Chemistry Test - 18

\(\mathrm{CH}_{3} \mathrm{NH}_{2}\) is primary alkyl amine and undergoes reaction with \(\mathrm{H N O}_{2}\) to give methanol in the following manner.

The reaction takes place at very low temperatures of about \(0^{0}-5^{\circ} \mathrm{C}\).

The intermediate is methyldiazonium, which upon hydrolysis gives methanol. Nitrogen is also produced as a side product.

Hence, Methyl amine reacts with nitrous acid to give ethanol.

The most non-metallic element among the following is \(1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{5}\).

A nonmetal is a chemical element that is mechanically weak in its most stable form, brittle if solid, and usually gains or shares electrons in chemical reactions. It represents halogen. In a period, on moving from left to right, the metallic character decreases and the non-metallic character increases. In any period, halogen is the most non-metallic element.

Ethyl alcohol is industrially prepared from ethylene by Absorbing in H₂SO₄ followed by hydrolysis.

Acid-catalyzed hydration of Alkene:

Alkenes react with water (hydrolysis) in the presence of acid as a catalyst to form alcohols.

In the case of unsymmetrical alkenes, the addition reaction occurs in accordance with Markovnikov’s rule.

Mechanism:
Step 1: Protonation of an alkene to form carbocation by an electrophilic attack of hydronium ion (H₃O⁺).

Step 2: Nucleophilic attack of water on carbocation.

Step 3: Deprotonation to form an alcohol.

Industrial preparation of Ethyl alcohol:

• Ethanol reacts with sulphuric acid to form an intermediate product.
• This is followed by hydrolysis to give ethanol and H₂SO₄ as byproducts.
• Here, the H₂SO₄ formed as a byproduct is reused for further reaction with ethyl alcohol.

The reaction is as follows:

\(CH _{2}= CH _{2} \stackrel{ H _{2} SO _{4}}{\longrightarrow} CH _{3}- CH _{2}- HSO _{4} \stackrel{\text { Hydrolysis }}{\longrightarrow} CH _{3} CH _{2}- OH + H _{2} SO _{4}\)

The production of dihydrogen obtained from coal gasification can be increased by reacting carbon monoxide of syngas mixture with steam in presence of a catalyst iron-chromium. This process is called a water-gas shift reaction.

When steam reacts with hydrocarbons at a very high temperature in the presence of some catalyst, then hydrogen is obtained. Now, the mixture of carbon monoxide and water is called water gas. This water gas is also known as syngas or synthesis gas. The reaction given below is known as the water-gas shift reaction.

\(CO ( g )+ H _{2} O \ce{->[673K][Catalyst]}CO _{2}( g )+ H _{2} O ( g )\)

Weight of empty glass vessel \(=40 \mathrm{gm}\)

Weight of glass vessel filled with liquid \(=135 \mathrm{gm}\)

\(\therefore\) Weight of liquid \(=135-40=95 \mathrm{gm}\)

Given density of liquid \(=0.95 \mathrm{gm} \mathrm{ml}^{-1}\)

\(\therefore\) Volume of liquid \(=\frac{95}{0.95}=100 \mathrm{ml}\)

Weight of glass filled with ideal gas \(=40.5 \mathrm{gm}\)

\(\therefore\) Weight of gas \(=40.5-40=0.5 \mathrm{~g}\)

Let the Molar mass \(=\mathrm{M}\)

\(\therefore\) Moles of gas \(=\frac{0.5}{\mathrm{M}}\)

\(\therefore\) Now applying ideal gas equation,

\(\mathrm{pV}=\mathrm{nRT} \)

\(\Rightarrow 0.82 \times \frac{100}{1000}=\frac{0.5}{\mathrm{M}} \times 0.082 \times 250 \)

\(\Rightarrow \mathrm{M}=0.5 \times 250=125 \mathrm{~g} / \mathrm{mol}\)

Here compound \(\mathrm{A}_{3}\left(\mathrm{BC}_{4}\right)_{2}\) is possible as in it sum of oxidation numbers is zero as follows:

\(3(+2)+2 \times 5+8(-2)=0\)

A nitrogen tank of fixed volume used where number of moles of nitrogen is fixed.

\(\therefore \mathrm{V}=\) constant

\(\mathrm{n}=\) constant

\(\mathrm{R}=\) constant

From ideal gas equation,

\(\mathrm{PV}=\mathrm{nRT} \)

\(\Rightarrow \mathrm{P} \propto \mathrm{T}[\mathrm{AsV}, \mathrm{n}, \mathrm{R}=\text { constant }]\)

Here, initially \(P_1=30 \mathrm{~atm}, T_1=300 \mathrm{~K}\)

Finally, \(\mathrm{P}_2=\) ?, \(\mathrm{T}_2=318 \mathrm{~K}\)

\(\therefore \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{T}_1}{\mathrm{~T}_2} \)

\(\Rightarrow \frac{30}{\mathrm{P}_2}=\frac{300}{318} \)

\(\Rightarrow \mathrm{P}_2=31.8 \simeq 32\)

\(\mathrm{P_{A}=P_{B} \times A \cdot P_{A}^{0}=X_{B} \cdot P_{B}^{0}}\)

\(\Rightarrow \mathrm{\frac{X_{A}}{X_{B}}=\frac{P_{B}^{0}}{P_{A}^{0}}}\)

\(\Rightarrow \mathrm{1+\frac{X_{\mathrm{A}}}{X_{\mathrm{B}}}=1+\frac{\mathrm{P}_{\mathrm{B}}^{0}}{\mathrm{P}_{\mathrm{A}}^{0}}}\)

\(\Rightarrow \mathrm{\frac{1}{X_{B}}=\frac{P_{\mathrm{A}}^{0}+\mathrm{P}_{\mathrm{B}}^{0}}{\mathrm{P}_{\mathrm{A}}^{0}}}\)

\(\Rightarrow \mathrm{X_{B}=\frac{P_{A}^{0}}{P_{A}^{0}+P_{B}^{0}}}\)

Similarly,

\(\mathrm{X_{A}=\frac{P_{B}^{0}}{P_{A}^{0}+P_{B}^{0}}}\)

\(\Rightarrow \mathrm{P_{T}=P_{A}^{0} \cdot X_{A}+P_{B}^{0} \cdot X_{B}}\)

\(=\mathrm{\frac{2 P_{A}^{0} P_{B}^{0}}{P_{A}^{0}+P_{B}^{0}}}\)