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Chemistry 5 Marks Questions

p- Block Elements · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 6 questions.

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Question 15 Marks
(a) Give reasons for the following:
(i) Sulphur in vapour state shows paramagnetic behaviour.
(ii) $N - N$ bond is weaker than $P - P$ bond.
(iii) Ozone is thermodynamically less stable than oxygen.
(b) Write the name of gas released when Cu is added to
(i) dilute $HNO _3$ and (ii) conc. $HNO _3$.
Answer
(a) (i) In gas phase, Sulphur exists mostly as diatomic molecule $S _2$. The bonding in this molecule is similar to that of dioxygen O2. According to molecular orbital theory it has two unpaired electrons in antibonding molecular ( $\pi$ ) orbitals.
(ii) The N-N sigma bond length is less than the P-P sigma bond length. The shorter separation means the nonbonding electrons (lone pairs) on the nitrogen atoms repel each other strongly, making its bond weaker than the P-P bond.
(iii) The enthalpy change for decomposition of ozone has a negative value, it results in liberation of heat :
$2 O _3 \rightarrow 3 O _2 \ldots \Delta H ^{\circ}=-142 kJ / mol$
The increase in number of moles of gas means the entropy change for the reaction is positive. This reinforces the effect of negative enthalpy change, and results in a large negative value of $\Delta G ^{\circ}$ for the decomposition reaction, making it less stable than oxygen.
(b) (i) Dilute $HNO _3:$ Nitric oxide, NO.
(ii) Conc. $HNO _3:$ Nitrogen dioxide, $NO _2$.
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Question 25 Marks
(a) (i) Write the disproportionation reaction of $H _3 PO _3$.
(ii) Draw the structure of $XeF _4$.
(b) Account for the following:
(i) Although fluorine has less negative electron gain enthalpy yet $F_2$ is a strong oxidizing agent.
(ii) Acidic character decreases from $N _2 O _3$ to $Bi _2 O _3$ in group 15.
(c) Write a chemical reaction to test Sulphur dioxide gas. Write the chemical equation involved.
Answer
(a) (i) $\quad 4 H _3 PO _3 \rightarrow 3 H _3 PO _4+ PH _3$
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(b) (i) The bond dissociation energy of $F _2$ is quite low, and the small size of the fluoride ion means it its hydration energy is high. These two compensate for the less negative value of electron gain energy, making $F_2$ a strong oxidizing agent.
(ii) As we move down a group, the atomic size and metallic character increases while electronegativity decreases. This leads to the acidic character of the oxide decreasing from top to bottom, $N _2 O _3$ to $Bi _2 O _3$.
(c) Sulphur dioxide turns an acidified potassium dichromate solution green :
$3 SO _2+ K _2 Cr _2 O _7+ H _2 SO _4 \rightarrow Cr _2\left( SO _4\right)_3+ H _2 O + H _2 SO _4$
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Question 35 Marks
Write the principle of manufacture of ammonia by Haber's process. How does it react with $ CuSO_{4} $ solution?
Answer
Industrial Preparation of Ammonia : Ammonia is manufactured by Haber's process industrially.
Principle : Nitrogen is made to combine with $H _2$ at a temperature of 700 K and 200 atm . pressure. Iron is used as a catalyst in the process and molybdenum is used as promoter.
Due to exothermic and reversible nature of the reaction a good yield of ammonia is obtained at low temperature and high pressure. So reaction is carried out under low temperature and high pressure.
$N _2+3 H _2 \underset{200 atm .}{\stackrel{ Fe ( Mo ), 770 K}{\rightleftharpoons}} 2 NH _3+92.2 kJ$
Process : Purified $N _2$ and $H _2$ gases are taken in the ratio of $1: 3$ and introduced in the compression unit where they are subjected to a pressure of 200 atm . After compression the gases are passed through ammonia converter, packed with finally divided iron as catalyst and molybdenum as promoter. Initially the temperature of the chamber is raised to 700 K . After the reaction starts the temperature is maintained, due to the exothermic nature of the reaction.
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A mixture of $Fe _2 O _3, K_2 O$ and $Al _2 O _3$ can also we used as catalyst in this process.
The gases coming out from the catalyst chamber contained $10-20 \%$ ammonia along with $H _2$ and $N _2$. This mixture is passed through a condensor to liquify the ammonium formed. Unreacted $H _2$ and $N _2$ are passed out.
The uncondensed gases are passed through the compression unit with fresh gases and whole process is repeated.
React with $CuSO _4$ Solution
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Question 45 Marks
(a) Write the principle of manufacture of sulphuric acid by contact process.###(b) Write the reaction of concentrated sulphuric acid with oxalic acid.
Answer
(a) The manufacturing of sulphuric acid by contact process is based on the following principle :
Sulphur dioxide with oxygen of air forms sulphur trioxide in presence of platinised asbestos. This $SO _3$ with water form $H _2 SO _4$.
$2 SO _2+ O _2 \leftrightharpoons 2 SO _3+45,200 Cal$
$SO _3+ H _2 O \leftrightharpoons H _2 SO _4$
For this required conditions are :
(i) Purity of gaseous mixture : The mixture of $SO _2$ and water must be pure because dust, assemious oxide etc. render the catalyst inactive.
(ii) Low temperature : Formations of $SO _3$ is an exothermic reaction. Hence low temperature favours its formation.
(iii) Use of a Catalyst :At low temperature, this reaction takes place at a slow rate hence a catalyst is required $V _2 O _5$ to be an important catalyst reaction the plotinised absesfos.
(iv) Excess of Oxygen : Since the reaction in reversible. Hence for converting whole $SO _2$ into $SO _3$ excess of oxygen should be used.
(v) High Pressure : The presence of high pressure $SO _3$ will be produced in high concentration.
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Question 55 Marks
Write only the principle for the manufacturing of sulphuric acid by contact process?
Answer
The manufacturing of sulphuric acid by contact process is based on the following principle :
Sulphur dioxide with oxygen of air forms sulphur trioxide in presence of platinised asbestos. This $SO _3$ with water form $H _2 SO _4$.
$2 SO _2+ O _2 \rightleftharpoons 2 SO _3+45,200 Cal$
$SO _3+ H _2 O \rightleftharpoons H _2 SO _4$
For this required conditions are :
(i) Purity of gaseous mixture : The mixture of $SO _2$ and water must be pure because dust, assemious oxide etc. render the catalyst inactive.
(ii) Low temperature : Formations of $SO _3$ is an exothermic reaction. Hence low temperature favours its formation.
(iii) Use of a Catalyst :At low temperature, this reaction takes place at a slow rate hence a catalyst is required $V _2 O _5$ to be an important catalyst reaction the plotinised asbesfos.
(iv) Excess of Oxygen : Since the reaction in reversible. Hence for converting whole $SO _2$ into $SO _3$ excess of oxygen should be used.
(v) High Pressure : The presence of high pressure $SO _3$ will be produced in high concentration.
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Question 65 Marks
How does nitric acid react with the following ? Give equations.
(i) Copper
(ii) Iron
Answer
(i) Reaction with hot and conc. $HNO _3$ :
$2 HNO _3 \xrightarrow{\Delta} 2 NO _2+ H _2 O +[ O ]$
$Cu +[ O ] \rightarrow CuO$
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Reaction with Iron (Fe) :
(i) Reaction with conc. $HNO _3$ : Iron becomes passive with pure conc. $HNO _3$ i.e. it does not react. As such it becomes due to deposited layer of ferroso ferric-oxide on the surface of iron.
(ii) Reaction with conc. $HNO _3$ :
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