Q: Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:\${N}_{2}(g)+{H}_{2}(g)\rightarrow 2{NH}_{3}(g)\$(i) Calculate the mass of ammonia produced if \$2.00\times {10}^{2}g\$ dinitrogen react with \$1.00\times {10}^{3}g\$ of dihydrogen(iii) Will any of the two reactants remain unreacted?(iii) If yes, which one and what would be its mass?

\$\underset {\underset {= 28.0\ g}{1\ mol}}{Na(g)} + \underset {\underset {= 6.0\ g}{3\ mol}}{3H_{2(g)}}\rightarrow \underset {\underset {= 34.0\ g}{2\ mol}}{2NH_{3(g)}}\$(i) \$28.0\ g\$ of \$N_{2}\$ require \$6.0\ g\$ of \$H_{2}\$ to produce\$= 34.0\ g\$ of \$NH_{3}\$\$2.00\times 10^{3}g\$ of \$ N_{2}\$ will produce\$= \dfrac {34}{28} \times 2.00\times 10^{3}g\$ of \$NH_{3}\$\$= 2.43\times 10^{3}g\$ of \$NH_{3}\$\$= 2430\ g\ NH_{3}\$(ii) Yes, dihydrogen will remain unreacted to some extent(iii) Amount of hydrogen that remains unreacted.\$28.0\ g\$ of \$N_{2}\$ require \$6.0\ g\$ of \$H_{2}\$\$2.00\ g\times 10^{3}g\$ of \$N_{2}\$ will require\$= \dfrac {6.0}{28.0}\times 2.00 \times 10^{3}\$ of \$H_{2} = 428.5\ g\$ of \$H_{2}\$\$\therefore\$ Amount of hydrogen that remains unreacted\$= [1.00\times 10^{3} - 428.5]_{g}\$\$= 571.5\ g\$.

Q: How is Nitric Acid manufactured by Ostwald's Process?

Ostwald process to manufacture nitric acid-Step \$1\$: Catalytic oxidation of AmmoniaA mixture of dry air and dry ammonia in the ratio of \$10 : 1\$ by volume is compressed and then passed into a platinum gauze which acts as catalyst at about \$800 ℃\$.\$4 N{H}_{3} + 5 {O}_{2} \longrightarrow 4 NO + 6 {H}_{2}O + \text{Heat}\$Step \$2\$: Oxidation of nitric oxideNitric oxide combines with oxygen to form nitrogen dioxide at about 50°C.\$2 NO + {O}_{2} \longrightarrow 2 N{O}_{2}\$Step \$3\$: Absorption of nitrogen dioxide in waterThe nitrogen dioxide and oxygen present in the air react with water to form nitric acid.\$4 N{O}_{2} + 2 {H}_{2}O + {O}_{2} \longrightarrow \underset{\text{Nitric acid}}{4 HN{O}_{3}}\$Nitric acid obtained is concentrated above \$50 \%\$. On further distillation, \$68 \%\$ nitric acid is produced.

Q: How is ammonia manufactured by Haber's process ? Explain.

On an industrial scale, ammonia is prepared by Haber’s process. The constituents of ammonia \${N}_{2}\$ and \${H}_{2}\$ combine in a ratio of \$1 : 3\$.\${N}_{2} + 3 {H}_{2} \rightleftharpoons 2N{H}_{3}\$ The reaction proceeds in the forward direction with a remarkable decrease in volume and thus the reaction is exothermic. In accordance with Le Chatelier’s principle, high pressure would favour the formation of ammonia. The optimum conditions for the production of ammonia are a pressure of approx \$200 \; atm\$, a temperature of approx \$700 \; K\$ and the use of a catalyst such as iron oxide with small amounts of \${K}_{2}O\$ and \${Al}_{2}{O}_{3}\$ to increase the rate of attainment of equilibrium.

Q: Two elements 'P' and 'Q' belong to the same period of the modern periodic table and are in Group 1 and Group 2, respectively. Compare their following characteristics in tabular form.(a) The number of electrons in their atoms(b) The sizes of their atoms(c) Their metallic character(d) Their tendencies to lose electrons(e) The formula of their oxides(f) The formula of their chlorides

Group 1 Group 2a) One electron in the valence shellTwo electrons in the valence shell b) Larger atomic size compared to Group 2Smaller atomic size compared to Group 1c) More metallic in natureComparatively less metallic d) Highly electropositive, readily loses electronComparatively less electropositive e) Oxide formula: \$X_2O\$Oxide formula: \$XO\$f) Chloride formula: \$XCl\$Chloride formula: \$XCl_2\$

Q: Arrange the following in order of property indicated for each set:a. \$F_{2}, Cl_{2}, Br_{2}, I_{2}\$ - increasing bond dissociation enthalpy.b. \$HF, HCl, HBr, HI\$ - increasing acid strength.c. \$NH_{3}, PH_{3}, AsH_{3}, SbH_{3}, BiH_{3}\$ - increasing base strength.

(i) \$\displaystyle I_2 < F_2 < Br_2 < Cl_2\$ The electron affinity of Chlorine is maximum in the periodic table and so the bond dissociation enthalpy. Flourine has lower bond dissociation enthalpy than \$Br_2\$ and \$Cl_2\$. Due to small size electronic repulsion is very high. \$I_2\$ has lowest bond dissociation enthalpy due to quite larger size it is easiest to break the bond.(ii) \$\displaystyle HF < HCl < HBr < HI \Rightarrow\$Acidic strength of Halogen acids increases down the group.(iii) \$\displaystyle BiH_3 < SbH_3 < AsH_3 < PH_3 < NH_3\Rightarrow\$ Basic strength decreases down the group because as we go down the group the size of the atom and thus the electron density decreases. These compounds are lewis base having 1 lone pair on the central atoms. As the electron density decrease the tendency to donate electron pair decreases.

Question 1

How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M $HNO_{3}$? The concentrated acid is 70% $HNO_{3}$.

Accepted Answer

The number of moles of nitric acid can be obtained by multiplying the molarity with volume$M\times V=$Moles of $\displaystyle HNO_{3}=\frac{250\times 2}{1000}=0.5$The mass of nitric acid can be obtained by multiplying the number of mols with molar mass and dividing with percentage concentration.$\therefore HNO_{3}$ required $=\displaystyle 0.5\times 63\times \frac{100}{70}=45\:g$Thus 45.0 g conc. $HNO_{3}$ of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M $HNO_{3}$.Hence, the correct option is $A$.

Question 2

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:${N}_{2}(g)+{H}_{2}(g)\rightarrow 2{NH}_{3}(g)$(i) Calculate the mass of ammonia produced if $2.00\times {10}^{2}g$ dinitrogen react with $1.00\times {10}^{3}g$ of dihydrogen(iii) Will any of the two reactants remain unreacted?(iii) If yes, which one and what would be its mass?

Accepted Answer

$\underset {\underset {= 28.0\ g}{1\ mol}}{Na(g)} + \underset {\underset {= 6.0\ g}{3\ mol}}{3H_{2(g)}}\rightarrow \underset {\underset {= 34.0\ g}{2\ mol}}{2NH_{3(g)}}$(i) $28.0\ g$ of $N_{2}$ require $6.0\ g$ of $H_{2}$ to produce$= 34.0\ g$ of $NH_{3}$$2.00\times 10^{3}g$ of $ N_{2}$ will produce$= \dfrac {34}{28} \times 2.00\times 10^{3}g$ of $NH_{3}$$= 2.43\times 10^{3}g$ of $NH_{3}$$= 2430\ g\ NH_{3}$(ii) Yes, dihydrogen will remain unreacted to some extent(iii) Amount of hydrogen that remains unreacted.$28.0\ g$ of $N_{2}$ require $6.0\ g$ of $H_{2}$$2.00\ g\times 10^{3}g$ of $N_{2}$ will require$= \dfrac {6.0}{28.0}\times 2.00 \times 10^{3}$ of $H_{2} = 428.5\ g$ of $H_{2}$$\therefore$ Amount of hydrogen that remains unreacted$= [1.00\times 10^{3} - 428.5]_{g}$$= 571.5\ g$.

Question 3

How is Nitric Acid manufactured by Ostwald's Process?

Accepted Answer

Ostwald process to manufacture nitric acid-Step $1$: Catalytic oxidation of AmmoniaA mixture of dry air and dry ammonia in the ratio of $10 : 1$ by volume is compressed and then passed into a platinum gauze which acts as catalyst at about $800 &#8451;$.$4 N{H}_{3} + 5 {O}_{2} \longrightarrow 4 NO + 6 {H}_{2}O + \text{Heat}$Step $2$: Oxidation of nitric oxideNitric oxide combines with oxygen to form nitrogen dioxide at about 50&#176;C.$2 NO + {O}_{2} \longrightarrow 2 N{O}_{2}$Step $3$: Absorption of nitrogen dioxide in waterThe nitrogen dioxide and oxygen present in the air react with water to form nitric acid.$4 N{O}_{2} + 2 {H}_{2}O + {O}_{2} \longrightarrow \underset{\text{Nitric acid}}{4 HN{O}_{3}}$Nitric acid obtained is concentrated above $50 \%$. On further distillation, $68 \%$ nitric acid is produced.

Question 4

How is ammonia manufactured by Haber's process ? Explain.

Accepted Answer

On an industrial scale, ammonia is prepared by Haber&#8217;s process. The constituents of ammonia ${N}_{2}$ and ${H}_{2}$ combine in a ratio of $1 : 3$.${N}_{2} + 3 {H}_{2} \rightleftharpoons 2N{H}_{3}$ The reaction proceeds in the forward direction with a remarkable decrease in volume and thus the reaction is exothermic. In accordance with Le Chatelier&#8217;s principle, high pressure would favour the formation of ammonia. The optimum conditions for the production of ammonia are a pressure of approx $200 \; atm$, a temperature of approx $700 \; K$ and the use of a catalyst such as iron oxide with small amounts of ${K}_{2}O$ and ${Al}_{2}{O}_{3}$ to increase the rate of attainment of equilibrium.

Question 5

What are amphoteric oxides? Give two examples of amphoteric oxides.

Accepted Answer

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Question 6

In one of the industrial processes used for the manufacture of sodium hydroxide, a gas $X$ is formed as a by-product. The gas $X$ reacts with lime water to give a compound $Y$ which is used as a bleaching agent in the chemical industry. Identify $X$ and $Y$.

Accepted Answer

Sodium chloride is used for manufacture of sodium hydroxide which is called chloralkali process. In this process chlorine and hydrogen gas are formed as byproducts along with sodium hydroxide. Chlorine gas gives bleaching power when reacts with lime water and used as bleaching agent in chemical industries.$2NaCl (aq) + 2H_{2}O(l) \rightarrow  2NaOH (aq) + Cl_{2} (g) + H_{2}(g)$$X \rightarrow  Cl_{2}$ (Chlorine gas)$Ca(OH)_2 +Cl_2 \longrightarrow     CaOCl_2 +  H_2O$Therefore,The gas $X$ is chlorine.Compound $Y$ is calcium oxychloride which is commonly known as bleaching powder and used as bleaching agent in chemical industries.

Question 7

Ammonia reacts with excess of chlorine to form :

Accepted Answer

Pure chlorine gas may react vigorously with ammonia gas. An&#10;excessive mix of the two gases in air can produce hazardous compounds such as nitrogen trichloride.The following reaction takes place:$N{ H }_{ 3 }+3{ Cl }_{ 2 }\longrightarrow N{ Cl }_{ 3 }+3HCl$Ammonia reacts with excess of chlorine to form $ N{ Cl }_{ 3 } $ and $ HCl $.

Question 8

Arrange the following in the increasing order of their basic character: $NH_{3}, PH_{3}, ASH_{3}, SbH_{3}, BiH_{3}$

Accepted Answer

The hydrides of nitrogen family contain one lone pair on the central atom. As we move down the group the size of the central atom increases thereby their orbital become more diffused, hence the nucleophilicity of the lone pair is less dense over the bigger atoms thereby decreasing the basicity. So, the increasing order of their basic character is,$BiH_3<SbH_3<AsH_3<PH_3<NH_3$

Question 9

Two elements 'P' and 'Q' belong to the same period of the modern periodic table and are in Group 1 and Group 2, respectively. Compare their following characteristics in tabular form.(a) The number of electrons in their atoms(b) The sizes of their atoms(c) Their metallic character(d) Their tendencies to lose electrons(e) The formula of their oxides(f) The formula of their chlorides

Accepted Answer

Group 1 Group 2a) One electron in the valence shellTwo electrons in the valence shell b) Larger atomic size compared to Group 2Smaller atomic size compared to Group 1c) More metallic in natureComparatively less metallic d) Highly electropositive, readily loses electronComparatively less electropositive e) Oxide formula: $X_2O$Oxide formula: $XO$f) Chloride formula: $XCl$Chloride formula: $XCl_2$

Question 10

Arrange the following in order of property indicated for each set:a. $F_{2}, Cl_{2}, Br_{2}, I_{2}$ - increasing bond dissociation enthalpy.b. $HF, HCl, HBr, HI$ - increasing acid strength.c. $NH_{3}, PH_{3}, AsH_{3}, SbH_{3}, BiH_{3}$ - increasing base strength.

Accepted Answer

(i) $\displaystyle I_2 < F_2 < Br_2 < Cl_2$ The electron affinity of Chlorine is maximum in the periodic table and so the bond dissociation enthalpy. Flourine has lower  bond dissociation enthalpy than $Br_2$ and $Cl_2$. Due to small size electronic repulsion is very high. $I_2$ has lowest bond dissociation enthalpy due to quite larger size it is easiest to break the bond.(ii) $\displaystyle HF < HCl < HBr < HI \Rightarrow$Acidic strength of Halogen acids increases down the group.(iii) $\displaystyle BiH_3 < SbH_3 < AsH_3 < PH_3 < NH_3\Rightarrow$ Basic strength decreases down the group because as we go down the group the size of the atom and thus the electron density decreases. These compounds are lewis base having 1 lone pair on the central atoms. As the electron density decrease the tendency to donate electron pair decreases.