Chemistry Test - 5

The process by which liquid hydrocarbons can be converted to gaseous hydrocarbons iscracking.

On cracking or pyrolysis, the hydrocarbon with higher molecular mass gives a mixture of hydrocarbons having lower molecular mass. Therefore, we can say that by cracking a liquid hydrocarbon can be converted into a mixture of gaseous hydrocarbons.

Separating the oxidation and reduction reaction from the redox reaction:

\(3 \mathrm{~N}_{2} \mathrm{H}_{4}+2 \mathrm{BrO}_{3}^{-} \rightarrow 3 \mathrm{~N}_{2}+2 \mathrm{Br}^{-}+6 \mathrm{H}_{2} \mathrm{O}\)

Assigning the oxidation number on central atom (N and Br) in each molecules by considering oxidation number of \(\mathrm{H}=+1, \mathrm{O}=-2\),we get oxidation state as:

\(3\overset {-2}{N}_2H_4 + 2[\overset{+5}{Br}O_3]^- \rightarrow 3\overset{0}{N_2} + 2\overset{-1}{Br^-} +6H_2O\)

As the oxidation number of \(N\) changes from \(-2\) to 0 as:

\(\mathrm{N}_{2} \mathrm{H}_{4} \rightarrow N_{2}\)

Its a oxidation reaction where \(N_{2} H_{4}\) gets oxidized.

Similarly, as the oxidation number of \(B r\) changes from \(+5\) to \(-1\) as:

\(\mathrm{BrO}_{3}^{-} \rightarrow B r^{-}\)

Its a reduction reaction where \(\mathrm{BrO}_{3}^{-}\) gets reduced.

The sulfur is the central atom in the molecule SO₃ and has six valence electrons. The oxygen atom given is the divalent therefore, the number of monovalent electrons is zero. There is no cationic charge or anionic charge present on the molecule hence the value is taken as zero. So, the hybridization of the SO₃ molecule is three which shows the hybridization as sp² hybridization which means there is no sigma bond present in the structure and there is a double bond present.

In SO₃ molecule, S atom remains sp² hybrid, so, it is trigonal planar structure.

Starting from alumina would obtain 'Aluminium nitride'.

Aluminium nitride can be synthesized by the Carbothermal reduction of an Aluminium oxide in the presence of gaseous nitrogen at \(1800^{\circ} \mathrm{C}\). The reaction is as follows:

\(\mathrm{Al}_{2} \mathrm{O}_{3}+3 C+\mathrm{N}_{2} \stackrel{1800^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{Al} \mathrm{N}+3 \mathrm{CO}\)

Aluminium nitride is a solid nitride of aluminium. It has a high thermal conductivity of up to \(321 \mathrm{~W} /(\mathrm{m} \cdot \mathrm{K})\) and is an electrical insulator. Its wurtzite phase has a bandgap of \(\sim 6 \mathrm{eV}\) at room temperature and has a potential application in optoelectronics operating at deep ultraviolet frequencies.

When C₂H₅OH reacts with sodium metal and forms C₂H₅ONa and on heating with H₂SO₄ giver diethyl ether.

C₂H₆ compound has 7 covalent bonds. The structure of the Ethane molecule is given below.

Given: Enthalpy of vaporisation of a substance is: \(8400{~J} {~mol}^{-1}\)

Temperature or boiling point of substance is: \(-173.15^{\circ}{C}\) 

We have Kelvin temperature\(=\) Celsius temperature+\(273.15^{\circ}{C}\)

\(=-173.15^{\circ}{C}+273.15^{\circ}{C}\)
\(=100{~K}\)

Change in entropy, \(\Delta S=\frac{q_rev}{T}\)

\(=\left(\frac{8400}{100}\right)\) \(=84~J m o l^{-1} K^{-1}\)                             

For the reaction,

\(\mathrm{m}_{1} \mathrm{~A}+\mathrm{m}_{2} \mathrm{~B} \rightarrow \mathrm{n}_{1} \mathrm{C}+\mathrm{n}_{2} \mathrm{D}\)

At equilibrium,

\(-\frac{1}{\mathrm{~m}_{1}} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=+\frac{1}{\mathrm{n}_{1}} \frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}\)

\(\Rightarrow -\frac{\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}}{\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}}=\frac{\mathrm{m}_{1}}{\mathrm{n}_{1}}\)

The increasing order of the first ionisation enthalpies of the elements B, P, S and \(\mathrm{F}\) (lowest first) is\(P

In general as we move from left to right in a period, the ionisation enthalpy increases with increasing atomic number. The ionisation enthalpy decreases as we move down a group. \(P\left(1 s^{2} 2 s^{2} 2 p^63 s^{2} 3 p^{3}\right)\) has a stable half filled electronic configuration than \(\mathrm{S}\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4}\right)\). For this reason, ionisation enthalpy of \(P\) is higher than S.

In the given options,

(A) Concentration of \(H^{+}\)in \(1.0 \mathrm{M} \mathrm{~HCl}=1 M\)

(B) Since, \(p H\) of solution is 4 .

So, concentration of \(H^{+}=10^{-4} M\)

(C) \(p H\) of pure water \(=7\)

So, concetration of \(H^{+}=10^{-7} M\)

(D) Concentration of \(H^{+}\)in \(1.0 \mathrm{M~NaOH}\) is much less than the all the above options.

Since, concentration of \(H^{+}\)is highest in option (A). So, potential of option (A) is maximum.