates and some acidic gases, such as SO2).

155 llected at different outletsC Hydrogen and oxygen (a by-product) produced are co

and they do not react explosively under room conditions.

156 ase in

would not decrease.

A Since the amount of Na+(aq) ions does not change and there is no incre

volume of solution, the concentration of Na+(aq) ions

157 B Chlorine is poisonous to most micro-organisms.

158 tually aims at allowing the anode and cathode reactions to proceed

he same time.

C The design ac

at t

159 A

160 diaphragm cell, titanium is used as the anode and steel is used as the

hode.

B In the

cat

161

H2(g) + Cl2(g)

B

2NaCl(aq) + 2H2O(l) 2NaOH(aq) +

No. of moles of HCl manufactured =0.15.35

36.5%0000100

mol = 10 000 mol

mass of NaCl needed = 10 000 × (23.0 + 35.5) g = 585 000 g = 585 kg

Since 1 mole NaCl produces 1 mole of HCl,

∴

162 C

163 D

164 thanol is a colourless liquid with molecular formula CH3OH. It is too toxic toD Me

drink!

165 D Ethene is used to produce polyethylene.

166 rming. ZnO/Cr 2O3 or Cu/ZnO/Al2O3 is used

he conversion of syngas to methanol.

D NiO is used in steam-methane refo

in t

167 A

168 D

169 (3) affect the ease of production and retail while (2) affects the running ofD (1) and

the plant.

170

le nearby since any leakage from the

D The oil field nearby can supply methane (main component of natural gas) to the

site. As the production of methanol needs steam for steam-methane reforming to give

synthesis gas, the lake nearby is a convenient supply of water. There should not be aresidential area with several thousands of peop


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plant may pose a serious threat to the people.

171 er = (12.0 × 6 + 1.0 × 12 + 16.0 × 6) × 5 B Relative molecular mass of the polym

(16.0 + 1.0 × 2) × 4 = 828

172 B ZnO cannot be used as a catalyst in the Haber process.

173 impurities because this will poison the catalyst and

reaction might be quenched.

D The set-up has to be free from

the

174 B

175 D Methanol is toxic.

176

t involve the use

t catalyst. Instead, E.coil is used to catalyse the reaction.##

##A (1) is correct because glucose which is obtained from corn is a renewable raw

material. (3) is incorrect because biosynthesis of adipic acid does no

of cobal

177 ##D##

178 ##A (3) is incorrect because natural gas is a non-renewable raw material.##

179 ##B (2) is incorrect because solar energy is a renewable energy source.##

180

me than that of the desired product.

##A (3) is incorrect. It is possible to achieve 100% yield but the reaction may

generate waste that is far greater in mass and volu

Thus, the atom economy is lower than 100%.##

181 ch as

ioxide and methane to the environment during decomposition.##

##C Some of the biodegradable plastics may release greenhouse gases su

carbon d

182 ##D##

183

tic steps. Otherwise, more reagents are needed and more waste

##D We should reduce the production of the derivatives of the desired product by

minimizing the synthe

will be generated.##

184 d be minimized in order to reduce

nts used and generate less wastes.##

##B The production of by-products shoul

reage

185

0 g

O6 reacts to form 2 moles of C 2H5OH.

atom econom tion

##C

Mass of atoms in C6H12O6 = (12.0 × 6 + 1.0 × 12 + 16.0 × 6) g = 18

Mass of atoms in C2H5OH = (12.0 × 2 + 1.0 × 6 + 16.0) g = 46.0 g

From the equation, 1 mole of C6H12

∴ y of the reac

= %100180

20.46

= 51.1%

##

186

.0 g

rin = (12.0 × 9 + 1.0 × 8 + 16.0 × 4) g = 180 g

atom economy

##B

Mass of atoms in C7H6O3 = (12.0 × 7 + 1.0 × 6 + 16.0 × 3) g = 138 g

Mass of atoms in CH3COOH = (12.0 × 2 + 1.0 × 4 + 16.0 × 2) g = 60

Mass of atoms in aspi

∴


16/16

= %100180

= 90.90.60138

%

##

187 ##B (2) is the fact only.##

188 ##C They are completely degradable and can be converted into harmless

substances.##

189 ##D The synthesis involves the use of cobalt catalysts which cannot be recovered

completely, so they would be disposed to the environment. The synthesis produces

.##dinitrogen oxide or nitrous oxide (N2O) gas which is a greenhouse gas

190 ##D The aim of applying the heat exchanger is to minimize the use of energy, so it

n chemistry practice.##is a gree

191 ##B The correct one is using less hazardous chemical syntheses.##

192 ##D##

193 ##D##

194 ##C##

195 ##D Butane, naphtha and methanol are the starting materials for manufactu

ethanoic acid in butane oxidation, naphtha oxidation and M

ring

onsanto process or

CATIVA process respectively.##

196 ##D Carbonlyation is the reaction that introduces a carbonyl group into a

compound. It is the way to manufacture ethanoic acid.##

197 ##B Ethene is used as the feedstock. Ethanal is produced during the reaction.##

198 sed

onsanto process.##

##C Iridium metal used in the CATIVA process is cheaper than rhodium metal u

in the M

199 ##C Two-stage fermentation process is used to produce vitamin C. Ester is

produced in esterification. Reduction of ethanal gives ethanol instead of ethanoic

acid.##

200 ##D (2) is correct because methanol obtained from biomass, wastes and sewages

community is a renewable raw material. (3) indicates that the atom economyfrom the

is 100%.##

201 ##B##

202 ##D##

203 ##C Even though the yield of the reaction is 100%, the reaction may generate waste

ethat is far greater in mass and volume than that of the desired product. Therefore, th

atom economy is lower than 100%.##

204 ##B##

205 ##C Monsanto process uses rhodium metal and hydrogen iodide as the catalyst.

Iridium metal is five times cheaper than rhodium metal.##

industrial chemistry mcq ans

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