Chemistry Test - 52

\(\begin{aligned} & \Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} \ldots \text {. (i) } \\ & \text { where, } \Delta \mathrm{G}^{\circ}=\text { standard Gibb's free energy change } \\ & \Delta \mathrm{H}^{\circ}=\text { standard change in enthalpy } \\ & \Delta \mathrm{S}^{\circ}=\text { standard entropy change } \\ & \text { Also, } \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\ & \text { where, } \mathrm{n}=\text { number of electrons } \\ & \mathrm{F}=\text { Faraday constant } \\ & \mathrm{E}^{\circ}=\text { standard cell potential } \\ & \therefore \text { Equation (i) is, } \\ & -\mathrm{nFE}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} \\ & -\mathrm{nFE}^{\circ}-\Delta \mathrm{H}^{\circ}=\mathrm{T} \Delta \mathrm{S}^{\circ} \\ & \Rightarrow \Delta \mathrm{S}^{\circ}=\frac{\mathrm{nFE}+\Delta \mathrm{H}^{\circ}}{\mathrm{T}}\left(2 \times 96487 \mathrm{Cmol}^{-1} \times 4.315 \mathrm{~V}\right) \\ & =\frac{+(-825.2 \mathrm{~kJ} / \mathrm{mol})}{298 \mathrm{~K}} \\ & =\frac{832.682 \times 10^3 \mathrm{~J} \mathrm{~mol}-825.2 \times 10^3 \mathrm{~J} / \mathrm{mol}}{298 \mathrm{~K}} \\ & =25.11 \mathrm{~J} / \mathrm{Kmol}\end{aligned}\)

Given that:

The density of the solution is \(1.41 \mathrm{~g} / \mathrm{mL}\).

Then,

\(1000 \mathrm{~g}\) of the solution will have a volume \(=\frac{\text { Mass }}{\text { Density }}=\frac{1000}{1.41}=709 \mathrm{~ml}\)

And, the mass per cent of nitric acid is \(69 \%\).

Then,

\(1000 \mathrm{~g}\) of the solution will have \(=1000 \times \frac{69}{100}\)

\(=690 \mathrm{~g}\) of nitric acid.

We know that:

The molarity of solution is given by:

\(=\frac{\text { Mass of nitric acid }}{\text { Molar mass of nitric acid } \times \text { Volume of solution }}\)

\(=\frac{690}{63 \times 709}\)

\(=0.0155 \mathrm{~M}\)

Arrhenius equation

\(\mathrm{K}=\mathrm{Ae}^{-\mathrm{Ea} / \mathrm{RT}}\)

\(\mathrm{As} \mathrm{T} \rightarrow \infty\)

\(\mathrm{RT} \rightarrow \infty\)

\(\frac{-\mathrm{Ea}}{\mathrm{RT}} \rightarrow 0\)

\(\mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}} \rightarrow 1\)

So, \(\mathrm{K} \rightarrow \mathrm{A}\) as \(\mathrm{T} \rightarrow \infty\)

\(\therefore\) Value of \(\mathrm{K}\) as \(\mathrm{T} \rightarrow \infty=6.0 \times 10^{14} \mathrm{~s}^{-1}\)

Disproportionation reaction are the reactions in which the same element/compound get oxidized and reduced simultaneously.

The reactions are as follows:

(I) \(\left.2 \mathrm{Cu}^{+1} \rightarrow \mathrm{Cu}^{+2}+ \mathrm{Cu}^{0}\right\}\) Disproportionation

(II) \(\left.3 \mathrm{MnO}_{4}^{2-}+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{MnO}_{4}^{-}+\mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\right\}\) Disproportionation

(III) \(\left.2 \mathrm{KMnO}_{4} \stackrel{\Delta}{\longrightarrow} \mathrm{K}_{2} \mathrm{MnO}_{4}+\mathrm{MnO}_{2}+\mathrm{O}_{2}^{+6}\right\}\) Not a Disproportionation

(IV) \(\left.2 \mathrm{MnO}_{4}^{-}+3 \mathrm{Mn}^{+2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 5 \mathrm{MnO}_{2}+4 \mathrm{H}^{+}\right\}\) Not a Disproportionation

The covalent charter of P−H bond depends on the formal charge distributed on each P−H bond

In PH₄⁺it is +1/4 = +0.25, in P₂H₅⁺it is +1/5 = +0.2 and in P₂H₆²⁺ it is +2/6 = +0.33.

The higher the formal charge the lesser the covalent character due to more polarisation. Thus the least covalent P−H bond is present in P₂H₆²⁺

Actual flame temperature is always lower than the adiabatic flame temperature because there isno possibility of obtaining complete combustion at a high temperature and there isalways a loss of heat from the flame.

The constant volume adiabatic flame temperature is the temperature that results from a complete combustion process that occurs without any work, heat transfer or changes in kinetic or potential energy. The constant pressure adiabatic flame temperature is the temperature that results from a complete combustion process that occurs without any heat transfer or changes in kinetic or potential energy. Its temperature is lower than the constant volume process because some of the energy is utilized to change the volume of the system (i.e., generate work).

The number of secondary hydrogens in \(2, 2\)-dimethyl butane is \(2\) because two hydrogen atoms are connected to the one carbon atom along with the methyl group.

\(\begin{array}{}\mathrm{\quad\ H\ \ \ CH_3}\\|\quad|\\\mathrm{H_3C-C-C-CH_3}\\|\quad |\\\mathrm{\quad H \ \ \ CH_3}\end{array}\)

Ammonia acts as a very good ligand but ammonium ion does not form complexes becauseNH₄⁺ion does not have any lone pair of electrons.

Complexes are formed by the donation of a pair of electrons from ligand to metal. In ammonia, N atom has one lone pair of electrons. So it can form complexes.Nitrogen donates this lone pair of electrons to the proton to form an ammonium ion.NH₄⁺ ion does not possess any lone pair of electrons which it can donate to central metal ion therefore it does not form complexes.Thus, Ammonia can be a very good ligand but ammonium ion does not form complexes.

The above reaction is known as Etard reaction.

• The Etard Reaction is a chemical reaction that involves the direct oxidation of an aromatic or heterocyclic bound methyl group to an aldehyde using chromyl chloride. 
• When toluene is reacted with Chromyl Chloride, then a chromium complex is formed (Etard Complex) whose hydrolysis gives Benzaldehyde.

\(p \pi-p \pi\) back-bonding is the main factor responsible for the weak acidic nature of B-F bonds in BF₃.