Surface Chemistry Test

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Surface Chemistry Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

Consider the following reactions at 300 K: X $$\rightarrow$$ Y (uncatalysed reaction) & X $$\rightarrow$$ Y (catalysed reaction). The energy of activation is lowered by 83.14 kJmol$$^{-1}$$ for the catalysed reaction.
The rate of catalysed reaction is:

A
$$3\times 10^{14}$$ times that of uncatalysed reaction

B
$$15\times 10^{14}$$ times that of uncatalysed reaction

C
25 times that of uncatalysed reaction

D
22 times that of uncatalysed reaction

Solution

The Arrhenius equation for the uncatalyzed reaction is $$\mathrm{K}_{\mathrm{m}\mathrm{c}\mathrm{a}\mathrm{t}}=\mathrm{A}\mathrm{e}^{-\mathrm{E}_{\iota \mathrm{R}}/\mathrm{R}\mathrm{T}}.$$

The Arrhenius equation for the catalyzed reaction is $$\mathrm{K}_{\mathrm{c}\mathrm{a}\mathrm{t}}=\mathrm{A}\mathrm{e}^{-\mathrm{E}_{ \mathrm{R}}/\mathrm{R}\mathrm{T}}.$$

Let A' be the activation energy for the uncatalyzed reaction.

The activation energy for the catalyzed reaction will be $$(\mathrm{A'}-83.14\times 10^{3}).$$

Divide the two equations.

$$\displaystyle \frac{\mathrm{K}_{\mathrm{c}\mathrm{a}\mathrm{t}}}{\mathrm{K}_{\mathrm{m}\mathrm{c}\mathrm{a}\mathrm{t}}}=\frac{\mathrm{e}^{-(\mathrm{A'}-83.14\times 10^{3})/\mathrm{R}\mathrm{T}}}{\mathrm{e}^{-\mathrm{A'}/\mathrm{R}\mathrm{T}}}=\mathrm{e}^{83.14\times 10^{3}/\mathrm{R}\mathrm{T}}=\mathrm{e}^{33.33} \approx 3 \times 10^{14}$$
Question 2
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A reaction is catalysed by $$H^{+}$$ ion in the presence of $$HA$$, rate constant is $$2\times 10^{-3}min^{-1}$$ and in presence of $$HB$$, rate constant is $$1\times 10^{-3}min^{-1}$$ . $$HA$$ and $$HB$$ (both strong acids) have relative strength as:

A
$$0.5$$

B
$$0.002$$

C
$$0.001$$

D
$$2$$

Solution

The relative strength is the ratio of the rate constants.
$$Relative\ strength = \dfrac {Rate\ constant\ of\ HA} {Rate\ constant \ of \ HB} = \dfrac {2\times 10^{-3}min^{-1}} {1\times 10^{-3}min^{-1}}=2.$$
Question 3
1 / -0

Gelatin protects

A
Gold sol

B
$$\mathrm{A}\mathrm{s}_{2}\mathrm{S}_{3}$$ sol

C
$$\mathrm{F}\mathrm{e}(\mathrm{O}\mathrm{H})_{3}$$ sol

D
All of these

Solution

Hint: Lyophilic sols protect other lyophilic sols.
Correct Answer: Option D
Explanation:
Lyophilic sols are produced by mixing substances directly in an effective, suitable dispersing medium. The dispersed phase particles in these sols have a strong affinity for the dispersion medium. These sols are very stable but also reversible in nature.
According to the given options, all the three mixtures - gold sol, $$As_2S_3$$ and $$Fe(OH)_3$$ sols are lyophilic in nature. Gelatin is also a lyophilic sol. So, gelatin protects all of these sols mentioned in the question.
Question 4
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Assertion: A negative catalyst retards the rate of chemical reaction.
Reason: Presence of negative catalyst boosted up the energy of activation.

A
Assertion is correct but Reason is wrong

B
Assertion is wrong but Reason is correct

C
Both Assertion and explanation are correct and Reason is correct explanation of Assertion

D
Both Assertion and explanation are correct but Reason is not correct explanation of Assertion

Solution

Negative catalysts (Inhibitors): A catalyst which decreases or retards the rate of reaction is called negative catalysts.
A negative catalysts retards rate of chemical reaction.
Working of negative catalyst is not related energy of activation.
Hence, statement is correct but explanation is wrong.
Please note that enzyme inhibitors acts as negative catalysts. They bind to the allosetric site of the enzyme and decrease the rate of reaction.
Question 5
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Consider the following statements and arrange in the order of true/false as given in the codes.
$$S_1: $$ The rate of the reaction A $$\rightarrow$$ B having the rate law $$- \displaystyle \frac{d[A]}{dt} = k [A] [B]$$ when plotted against time will exhibit a maximum at some time.
$$S_2: $$ A catalyst in a chemical reaction increases the forward E$$_a$$ and decreases the backward $$E_a$$
$$S_3: $$ A catalyst in a chemical reaction decreases both forward and backward $$E_a$$
$$S_4: $$ For a first-order reaction, the time required to reduce successively the concentration of reactant by a constant fraction is always same.

A
TTTT

B
FFFF

C
FTFT

D
TFTT

Solution

The rate of the reaction A $$\rightarrow$$ B having the rate law $$-\cfrac{d[A]}{dt}=k[A][B]$$ when plotted against time will exhibit a maximum at some time.
This is because, rate = 0 at t = 0 and t = t$$_\inf$$ . Hence statement 1 is true.

A catalyst in a chemical reaction decreases the forward Ea and decreases the backward Ea.
Note: It decreases both by an equal amount. Hence, statement 2 is false.

A catalyst in a chemical reaction either increases or decreases both forward and backward Ea by an equal amount. Hence statement 3 is true.

For a first-order reaction, the time required to reduce successively the concentration of reactant by a constant fraction is always the same. This is because the half-life period is constant for the first-order reactions. Hence, statement (4) is correct.
Question 6
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Select the incorrect statement regarding the given reaction.

$$RCOCl+H_2\xrightarrow{Pd-BaSO_4} RCHO+HCl$$

A
$$Pd$$ acts as a positive catalyst.

B
It is Rosenmund reaction.

C
$$BaSO_4$$ acts as poison for $$Pd$$.

D
$$BaSO_4$$ acts as promoter for $$Pd$$.

Solution

The Rosenmund reaction is catalyzed by palladium on barium sulphate. Barium sulphate reduces the activity of palladium due to its low surface area meaning it decreases the reducing power of palladium in order to prevent over-reduction of the acid.

So, $$BaSO_4$$ acts as a poison for Pd, not as a promoter for Pd

So, D is here correct answer
Question 7
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Which of the following cannot be used to prepare an emulsion of benzene in water?

A
$$CH_3-(CH_2)_9-(SO_3)^-Na^+$$

B
$$CH_3-(CH_2)_9-COO^-Na^+$$

C
$$CH_3-(CH_2)_9-NH_3^+Cl^-$$

D
$$(CH_3)_4 \overset{+}{N}Br^-$$

Solution

Surface active agents are used as emulsifiers. The surface-active agents must have a hydrophobic chain with carbon atoms $$\geq$$ 5. So, D option is correct.
Question 8
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Smoke screens consist of:

A
fine particles of $$TiO_2$$ dispersed in air by aeroplanes

B
fine particles of $$AgI$$ dispersed in air by aeroplanes

C
fine particles of $$Al_2O_3$$ dispersed in air by aeroplanes

D
none of the above

Solution

Smoke screen is a cloud of smoke. It consists of fine particles of $$TiO_2$$ which are dispersed in air by aeroplanes.
Question 9
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Directions For Questions

The catalytic activity and collodial nature of a substance are surface phenomenon. Both these properties depend upon the property of adsorption. Adsorption may be physisorption or chemisorption. Adsorption is spontaneous and always leads to decrease in entropy and evolution of heat. Chemisorption is irreversible with temperature, unilayer, specific and directional. Adsorbate molecules adsorbs on cataylst surface and thus, lower energy of activation of reaction to provide a new path way for reaction. In colloidal state, dispersed phase particles possess the adsorption characteristics at the interface.

...view full instructions

Which statement is incorrect?

A
Catalytic decomposition of $$KClO_3$$ by $$MnO_2$$ is an example of homogeneous catalysis.

B
Only the surface atoms in an adsorbent play an active role in adsorption.

C
The rate of adsorption is very rapid in the beginning but decreases gradually until equilibrium is attained.

D
Absorption proceeds at a steady rate.

Solution

Catalytic decomposition of $$KClO_3$$ by $$MnO_2$$ is an example of heterogeneous catalysis.
$$2KClO_3(s) \xrightarrow {MnO_2 (s)} 2KCl(s)+3O_2(g)$$
It contains two phase solid and gas.
Homogeneous catalysis involves only one phase whereas heterogeneous catalysis involves more than one phase.
Question 10
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Directions For Questions

$$1$$ g of activated charcoal has a surface area $${10}^{3}\ {m}^{2}$$. One molecule of $$N{H}_{3}$$ having diameter $$0.3$$ nm is supposed to be adsorbed on complete surface of charcoal forming unilayer adsorption. The activated charcoal is brought in contact with $$100$$ mL of $$2\ M$$ $$N{H}_{3}$$ solution.

...view full instructions

How many $$N{H}_{3}$$ molecules are adsorbed on charcoal surface?

A
$$1.4\times {10}^{22}$$

B
$$1.4\times {10}^{23}$$

C
$$1.4\times {10}^{21}$$

D
$$1.4\times {10}^{20}$$

Solution

Surface area occupied by one $$N{H}_{3}$$ molecule $$=\pi{r}^{2}$$.

$$=3.14\times {(\displaystyle\frac{0.3\times {10}^{-9}}{2})}^{2} =7.065\times {10}^{-20}\ {m}^{2}$$

Total surface area $$= {10}^{3}\ {m}^{2}$$.

Number of $$N{H}_{3}$$ molecules adsorbed $$= \displaystyle\frac{{10}^{3}}{7.065\times {10}^{-20}}$$

$$= 1.415\times {10}^{22}$$ molecules.

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Surface Chemistry Test - 51

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Surface Chemistry Test - 51

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Question 1

$$3\times 10^{14}$$ times that of uncatalysed reaction

$$15\times 10^{14}$$ times that of uncatalysed reaction

25 times that of uncatalysed reaction

22 times that of uncatalysed reaction

Solution

Question 2

$$0.5$$

$$0.002$$

$$0.001$$

$$2$$

Solution

Question 3

Gold sol

$$\mathrm{A}\mathrm{s}_{2}\mathrm{S}_{3}$$ sol

$$\mathrm{F}\mathrm{e}(\mathrm{O}\mathrm{H})_{3}$$ sol

All of these

Solution

Question 4

Assertion is correct but Reason is wrong

Assertion is wrong but Reason is correct

Both Assertion and explanation are correct and Reason is correct explanation of Assertion

Both Assertion and explanation are correct but Reason is not correct explanation of Assertion

Solution

Question 5

TTTT

FFFF

FTFT

TFTT

Solution

Question 6

$$Pd$$ acts as a positive catalyst.

It is Rosenmund reaction.

$$BaSO_4$$ acts as poison for $$Pd$$.

$$BaSO_4$$ acts as promoter for $$Pd$$.

Solution

Question 7

$$CH_3-(CH_2)_9-(SO_3)^-Na^+$$

$$CH_3-(CH_2)_9-COO^-Na^+$$

$$CH_3-(CH_2)_9-NH_3^+Cl^-$$

$$(CH_3)_4 \overset{+}{N}Br^-$$

Solution

Question 8

fine particles of $$TiO_2$$ dispersed in air by aeroplanes

fine particles of $$AgI$$ dispersed in air by aeroplanes

fine particles of $$Al_2O_3$$ dispersed in air by aeroplanes

none of the above

Solution

Question 9

Catalytic decomposition of $$KClO_3$$ by $$MnO_2$$ is an example of homogeneous catalysis.

Only the surface atoms in an adsorbent play an active role in adsorption.

The rate of adsorption is very rapid in the beginning but decreases gradually until equilibrium is attained.

Absorption proceeds at a steady rate.

Solution

Question 10

$$1.4\times {10}^{22}$$

$$1.4\times {10}^{23}$$

$$1.4\times {10}^{21}$$

$$1.4\times {10}^{20}$$

Solution

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