Chemistry Test-2

Iso butane reacts with chlorine in the presence of light to form t-butyl chloride is an exothermic reaction.

This reaction occur through most stable tertiary free radical, then C − Cl bond is formed. C − Cl bond is more stronger than C−Br bond. So this reaction is most exothermic reaction.

CH₃CH(CH₃)₂ + Cl₂/hv → CH₃C^.(CH₃)₂ → CH₃ClC(CH₃)₂

The IUPAC of dexamethasone is \(9 \alpha\)-Fluoro- \(11 \beta, 17 \alpha, 21\)-trihydroxy- \(16 \beta\)-methylpregna- 1,4 -diene- 3,20 -dione.

Dexamethasone is a glucocorticoid medication used to treat rheumatic problems, a number of skin diseases, severe allergies, asthma, chronic obstructive lung disease, croup, brain swelling, eye pain following eye surgery.

Given:

Mole fraction of solute in first solution \(= 0.2\)

According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of solute, i.e.,

\(\frac{p^{\circ}-p}{p^{\circ}}=\frac{n}{n+N}\)

or, \(\frac{\Delta p}{p^{\circ}}=\frac{n}{n+N} \)

The decrease in vapour pressure is \(10\) mm Hg:

\(\frac{10}{p^{\circ}}=0.2\)

\(\therefore p^{\circ}=50 \mathrm{~mm}\) ....(1)

For other solution of same solvent, the decrease in vapor pressure is \(20\) mm Hg:

\(\frac{20}{p^{\circ}}=\frac{n}{n+N}\)

or, \(\frac{20}{50}=\frac{n}{n+N}\) (from (1))

\(0.4=\frac{n}{n+N}\) (mole fraction of solute)

\(\because\) Mole fraction of solvent \(+\) mole fraction of solute \(=1\)

So, mole fraction of solvent \(=1-0.4=0.6\)

Freons are chlorofluorocarbons and are responsible for the depletion of the ozone layer in the upper strata of the atmosphere. They are used as propellants in aerosol spray cans, refrigerators, fire fighting reagents, etc. They are stable compounds and are chemically inert. They do not react with any substance with which they come in contact and thus float through the atmosphere unchanged and eventually enter the stratosphere. There they absorb UV radiations and break down liberating free atomic chlorine which causes decomposition of ozone. This results in the depletion of the ozone layer.

\(\dot{ Cl }+ O _{3} \rightarrow Cl \dot{ O }+ O _{2} \)

\( ClO + O _{3} \rightarrow \dot{ Cl } +2 O _{2}\)

Genius Reaction is,

N₂O₄(g) + 3CO(g) -----> N₂O(g) + 3CO₂(g)

We know,

Heat of reaction ( ∆rH°) = ( sum of heat of formation of products ) - ( sum of heat of formation of reactants )

e.g., ∆rH° = [∆fH°(N₂O) + 3fH°(CO₂) ] - [∆fH°(N₂O₄) + 3fH°(CO) ]

Given, enthalpy of formation of CO, CO₂ , N₂O, and N₂O₄ are -110, -393 , 81 and 9.7 KJ/mol respectively .

Now, ∆rH° = [81 + 3(-393)] - [9.7 + 3(-110)]

= -777.7 KJ

According to the question,

(i)\(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; \Delta_{\mathrm{c}} \mathrm{H}^{\Theta}=-890.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

(ii) \(\mathrm{C}\) (s) \(+2 \mathrm{O}_{2}\) (g) \(\rightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \Delta_{\mathrm{c}} \mathrm{H}^{\Theta}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

(iii) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}\) (g) \(\rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; \Delta_{\mathrm{c}} \mathrm{H}^{\Theta}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Thus, the desired equation is the one that represents the formation of \(\mathrm{CH}_{4}(\mathrm{~g})\) that is as follows:

\(\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g}) ; \Delta_{\mathrm{f}} \mathrm{H}_{\mathrm{CH}_{4}}=\Delta_{\mathrm{c}} \mathrm{H}_{\mathrm{c}}+2 \Delta_{\mathrm{c}} \mathrm{H}_{\mathrm{H}_{2}}-\Delta_{\mathrm{c}} \mathrm{H}_{\mathrm{CO}_{2}}\)

Substituting the values in the above formula :

Enthalpy of formation \(\mathrm{CH}_{4}(\mathrm{g})=(-393.5)+2 \times(-285.8)-(-890.3)=-74.8 \mathrm{kJmol}^{-1}\)

As given, the temperature is increased from \(\mathrm{T}_{1}\) to \(\mathrm{T}_{2}\), then new constants become \(\mathrm{k}_{3}\) and \(\mathrm{k}_{4}\) respectively.

Rate constant increases with increase in temperature.

So, \(k_{1}<\mathrm{k}_{3}\) and \(\mathrm{k}_{2}<\mathrm{k}_{4}\)

\(\delta=\frac{\text { Dipole moment }}{\mathrm{d}}\)

\(=\frac{1.2 \mathrm{D}}{1.0 \times 10^{-8} \mathrm{~cm}}\)

\(=\frac{1.2 \times 10^{-8} \text { esu } \mathrm{cm}}{1.0 \times 10^{-8} \mathrm{~cm}}\)

\(=1.2 \times 10^{-8}\) esu

The fraction of electronic charge, e is

\(=\frac{1.2 \times 10^{-10} \mathrm{esu}}{4.8 \times 10^{-10} \mathrm{esu}} \)

\(=0.25 \mathrm{e}\)

\(=25 \%\) of e

Argon element has three shells that are completely filled with electrons.

Third-most abundant gas in the Earth's atmosphere. It is produced industrially by the fractional distillation of liquid air. It is commonly used to fill incandescent light bulbs. It is used in tungsten filament. All Helium, Neon, and Argon have completely filled outermost shells. But it is asked which element have all three shells filled. So, Argon-Ar (2,8,8) has all three shells filled completely filled with electrons.

Given elements are:

Magnesium \((Mg)\), Aluminium \((Al)\) and Sodium\((Na)\)

Electrode reactions of these elements involved are:

\(M g^{+2}+2 e^{-} \rightarrow M g\)

\(A l^{+3}+3 e^{-} \rightarrow A l\)

\(N a^{+}+e^{-} \rightarrow N a\)

From the above reactions, we can see that:

1 mole of \(M g\) requires 2 Faradays.

1 mole of \(A l\) requires 3 Faradays.

1 mole of sodium requires 1 Faraday.

Ratio of number of Faraday \(=2: 3: 1\)

In the decomposition of hydrogen peroxide, two successive reactions of H₂O₂, with bromide and then with bromine, take place. Because these two reactions together serve as a catalytic pathway for hydrogen peroxide decomposition, both of them must have significantly lower activation energies than the uncatalyzed decomposition, as shown schematically in Figure.

Bromine acts as negative catalyst in the reaction. Bromine forms an intermediate (Br−) on reacting with H₂O₂ following reduction of bromine and then to show oxidation of Br− to Br₂ by H₂O₂.

A complex has the molecular formula \(\mathrm{Co} .5 \mathrm{NH}_{3} . \mathrm{NO}_{2} . \mathrm{Cl}_{2}\). One mole of this complex produces three moles of ions in an aqueous solution. On reacting this solution with excess of \(\mathrm{AgNO}_{3}\) solution, we get 2 moles of white ppt. The complex is\(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right] \mathrm{Cl}_{2}\).

\(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right] \mathrm{Cl}_{2}\) can form three ions, one \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right]^{2+}\) and two \(\mathrm{Cl}^{-}\)ions.

As 1 mole of complex gives 2 moles of \(\mathrm{Cl}^{-}\)ions, 2 moles of AgCI will be precipitated.

Nitrogen molecule \(\left(\mathrm{N}_{2}\right)\) is relatively more stable. Each \(\mathrm{N}\) atom has completed its octet and there is a triple bond between two \(\mathrm{N}\) atoms. In other molecules, there is either a single bond or a double bond between two atoms. A triple bond is stronger than a single or double bond.

Ellingham diagram represents the change in free energy (ΔG) with temperature.

An Ellingham diagram is a graph showing the temperature dependence of the stability for compounds. This analysis is usually used to evaluate the ease of reduction of metal oxides and sulphides.

The diagrams are useful in predicting the conditions under which an ore will be reduced to its metal. The analysis is thermodynamic in nature and ignores reaction kinetics. Thus, processes that are predicted to be favourable by the Ellingham diagram can still be slow.