NEB Grade 12 Chemistry 2082 Question Paper Solutions

NEB Grade 12 Chemistry 2082 Question Paper Solutions

This post contains the Grade 12 Chemistry 2082 Question Papers solutions. Includes Multiple Choice questions with answers, Short Questions with answers, Long Questions with answers, numerical problems and solutions. NEB Chemistry Past Question papers.

Group ‘A’ [11×1=11]

Rewrite the correct option of each question in your answer sheet.

1. To estimate the chloride ion present in tap water, volumetric analysis has been commonly used in the laboratory. Which of the following solutions can be used for the above mentioned titration?

A) BaCl₂
B) NaOH
C) K₂Cr₂O₇
D) AgNO₃

2. The half-life of a certain reaction is 50 seconds, and it remains constant even if the initial concentration of the reaction is doubled. What is the order of the reaction?

A) Zero-order reaction
B) First-order reaction
C) Second-order reaction
D) Pseudo-first order reaction

3. Gibbs-Helmholtz equation is represented as ∆G = ∆H-T∆S. For what value of ∆G will the reaction be spontaneous?

A) Positive
B) Zero
C)Negative
D) Infinite

4. Standard reduction potential values for P, Q, R and S are +0.34V, -0.76V, +0.0V and -3.05V, respectively. What is the correct order of their reducing power?

A) P>R>Q>S
B) S>R>Q> P
C) Q>P>S>R
D) S>Q>R>P

5. A deep blue complex is formed when conc. ammonium hydroxide is added to an aqueous solution of CuSO4. Which of the following formulae represents the complex?

A) [Cu(NH₃)₄]²⁺
B) [Cu(NH₃)₄]⁺
C) [Cu(NH₃)₂]²⁺
D) [Cu(NH₃)₆]²⁺

6. Which of the following is the maximum possible oxidation state of the transition metals in the 3d series?

A) +3
B) +6
C) +7
D) +8

7. Which of the following are organometallic compounds?
i) CH₃MgI
ii) Ni(CO)₄
iii) C₄H₉Li
iv) C₂H₅ONa
v) (C₄H₉)2SnCl₂

A) i)and ii) only
B) i), ii) and iii) only
C) i), ii), iii) and iv) only
D) i), ii), iii) and v) only

8. The organic compound (X) boiled with water gives phenol. What is the compound (X)?

A) Chlorobenzene
B) Benzenediazonium chloride
C) Benzenesulphonic acid
D) Aniline

9. Which of the following compounds gives a positive carbylamine reaction?

A) (CH₃)₂NH
B) (CH₃)₃N
C) (CH₃)₄N⁺
D) CH₃NH₂

10. A compound ‘Z’ with molecular formula C₃H₄O₄ is heated to produce compound ‘Y’ and CO₂ gas. Heating of ‘Y’ with P₂O₅ forms compound ‘X’ and a water molecule. Identify the compound ‘X’

A) Ethanoic acid
B) Ethanoic anhydride
C) Methanoic anhydride
D) Ethanal

11. A sequence of reactions is represented as:

Predict the name of the reaction that converts (B) to (C).

A) Benzoin condensation
B) Perkins’ condensation
C) Aldol condensation
D) Claisen condensation

Group ‘B’

12. Define instantaneous rate of reaction. Write down the rate expressions of the following reaction.

2N₂O₅(g) → 4NO₂(g) + O₂(g)

If 2.24 litres of O₂ at NTP is produced in 30 minutes, what is the rate of disappearance of N₂O₅(g)?

Answer: Instantaneous rate of reaction is defined as the change in concentration of reactant or product at a particular instant or time. It is denoted by dx/dt.

For a general reaction:

aA + bB → cC + dD

\displaystyle
\frac{dx}{dt}
=
-\frac{1}{a}\frac{d[A]}{dt}
=
-\frac{1}{b}\frac{d[B]}{dt}
= 
+\frac{1}{c}\frac{d[C]}{dt}
=
+\frac{1}{d}\frac{d[D]}{dt}

Now, Rate expression:

\displaystyle
\frac{dx}{dt}
=
-\frac{1}{a}\frac{d[A]}{dt}
=
-\frac{1}{b}\frac{d[B]}{dt}
=
+\frac{1}{c}\frac{d[C]}{dt}
=
+\frac{1}{d}\frac{d[D]}{dt}

\text{Rate expression:}\\ 
\text{Rate} = -\frac{1}{2}\frac{d[N_2O_5]}{dt}
= +\frac{1}{4}\frac{d[NO_2]}{dt}= \frac{d[O_2]}{dt}

\displaystyle
\begin{aligned}
\frac{d[O_2]}{dt}
&= \frac{2.24}{22.4 \times 30 \times 60}\ \text{mol s}^{-1} \\
&= 5.5 \times 10^{-5}\ \text{mol s}^{-1}
\end{aligned}

\displaystyle
\begin{aligned}
\text{We have:}\\ \quad
\frac{1}{2}\frac{d[N_2O_5]}{dt}
&= \frac{d[O_2]}{dt} \\
\text{or,}\quad
\frac{d[N_2O_5]}{dt}
&= (5.5 \times 10^{-5})\times 2 \\
&= 1.1 \times 10^{-4}\ \text{mol s}^{-1}
\end{aligned}

The rate of disappearance of N₂O₅(g) is 1.1 x 10^-4 mol sec^-1.

13. Define standard heat of formation. Calculate the heat of formation of acetic acid from the following data:

C(s) + O₂(g) → CO₂(g), ∆ H = -99 Kcal/mol
H₂(g) + 1/2O₂(g) → H₂O(l), ∆ H = -68 Kcal/mol
CH₃COOH(l) + 2O₂(g) → 2CO₂(g) + 2H₂O(l), ∆ H = -208 Kcal/mol

Answer: The enthalpy change (ΔH⁰f) when one mole of a compound is formed from its elements in their standard states under standard conditions (298 K, 1 atm, and 1 M concentration for solutions) is called the standard heat of formation.

2C(s) + 2H₂(g) + O₂(g) → CH₃COOH(l), ∆ H = ?

We know that,

∆H = ∆H(product) – ∆H(reactant)
or, -208 = [2 x ∆H(CO₂) + 2 x ∆H(H₂O)] – [∆H(CH₃COOH) + ∆H(O₂)]
or, -208 = [2 x (-99) + 2 x (-68)] – [∆H(CH₃COOH) + 0]
or, -208 = [-198 – 136] – ∆H(CH₃COOH)
∆H(CH₃COOH) = -126 Kcal/mol

‘OR’

You are given the standard reduction potentials:

EºZn²⁺/ Zn = – 0.76V,
EºMg²⁺/ Mg = – 2.38 V

Answer the following questions:

1. What is the standard electrode potential?
   Ans: The potential difference between a given half-cell and the standard hydrogen electrode (SHE), when all components are in their standard states (1 M concentration, 1 atm pressure, 25°C or 298 K), is called the standard electrode potential.
   
2. Construct the galvanic cell notation for the given electrodes, indicating the anode and cathode.
   Ans: Anode = Mg²⁺/ Mg 
   Cathode = Zn²⁺/ Zn
   
   Galvanic cell: Mg(s)/Mg²⁺(aq.)//Zn²⁺(aq.)/Zn(s)
   
3. Write the complete cell reaction for the given cell.
   Ans:
   At anode: Mg → Mg²⁺ + 2e^– 
   At cathode: Zn²⁺ + 2e^– → Zn
   
   Complete reaction: Mg + Zn²⁺ → Mg²⁺ + Zn
   
4. Calculate the cell potential for the given cell at 25ºC.
   Ans: E^ocell = E^ocathode – E^oanode = -0.76 – (- 2.38) = +1.62V

14) A chemical reaction occurs as follows:

\underset{(A)}{CH_{2}CH_{2}CH_{2}Br}+\underset{aq.}{KOH}\rightarrow B+KBr

a) Identify the compound ‘B’ in the above reaction.

B=\underset{Propan-1-ol}{CH_{3}CH_{2}CH_{2}OH}

b) Which substitution reaction mechanism occurs in the above reaction and why?
Ans: Bimolecular nucleophilic substitution reaction (SN2) mechanism occurs because it’s a 1° alkyl halide and OH⁻ is a strong nucleophile.

c) What is the product formed when alc. KOH solution is used instead of aq. KOH in the above reaction?

CH₂CH₂CH₂Br + alc. KOH → CH₃-CH=CH₂ (Propene) + KBr + H₂O

d) How can you obtain nitropropane from compound ‘A’?

CH₂CH₂CH₂Br + alc. AgNO₂ → CH₂CH₂CH₂NO₂(1-Nitropropane) + alc. AgBr

15. A list of compounds is given as follows:
P-aminoazobenze, Benzenediazonium chloride, Aniline, Nitrobenzene, Phenol and Benzene

From the above list of compounds, prepare a sequence of reaction chain with suitable conditions and reagents.

‘OR’

An aliphatic compound (A) reacts with SOCI₂ to give (B). The compound (B) is heated with ammonia to produce (C). The compound (C) is further heated with Br₂/KOH to yield (D). The compound (D) gives (E) when treated with NaNO₂ /HCI at low temperature. The compound (E) is a primary alcohol which gives positive iodoform test. Identify compounds (A) to (E) with the reactions involved. 

Ans:

16. Write the chemical equations to illustrate the following reactions:

a. 2,4-DNP test

b. Reimer-Tiemann Reaction

c. Williamson’s reaction

d. Perkin Condensation reaction

e. Wurtz-Fitting reaction

17. You are given the mixture of two compounds A and B.

A = CH₃CH₂CH₂NH₂ and B = CH₃CH₂NHCH₃

How would you separate their mixture by applying Hoffmann’s method ? Give the nitrous acid test to distinguish between compounds A and B.
Ans: Hoffman’s method of separation of primary and secondary amine:

CH₃CH₂CH₂NH₂
Propan-1-amine (Primary amine)

CH₃CH₂NHCH₃
N-methyl etanamine (secondary amine)

1. Treatment with diethyl oxalate

a. Primary amine reacts with diethyl oxalate to form oxamide which is solid.

b. Secondary amine reacts with diethyl oxalate to form oxamic ester which is liquid.

2. Separation of oxamide and oxamic ester: Solid oxamide and liquid oxamic ester are separated by filtration.

3. Regeneration of primary and secondary amine: Oxamide and oxamic ester are separately boiled with KOH to regenerate respective amine.

Nitrous acid test to distinguish A and B:

i. A reacts with nitrous acid to form alcohol with the effervescence of nitrogen gas.

ii. B reacts with nitrous acid to form nitroso-amine, which is a yellow liquid.

18. The metal ‘M’ with electronic configuration [Ar] 3d¹⁰4s² belongs to group IIB in the periodic table and is commonly called ‘Jasta’.

a) Identify metal ‘M’ and write its main ore.
Ans: Metal (M) = Zinc (Zn)
main ore = Zinc Blende (ZnS)

b) Why is metal ‘M’ not considered a transition element?
Ans: Zinc is not considered a transition element due to the following reasons;

• Zinc forms colourless compounds, e.g., ZnSO₄, ZnCl₂, etc.
• Zinc does not show variable valency and oxidation state. (Only +2 oxidation state).
• Its d-electrons are not involved in metallic bonding. It possesses a weak metallic bond.
• Its d- orbital is completely filled and does not show the catalytic property.

c) Draw the vertical retort. Write its reduction reaction.

Reduction reaction: ZnO + C → Zn + CO

d) What happens when metal ‘M’ is exposed to air for a long time?
Ans: When zinc is exposed to air for a long time, basic zinc carbonate is formed.

2Zn + H₂O + O₂ + CO₂ → ZnCO₃.Zn(OH)₂ (Basic Zinc Carbonate)

19. What is rusting? Describe the electrochemical theory of the rusting of iron. List any two methods for the prevention of rusting.
Ans: When iron is exposed to moist air for a long time, a reddish-brown layer of hydrated ferric oxide is formed on the surface of iron, which is called rust, and the process of forming rust is called rusting.

4Fe+ 3O_{2} + 2xH_{2}O\rightarrow \underset{\underset{(Rust)}{Hydrated\ ferric\ oxide}}{2Fe_{2}O_{3}.xH_{2}O}

Electrochemical theory of rusting of iron

This is modern theory. According to this theory, when impure iron comes in contact with water containing dissolved oxygen and carbon dioxide, a galvanic cell is set up. Iron acts as the anode, and the impurities act as the cathode. Water containing dissolved oxygen and carbon dioxide acts as an electrolyte.

At the anode, the Fe atom passes into solution as a ferrous ion.

Fe → Fe⁺⁺ + 2e^–

At the cathode, electrons are used to form hydroxide ions.

2H₂O + O₂ + 4e^– → 4OH^–

Fe⁺⁺ obtained from the anode and OH^– obtained from the cathode combine to form ferrous hydroxide.

Fe⁺⁺ + 2OH^– → Fe(OH)₂

In excess of oxygen and water, ferrous hydroxide is oxidized to ferric hydroxide.

4Fe(OH)₂ + O₂ + 2H₂O → 4Fe(OH)₃

Ferric hydroxide is unstable and loses water to give hydrated ferric oxide.

2Fe(OH)₃ → Fe₂O₃ + 3H₂O

Ferric oxide absorbs moisture from the air and gives hydrated ferric oxide, known as rust.

Fe_{2}O_{3} + xH_{2}O \rightarrow \underset{Rust}{Fe_{2}O_{3}.xH_{2}O}

Prevention of rusting

1. Protective coating

• A coating of oil and grease
• A thin coating of paints and enamels
• By passing steam on red-hot iron

2. Application of corrosion inhibitor: Treating iron with certain solutions like potassium chromate, phosphoric acid, and conc. Nitric acid inhibits rusting.

3. Galvanization (cathodic protection): The process of coating zinc (electropositive metal) over the surface of iron is called galvanization.

Zinc becomes the anode, and iron becomes the cathode. The anode sacrifices itself and prevents the cathode from rusting.

4. Allowing: Iron can be prevented from rusting by allowing it to be coated with chromium, nickel, etc.