General Principles and Processes of Isolation of Elements Test

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General Principles and Processes of Isolation of Elements Test - 54

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Question 1
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Directions For Questions

Questions given below are based on the given diagram for extraction metallurgy.
The points noted by arrows are the melting and boiling points of the metals zinc and magnesium. $$\Delta G^{\small\circ}$$ is a function of temperature for some reactions and extractive metallurgy.

...view full instructions

At $$1000^oC$$, $$\Delta G^{\small\circ}$$ of the reaction is:
$$ZnO + C \longrightarrow Zn + CO$$

A
$$-ve$$

B
$$+ve$$

C
zero

D
None of these

Solution

As they both oxides met at $$1000^oC$$ in graph so at this temperature $$\Delta G$$ of the reaction will be zero.
Question 2
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Directions For Questions

Questions given below are based on the given diagram for extraction metallurgy.
The points noted by arrows are the melting and boiling points of the metals zinc and magnesium. $$\Delta G^{\small\circ}$$ is a function of temperature for some reactions and extractive metallurgy.

...view full instructions

To make the following reduction process spontaneous, temperature should be
$$ZnO + C \longrightarrow ZN + CO$$

A
$$<1000^{\small\circ}C$$

B
$$>1100^{\small\circ}C$$

C
$$<500^{\small\circ}C$$

D
$$>500^{\small\circ}C$$ but $$<1000^{\small\circ}C$$

Solution

above $$ 1000^oC$$ ZnO in graph is above than CO so above this temp., difference between free energy of both will be negative and reaction will be spontaneous.
So option B is correct.
Question 3
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Directions For Questions

The Ellingham diagram for a number of metallic sulphides is shown below :

...view full instructions

Which sulphide occurs to minimum extent in nature?

A
$$HgS$$

B
$$H_2S$$

C
$$Bi_2S_3$$

D
$$CS_2$$

Solution

As at natural temperature $$\triangle G^{\small\circ}$$ of $$CS_2$$ is least negative so it is most unstable and has a minimum extent in nature.

Option D is correct.
Question 4
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Directions For Questions

The Ellingham diagram for a number of metallic sulphides is shown below :

...view full instructions

Formation of which of the sulphides is most spontaneous?

A
$$HgS$$

B
$$Bi_2S_3$$

C
$$PbS$$

D
$$CS_2$$

Solution

As at all temperature $$\triangle G^{\small\circ}$$ of $$PbS$$ is highest negative so it is the most spontaneous.
Question 5
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Directions For Questions

For a spontaneous reaction, the free energy change must be negative.

$$\Delta G=\Delta H- T $$ $$\Delta S$$

$$\Delta$$$$H$$ is the enthalpy change during the reaction. $$T$$ is the absolute temperature, and $$\Delta$$$$S$$ is the change in entropy during the reaction. Consider a reaction such as the formation of an oxide.

$$M+O_{2}$$ $$\rightarrow$$ $$MO$$

Dioxygen is used up in the course of this reaction. Gases have a more random structure (less ordered) than liquid or solids. Consequently, gases have a higher entropy than liquids and solids. In this reaction, $$S$$ (entropy or randomness) decreases, hence $$\Delta$$S is negative. Thus, if the temperature is raised then $$T\Delta$$S becomes more negative. Since $$T\Delta$$S is subtracted in the equation, then $$\Delta$$G becomes negative.

Thus, the free energy change increases with the increase in temperature. The free energy changes that occur when one mole of common reactant (in this case dioxygen) is used may be plotted graphically against temperature for a number of reactions of metals to their oxides. The following plot is called an Ellingham diagram for metal oxide. Understanding of Ellingham diagram is extremely important for the efficient extraction of metals

...view full instructions

As per the Ellingham diagram of oxides which of the following conclusion is true?

A
Al reduces $$Fe_{2}O_{3}$$, whereas $$MgO$$ cannot be reduced by Al at $$1500^{o}C$$

B
Fe reduces $$Al_{2}O_{3}$$, whereas $$MgO$$ cannot be reduced by Al at $$1500^{o}C$$

C
Al reduces $$Fe_{2}O_{3}$$, whereas $$MgO$$ cannot be reduced by Ca at $$1500^{o}C$$

D
Al can reduce both $$Fe_{2}O_{3}$$ and $$MgO$$ to the corresponding metal at $$1500^{0}C$$
Question 6
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When $$FeCr_2O_4$$ (chromite) is reduced with Carbon in an electric-arc furnace:

A
Cr and $$Fe_2O_3$$ are formed

B
Fe and $$Cr_2O_3$$ are formed

C
Fe and Cr (ferrochrome) are formed

D
$$FeCrO_4$$ is formed

Solution

When chromite $$(FeCr_2O_4)$$ is reduced with carbon in an electric-arc furnace, ferrochrome (Fe and Cr) is obtained.
Question 7
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Directions For Questions

The Ellingham diagram for a number of metallic sulphides is shown below :

...view full instructions

Which of the following sulphides can not be reduced to metal by $$H_2$$ at about $$1000^{\small\circ}C$$?

A
$$HgS$$

B
$$PbS$$

C
$$Bi_2S_3$$

D
All of these

Solution

Hydrogen will reduce sulphide of other metals which lie above it in the Ellingham diagram because free energy change will become more negative and more the difference & more energy will be released and reaction will be more spontaneous.

So it can reduce others but can not reduce $$PbS$$ as it lies below in the graph.
Question 8
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Directions For Questions

For a spontaneous reaction, the free energy change must be negative.

$$\Delta G=\Delta H- T $$ $$\Delta S$$

$$\Delta$$$$H$$ is the enthalpy change during the reaction. $$T$$ is the absolute temperature, and $$\Delta$$$$S$$ is the change in entropy during the reaction. Consider a reaction such as the formation of an oxide.

$$M+O_{2}$$ $$\rightarrow$$ $$MO$$

Dioxygen is used up in the course of this reaction. Gases have a more random structure (less ordered) than liquid or solids. Consequently, gases have a higher entropy than liquids and solids. In this reaction, $$S$$ (entropy or randomness) decreases, hence $$\Delta$$S is negative. Thus, if the temperature is raised then $$T\Delta$$S becomes more negative. Since $$T\Delta$$S is subtracted in the equation, then $$\Delta$$G becomes negative.

Thus, the free energy change increases with the increase in temperature. The free energy changes that occur when one mole of common reactant (in this case dioxygen) is used may be plotted graphically against temperature for a number of reactions of metals to their oxides. The following plot is called an Ellingham diagram for metal oxide. Understanding of Ellingham diagram is extremely important for the efficient extraction of metals

...view full instructions

Which of the following elements can be prepared by heating the oxide above $$400^{o}C $$?

A
Hg

B
Mg

C
Fe

D
Al

Solution

As $$\Delta G$$ is least negative for Hg among all metals above $$400^0C$$ temp. So, Hg can be prepared by heating the oxide above $$400^0C$$.
Question 9
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Si of high purity to be used in semiconductor can be prepared by following methods :
I. $$SiO_2\, +\, 2C\, \rightarrow\, Si\, +\, 2CO$$
II. $$Si\, +\, 2Cl_2\, \rightarrow\, SiCl_4$$
$$SiCl_4\, +\, 2Mg\, \rightarrow\, Si\, +\, 2MgCl_2$$
Better method is :

A
I

B
II

C
both (A) and (B)

D
None of these

Solution

Si of high purity to be used in semiconductor can be prepared by the reaction of Si with chlorine to form $$SiCl_4$$ followed by fractional distillation (to purify $$SiCl_4$$) and reduction with hydrogen. Zone refining is used for the further purification.
$$Si\, +\, 2Cl_2\, \rightarrow\, SiCl_4(l) \xrightarrow {2H_2(g)} Si(s) +4HCl (g) \xrightarrow {Zone refining} Ultra pure silicon $$
Question 10
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In electrtolysis of $$Al_2O_3$$ by Hall-Heroult process:

A
Cryolite $$Na_3[AIF_6]$$ lowers the melting point of $$Al_2O_3$$ and increases its electrical conductivity.

B
$$Al$$ is obtained at cathode and probably $$CO_2$$ at anode

C
both (a) and (b) are correct

D
none of the above is correct

Solution

The electrolysis of alumina by Hall and Heroult's process is carried by using a fused mixture of alumina and cryolite along with minor quantities of aluminum fluoride and fluorspar. The addition of cryolite and fluorspar increases the electrical conductivity of alumina and lowers the fusion temperature. Thus, the option A is correct. Also at anode, alumina reacts with fluorine to give oxygen. The liberated oxygen reacts with carbon to form $$CO$$ and $$CO_2$$. These gases are liberated at anode.