hydrogenWHERE cd.courseId=2 AND cd.subId=9 AND chapterSlug='hydrogen' and status=1SELECT ex_no,page_number,question,question_no,id,chapter,solution FROM question_mgmt as q WHERE courseId='2' AND subId='9' AND chapterId='50' AND ex_no!=0 AND status=1 ORDER BY ex_no,CAST(question_no AS UNSIGNED) CBSE Class 11 Free NCERT Book Solution for Chemistry

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Chapter 9 : Hydrogen


From this chapter, you will be able to present informed opinions on the position of hydrogen in the periodic table. You can also identify the modes of occurrence and preparation of dihydrogen on a small and commercial scale. Description of isotopes of hydrogen and how different elements combine with hydrogen to form ionic, molecular and non stoichiometric compounds. You will also be able to understand the structure of water and use the knowledge for explaining physical and chemical properties.Explanation of how environmental water quality depends on a variety of dissolved substances and difference between hard and soft water is also given.You will also acquire the knowledge about heavy water and its importance. You will also understand the structure of hydrogen peroxide,learn its preparatory methods and properties leading to the manufacture of useful chemicals and cleaning of the environment. You will also understand the use of certain terms like electron-deficient, electron-precise, electron-rich, hydrogen economy, hydrogenation etc.

Exercise 1 ( Page No. : 298 )
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Hydrogen is the first element of the periodic table & often called protium. Its electronic configuration is [1s1]. Due to the presence of only one electron in its 1s shell, hydrogen exhibits a dual behaviour, i.e., it resembles both alkali metals and halogens. As a result it is placed at the top of the alkali metals in group 1 & also along the halogens in group 17 since just like halogens,it also requires any one electron to have the configuration of the nearest noble gas element..The dual nature of hydrogen can be justified by following points:

Resemblance with alkali metals:

1. Like alkali metals, hydrogen contains one valence electron in its valence shell.

H : 1s1          

Li : 1s2 2s1

Na : 1s2 2s2 2p6 3s1

Hence, it can lose one electron to form a unipositive ion.

2 Both hydrogen & alkali metals form monovalent cations by losing the electrons in the valence shell,showing hydrogens electropositive character similar to alkali metals.

H --------> H+  + e- , Na  ---------->Na+   +   e-

3 In their compounds both hydrogen & alkali metals show +1 oxidation states. Eg Hcl  & Nacl

4 Both hydrogen & alkali metals are electropositive in nature,they have affinity for the electronegative elements & readily combine with them & form oxides, halides, and sulphides.

Resemblance with halogens:

1. Both hydrogen and halogens require one electron to complete their octets.

H : 1s1

F : 1s2 2s2 2p5

Cl : 1s2 2s2 2p6 3s2 3p5

Hence, hydrogen can gain one electron to form a uninegative ion.

2. Like halogens, it forms a diatomic molecule and several covalent compounds.

Though hydrogen shows some similarity with both alkali metals and halogens, it differs from them on some grounds. Unlike alkali metals, hydrogen does not possess metallic characteristics. On the other hand, it possesses a high ionization enthalpy. Also, it is less reactive than halogens.

Owing to these reasons, hydrogen cannot be placed with alkali metals (group I) or with halogens (group VII). In addition, it was also established that H+ ions cannot exist freely as they are extremely small. H+ ions are always associated with other atoms or molecules. Hence, hydrogen is best placed separately in the periodic table.


Exercise 1 ( Page No. : 298 )
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A:

Hydrogen has three isotopes.

They are:

1. Protium, 1H1,

2. Deuterium, 2Hor D, and

3. Tritium, 3H1 or T

The mass ratio of protium, deuterium and tritium is 1:2:3.


Exercise 1 ( Page No. : 298 )
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The ionization enthalpy of hydrogen atom is very high (1312 kJ mol-1). Hence, it is very hard to remove its only electron. As a result, its tendency to exist in the monoatomic form is rather low. Instead, hydrogen forms a covalent bond with another hydrogen atom and exists as a diatomic (H2) molecule.


Exercise 1 ( Page No. : 298 )
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A:

Dihydrogenis produced by coal gasification method as:

                                          1270K

C(s)      +      H2O(g)      →     CO(g)    +   H2(g)        

(Coal)

The yield of dihydrogen (obtained from coal gasification) can be increased by reacting carbon monoxide (formed during the reaction) with steam in the presence of iron chromate as a catalyst.

This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it with a solution of sodium arsenite.


Exercise 1 ( Page No. : 298 )
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Dihydrogen is prepared by the electrolysis of acidified or alkaline water using platinum electrodes. Generally, 15 - 20% of an acid (H2SO4) or a base (NaOH) is used.

Reduction of water occurs at the cathode as:

2H2O   +    2e-       →       2H2    +     2OH-

At the anode, oxidation of OH-ions takes place as:

2OH-      →       H2O   +   ½ O2   +   2e-

 Net reaction can be represented as:

H2O(I)     →     H2(g)     +   ½ O2(g)

Electrical conductivity of pure water is very low owing to the absence of ions in it. Therefore, electrolysis of pure water also takes place at a low rate. If an electrolyte such as an acid or a base is added to the process, the rate of electrolysis increases. The addition of the electrolyte makes the ions available in the process for the conduction of electricity and for electrolysis to take place.


Exercise 1 ( Page No. : 298 )

Exercise 1 ( Page No. : 298 )
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A:

The ionization enthalpy of H-H bond is very high (1312 kJ mol-1). This indicates that hydrogen has a low tendency to form H+ions. Its ionization enthalpy value is comparable to that of halogens. Hence, it forms diatomic molecules (H2), hydrides with elements, and a large number of covalent bonds. Since ionization enthalpy is very high, hydrogen does not possess metallic characteristics (lustre, ductility, etc.) like metals.


Exercise 1 ( Page No. : 298 )
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A:

Molecular hydrides are classified on the basis of the presence of the total number of electrons and bonds in their Lewis structures as:

1. Electron-deficient hydrides

2. Electron-precise hydrides

3. Electron-rich hydrides

An electron-deficient hydride has very few electrons, less than that required for representing its conventional Lewis structure e.g. diborane (B2H6). In B2H6, there are six bonds in all, out of which only four bonds are regular two centered-two electron bonds. The remaining two bonds are three centered-two electron bonds i.e., two electrons are shared by three atoms. Hence, its conventional Lewis structure cannot be drawn.

An electron-precise hydride has a sufficient number of electrons to be represented by its conventional Lewis structure e.g. CH4. The Lewis structure can be written as:

 

Four regular bonds are formed where two electrons are shared by two atoms.

An electron-rich hydride contains excess electrons as lone pairs e.g. NH3.

There are three regular bonds in all with a lone pair of electrons on the nitrogen atom.


Exercise 1 ( Page No. : 298 )
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A:

An electron-deficient hydride does not have sufficient electrons to form a regular bond in which two electrons are shared by two atoms e.g., B2H6, Al2H6 etc.

These hydrides cannot be represented by conventional Lewis structures. B2H6, for example, contains four regular bonds and two three centered-two electron bond. Its structure can be represented as:

Since these hydrides are electron-deficient, they have a tendency to accept electrons. Hence, they act as Lewis acids.

B2H  +  2NMe    →    2BH3 . NMe3

B2H  +  2CO    →    2BH3 . CO


Exercise 1 ( Page No. : 298 )
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A:

For carbon hydrides of type CnH2n+2, the following hydrides are possible for

n = 1 ⇒ CH4

n = 2 ⇒ C2H6

n = 3 ⇒ C3H8 

:            :

For a hydride to act as a Lewis acid i.e., electron accepting, it should be electron-deficient. Also, for it to act as a Lewis base i.e., electron donating, it should be electron-rich.

Taking C2H6 as an example, the total number of electrons are 14 and the total covalent bonds are seven. Hence, the bonds are regular 2e- -2 centered bonds.

Hence, hydride C2H6 has sufficient electrons to be represented by a conventional Lewis structure. Therefore, it is an electron-precise hydride, having all atoms with complete octets. Thus, it can neither donate nor accept electrons to act as a Lewis acid or Lewis base.


Exercise 1 ( Page No. : 298 )
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A:

Non-Stoichiometric hydrides are hydrogen-deficient compounds formed by the reaction of dihydrogen with d-block and f-block elements. These hydrides do not follow the law of constant composition. For example: LaH2.87, YbH2.55, TiH1.5 - 1.8 etc.

Alkali metals form stoichiometric hydrides. These hydrides are ionic in nature. Hydride ions have comparable sizes (208 pm) with alkali metal ions. Hence, strong binding forces exist between the constituting metal and hydride ion. As a result, stoichiometric hydrides are formed.

Alkali metals will not form non-stoichiometric hydrides.


Exercise 1 ( Page No. : 298 )
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Metallic hydrides are hydrogen deficient, i.e., they do not hold the law of constant composition. It has been established that in the hydrides of Ni, Pd, Ce, and Ac, hydrogen occupies the interstitial position in lattices allowing further absorption of hydrogen on these metals. Metals like Pd, Pt, etc. have the capacity to accommodate a large volume of hydrogen. Therefore, they are used for the storage of hydrogen and serve as a source of energy.


Exercise 1 ( Page No. : 298 )
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A:

Atomic hydrogen atoms are produced by the dissociation of dihydrogen with the help of an electric arc. This releases a huge amount of energy (435.88 kJ mol-1). This energy can be used to generate a temperature of 4000 K, which is ideal for welding and cutting metals. Hence, atomic hydrogen or oxy-hydrogen torches are used for these purposes. For this reason, atomic hydrogen is allowed to recombine on the surface to be welded to generate the desired temperature.


Exercise 1 ( Page No. : 298 )
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A:

The extent of hydrogen bonding depends upon electronegativity and the number of hydrogen atoms available for bonding. Among nitrogen, fluorine, and oxygen, the increasing order of their electronegativities are N < O < F.

Hence, the expected order of the extent of hydrogen bonding is HF > H2O > NH3.

But, the actual order is H2O > HF > NH3.

Although fluorine is more electronegative than oxygen, the extent of hydrogen bonding is higher in water. There is a shortage of hydrogens in HF, whereas there are exactly the right numbers of hydrogens in water. As a result, only straight chain bonding takes place. On the other hand, oxygen forms a huge ring-like structure through its high ability of hydrogen bonding.

In case of ammonia, the extent of hydrogen bonding is limited because nitrogen has only one lone pair. Therefore, it cannot satisfy all hydrogens.


Exercise 1 ( Page No. : 298 )
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A:

Saline hydrides (i.e., NaH, LiH, etc.) react with water to form a base and hydrogen gas. The chemical equation used to represent the reaction can be written as:

MH(s)   +   H2O(aq)   →     MOH(aq)  + H2(g)

The reaction is violent and produces fire.

CO2 is heavier than dioxygen. It is used as a fire extinguisher because it covers the fire as a blanket and inhibits the supply of dioxygen, thereby dousing the fire.

CO2 can be used in the present case as well. It is heavier than dihydrogen and will be effective in isolating the burning surface from dihydrogen and dioxygen.


Exercise 1 ( Page No. : 298 )
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A:

(i) The electrical conductance of a molecule depends upon its ionic or covalent nature. Ionic compounds conduct, whereas covalent compounds do not.

BeH2 is a covalent hydride. Hence, it does not conduct. CaH2 is an ionic hydride, which conducts electricity in the molten state. Titanium hydride, TiH2 is metallic in nature and conducts electricity at room temperature. Hence, the increasing order of electrical conductance is as follows:

BeH2 < CaH2 < TiH2

 

(ii) The ionic character of a bond is dependent on the electronegativities of the atoms involved. The higher the difference between the electronegativities of atoms, the smaller is the ionic character.

Electronegativity decreases down the group from Lithium to Caesium. Hence, the ionic character of their hydrides will increase (as shown below).

LiH < NaH < CsH

 

(iii) Bond dissociation energy depends upon the bond strength of a molecule, which in turn depends upon the attractive and repulsive forces present in a molecule.

The bond pair in D-D bond is more strongly attracted by the nucleus than the bond pair in H-H bond. This is because of the higher nuclear mass of D2. The stronger the attraction, the greater will be the bond strength and the higher is the bond dissociation enthalpy. Hence, the bond dissociation enthalpy of D-D is higher than H-H.

However, bond dissociation enthalpy is the minimum in the case of F-F. The bond pair experiences strong repulsion from the lone pairs present on each F-centre.

Therefore, the increasing order of bond dissociation enthalpy is as follows:

F-F < H-H < D-D

 

(iv) Ionic hydrides are strong reducing agents. NaH can easily donate its electrons. Hence, it is most reducing in nature.

Both, MgH2 and H2O are covalent hydrides. H2O is less reducing than MgH2 since the bond dissociation energy of H2O is higher than MgH2.

Hence, the increasing order of the reducing property is

H2O < MgH2 < NaH.


Exercise 1 ( Page No. : 298 )
Q:
A:

In gaseous phase, water molecule has a bent form with a bond angle of 104.5°. The O-H bond length is 95.7 pm. The structure can be shown as:

Hydrogen peroxide has a non-planar structure both in gas and solid phase. The dihedral angle in gas and solid phase is 111.5° and 90.2° respectively.

 


Exercise 1 ( Page No. : 298 )
Q:
A:

Auto-protolysis (self-ionization) of water is a chemical reaction in which two water molecules react to produce a hydroxide ion (OH-) and a hydronium ion (H3O+).

The reaction involved can be represented as:

Auto-protolysis of water indicates its amphoteric nature i.e., its ability to act as an acid as well as a base. The acid-base reaction can be written as:


Exercise 1 ( Page No. : 298 )
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A:

The reaction between fluorine and water can be represented as:

2F2(g)     +    2H2O(I)    →     4H+(aq)  +  4F-(aq)   +  O2(g)

This is an example of a redox reaction as water is getting oxidized to oxygen, while fluorine is being reduced to fluoride ion.

The oxidation numbers of various species can be represented as:

Fluorine is reduced from zero to (-1) oxidation state. A decrease in oxidation state indicates the reduction of fluorine. Water is oxidized from (-2) to zero oxidation state. An increase in oxidation state indicates oxidation of water.

 


Exercise 1 ( Page No. : 298 )
Q:
A:

(i) PbS(s)  +  H2O2(aq)   →   PbSO4(s)  +  H2O(l)

H2O2 is acting as an oxidizing agent in the reaction. Hence, it is a redox reaction.

 

(ii) 2MnO-4(aq)  + 5H2O2(aq)   →  6H+(aq)    →  2Mn2+(aq)  +    8H2O(l)   +   5O2(g)

H2O2(aq) is acting as a reducing agent in the acidic medium, thereby oxidizing MnO-4(aq). Hence, the given reaction is a redox reaction.

 

(iii) CaO(s)  +  H2O(g)  →   Ca(OH)2aq

The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. The given reaction is hydrolysis.

 

(iv) 2AlCl3(g)   +   3H2O(I)   →  Al2O3(s)   + 6HCl(aq)

The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. The given reaction represents hydrolysis of AlCl3.

 

(v) Ca3N2(s)   +  6H2O(I)  →     3Ca(OH)2(aq)  + 2NH3(g)

The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. The given reaction represents hydrolysis of Ca3N2.

 


Exercise 1 ( Page No. : 298 )
Q:
A:

Ice is the crystalline form of water. It takes a hexagonal form if crystallized at atmospheric pressure, but condenses to cubic form if the temperature is very low. The three-dimensional structure of ice is represented as:

The structure is highly ordered and has hydrogen bonding. Each oxygen atom is surrounded tetrahedrally by four other oxygen atoms at a distance of 276 pm. The structure also contains wide holes that can hold molecules of appropriate sizes interstitially.


Exercise 1 ( Page No. : 298 )
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A:

Temporary hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of hydrogen carbonates (MHCO3, where M = Mg, Ca) in water.

Permanent hardness of water is because of the presence of soluble salts of calcium and magnesium in the form of chlorides in water.


Exercise 1 ( Page No. : 298 )
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A:

The process of treating permanent hardness of water using synthetic resins is based on the exchange of cations (e.g., Na+, Ca2+, Mg2+ etc) and anions (e.g., Cl-, SO2-4, HCO-3 etc) present in water by H+ and OH- ions respectively.

Synthetic resins are of two types:

1) Cation exchange resins

2) Anion exchange resins

Cation exchange resins are large organic molecules that contain the -SO3H group. The resin is firstly changed to RNa (from RSO3H) by treating it with NaCl. This resin then exchanges Na+ ions with Ca2+ and Mg2+ ions, thereby making the water soft.

2RNa   +   M2+ (aq)  →  R2M(s) +  2Na + (aq)

There are cation exchange resins in H+ form. The resins exchange H+ ions for Na+, Ca2+, and Mg2+ ions.

Anion exchange resins exchange OH- ions for anions like Cl-,HCO-3, and SO42- present in water.

During the complete process, water first passes through the cation exchange process. The water obtained after this process is free from mineral cations and is acidic in nature.

This acidic water is then passed through the anion exchange process where OH- ions neutralize the H+ ions and de-ionize the water obtained.


Exercise 1 ( Page No. : 298 )
Q:
A:

The amphoteric nature of water can be described on the basis of the following reactions:

1) Reaction with H2S

The reaction takes place as:

In the forward reaction, H2O(l) accepts a proton from H2S(aq). Hence, it acts as a Lewis base.

 

2) Reaction with NH3

The reaction takes place as:

In the forward reaction, H2O(l) denotes its proton NH3(aq). Hence, it acts as a Lewis acid.

 

3) Self-ionization of water

In the reaction, two water molecules react as:


Exercise 1 ( Page No. : 298 )
Q:
A:

Hydrogen peroxide, H2O2 acts as an oxidizing as well as a reducing agent in both acidic and alkaline media.

Reactions involving oxidizing actions are:

1)  2Fe2+ + 2H+  + H2O2  →  2Fe3+  +  2H2O

2) Mn2+  +  H2O2   →   Mn4+   +  2OH-

3) PbS  +  4H2O2    →   PbSO4  +  4H2O

4) 2Fe2+  + H2O2   →  2 Fe3+  + 2OH-

 

Reactions involving reduction actions are:
1)  2MnO-4  + 6H+  +  5H2O2  →  2Mn2+  +  8H2O  + 5O2

2) I2  + H2O2  +  2OH-   →   2l- + 2H2O + O2

3) HOCl  +  H2O2   →  H3O+  + Cl-  +  O2

4) 2MnO- +  3H2O2   →  2MnO2  +  3O2  + 2H2O  + 2OH-


Exercise 1 ( Page No. : 298 )
Q:
A:

Demineralised water is free from all soluble mineral salts. It does not contain any anions or cations.

Demineralised water is obtained by passing water successively through a cation exchange (in the H+ form) and an anion exchange (in the OH- form) resin.

During the cation exchange process, H+exchanges for Na+, Mg2+, Ca2+, and other cations present in water.

In the anion exchange process, OH- exchanges for anions such as CO2-3, SO2-4 , Cl- , HCO-3 etc. present in water.

OH- ions liberated in reaction (2) neutralize H+ ions liberated in reaction (1), thereby forming water.

H+ (aq)  +  OH- (aq)  →   H2O(l)


Exercise 1 ( Page No. : 298 )
Q:
A:

Water is an important part of life. It contains several dissolved nutrients that are required by human beings, plants, and animals for survival. Demineralised water is free of all soluble minerals. Hence, it is not fit for drinking.

It can be made useful only after the addition of desired minerals in specific amounts, which are important for growth.


Exercise 1 ( Page No. : 298 )
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A:

Wateris essential for all forms of life. It constitutes around 65% of the human body and 95% of plants. Water plays an important role in the biosphere owing to its high specific heat, thermal conductivity, surface tension, dipole moment, and dielectric constant.

The high heat of vapourization and heat of capacity of water helps in moderating the climate and body temperature of all living beings.

It acts as a carrier of various nutrients required by plants and animals for various metabolic reactions.


Exercise 1 ( Page No. : 298 )
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A:

A high value of dielectric constants (78.39 C2/Nm2) and dipole moment make water a universal solvent.

Water is able to dissolve most ionic and covalent compounds. Ionic compounds dissolve in water because of the ion-dipole interaction, whereas covalent compounds form hydrogen bonding and dissolve in water.

Water can hydrolyze metallic and non-metallic oxides, hydrides, carbides, phosphides, nitrides and various other salts. During hydrolysis, H+ and OH- ions of water interact with the reacting molecule.

Some reactions are:

CaO  +  H2O  →  Ca(OH)2

NaH  +  H2O →  NaOH + H2

CaC2  + H2O → C2H2 +  Ca(OH)2


Exercise 1 ( Page No. : 298 )
Q:
A:

Heavy water (D2O) acts as a moderator, i.e., it slows the rate of a reaction. Due to this property of D2O, it cannot be used for drinking purposes because it will slow down anabolic and catabolic reactions taking place in the body and lead to a casualty.


Exercise 1 ( Page No. : 298 )
Q:
A:

Hydrolysis is defined as a chemical reaction in which hydrogen and hydroxide ions (H+ and OH- ions) of water molecule react with a compound to form products. For example:

NaH + H2O  →  NaOH  + H2

Hydration is defined as the addition of one or more water molecules to ions or molecules to form hydrated compounds. For example:

CuSO +  5H2O →  CuSO4 .  5H2O


Exercise 1 ( Page No. : 298 )
Q:
A:

Saline hydrides areionic in nature. They react with water to form a metal hydroxide along with the liberation of hydrogen gas. The reaction of saline hydrides with water can be represented as:

AH(s) +  H2O(l)  →  AOH(aq) + H2(g)

(where, A = Na, Ca,……)

When added to an organic solvent, they react with water present in it. Hydrogen escapes into the atmosphere leaving behind the metallic hydroxide. The dry organic solvent distills over.


Exercise 1 ( Page No. : 298 )
Q:
A:

The elements of atomic numbers 15, 19, 23, and 44 are nitrogen, potassium, vanadium, and ruthenium respectively.

1) Hydride of nitrogen

Hydride of nitrogen (NH3) is a covalent molecule. It is an electron-rich hydride owing to the presence of excess electrons as a lone pair on nitrogen.

 

2) Hydride of potassium

Dihydrogen forms an ionic hydride with potassium owing to the high electropositive nature of potassium. It is crystalline and non-volatile in nature.

 

3) Hydrides of Vanadium and Ruthenium

Both vanadium and ruthenium belong to the d-block of the periodic table. The metals of d-block form metallic or non-stoichiometric hydrides. Hydrides of vanadium and ruthenium are therefore, metallic in nature having a deficiency of hydrogen.

 

4) Behaviour of hydrides towards water

Potassium hydride reacts violently with water as: KH(s) + H2O(aq)  →  KOH(aq) + H2(g)

Ammonia (NH3) behaves as a Lewis base and reacts with water as:

H2O(l)  +  NH3(aq)  ↔   OH-(aq)  +  NH+ 4(aq)

Hydrides of vanadium and Ruthenium do not react with water. Hence, the increasing order of reactivity of the hydrides is (V, Ru) H < NH3 < KH.


Exercise 1 ( Page No. : 298 )
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A:

Potassium chloride (KCl) is the salt of a strong acid (HCl) and strong base (KOH). Hence, it is neutral in nature and does not undergo hydrolysis in normal water. It dissociates into ions as follows:

In acidified and alkaline water, the ions do not react and remain as such.

Aluminium (III) chloride is the salt of a strong acid (HCl) and weak base [Al(OH)3]. Hence, it undergoes hydrolysis in normal water.

In acidified water, H+ions react with Al(OH)3 forming water and giving Al3+ions. Hence, in acidified water, AlCl3 will exist as Al3+(aq) and Cl-(aq) ions.


Exercise 1 ( Page No. : 298 )
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A:

H2O2 or hydrogen peroxide acts as a strong oxidizing agent both in acidic and basic media.

When added to a cloth, it breaks the chemical bonds of the chromophores (colour producing agents). Hence, the visible light is not absorbed and the cloth gets whitened.


Exercise 1 ( Page No. : 298 )
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A:

(i) Hydrogen economy Hydrogen economy is a technique of using dihydrogen in an efficient way. It involves transportation and storage of dihydrogen in the form of liquid or gas. Dihydrogen releases more energy than petrol and is more eco-friendly. Hence, it can be used in fuel cells to generate electric power. Hydrogen economy is about the transmission of this energy in the form of dihydrogen.

 

(ii) Hydrogenation Hydrogenation is the addition of dihydrogen to another reactant. This process is used to reduce a compound in the presence of a suitable catalyst. For example, hydrogenation of vegetable oil using nickel as a catalyst gives edible fats such as vanaspati, ghee etc.

 

(iii) Syngas Syngas is a mixture of carbon monoxide and dihydrogen. Since the mixture of the two gases is used for the synthesis of methanol, it is called syngas, synthesis gas, or water gas.

Syngas is produced on the action of steam with hydrocarbons or coke at a high temperature in the presence of a catalyst.

 

(iv) Water shift reaction It is a reaction of carbon monoxide of syngas mixture with steam in the presence of a catalyst as:

This reaction is used to increase the yield of dihydrogen obtained from the coal gasification reaction as:

C(s)  + H2O(g) →  CO(g) + H2(g)

(v) Fuel cells Fuel cells are devices for producing electricity from fuel in the presence of an electrolyte. Dihydrogen can be used as a fuel in these cells. It is preferred over other fuels because it is eco-friendly and releases greater energy per unit mass of fuel as compared to gasoline and other fuels.