Chemistry Test - 44

'On reaction with nitrous acid arylamines produce phenol'is incorrect for primary amines.

Primary amines are those amines whose one hydrogen is substituted by some other groups like \(\mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2}\) etc or those amines which have one \(\mathrm{C}-\mathrm{N}\) bond present in them. Example:

\(\mathrm{H}_{3} \mathrm{C}-\mathrm{NH}_{2}, \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{NH}_{2}\)

Aryl amines (if it is primary) on reaction with nitrous acid gives diazonium salt.

The Element present in third period and seventeenth group of the periodic table is Chlorine and atomic number of Chlorine is 17.

Chlorine is a chemical element with the symbol Cl and atomic number 17. The second-lightest of the halogens, it appears between fluorine and bromine in the periodic table and its properties are mostly intermediate between them. Chlorine is a yellow-green gas at room temperature.

Phenol is also known as Carbolic acid. In 1865, the British surgeon Joseph Lister used phenol as an antiseptic to sterilize his operating field.

Phenol is an aromatic organic compound characterized by a hydroxyl (―OH) group attached to a carbon atom that is part of an aromatic ring.

Besides serving as the generic name for the entire family, the term phenol is also the specific name of its simplest member, monohydroxybenzene (C₆H₅OH).

\(\begin{aligned} & \mathrm{CaCO}_3(\mathrm{~s}) \xrightarrow{\Delta} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \\ & \mathrm{MgCO}_3(\mathrm{~s}) \xrightarrow{\Delta} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \\ & \end{aligned}\)

Let the weight of \(\mathrm{CaCO}_3\) be \(\mathrm{xgm}\)

\(\therefore\) weight of \(\mathrm{MgCO}_3=(2.21-\mathrm{x}) \mathrm{gm}\)

Moles of \(\mathrm{CaCO}_3\) decomposed \(=\) moles of \(\mathrm{CaO}\) formed

\(\frac{x}{100}=\) moles of \(\mathrm{CaO}\) formed

\(\therefore\) weight of \(\mathrm{CaO}\) formed \(=\frac{\mathrm{x}}{100} \times 56\)

Moles of \(\mathrm{MgCO}_3\) decomposed = moles of \(\mathrm{MgO}\) formed

\(\frac{(2.21-\mathrm{x})}{84}=\) moles of \(\mathrm{MgO}\) formed

\(\therefore\) weight of \(\mathrm{MgO}\) formed \(=\frac{2.21-\mathrm{x}}{84} \times 40\)

\(\Rightarrow \frac{2.21-\mathrm{x}}{84} \times 40+\frac{\mathrm{x}}{100} \times 56=1.152\)

\(\therefore \mathrm{x}=1.1886 \mathrm{~g}=\) weight of \(\mathrm{CaCO}_3\)

& weight of \(\mathrm{MgCO}_3=1.0214 \mathrm{~g}\)

Molality of a solution is defined as the number of moles of solute dissolved per kg of solvent.

\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}=10 \%\) by weight.

Thus, in 100 g solution mass of glucose \(=10 \mathrm{~g}\)

and, mass of water \(=90 \mathrm{~g}\)

Number of moles of glucose \(=\frac{\text { mass }}{\text { molar mass }}\)

\(n=\frac{10}{180}\)

\(=\frac{1}{18} ~\mathrm{m o l}\)

Mass of solvent \(=90 \mathrm{~g}=0.09 \mathrm{~kg}\)

Molality, \(m=\frac{\left(\frac{1}{18}\right)}{0.09}\)

\(m=\frac{100}{18 \times 9}\)

\(=0.617 \mathrm{~molal}\)

\( \approx 0.6 \mathrm{~molal}\)

Yes, the potential of mercury cell does not change during life time beacuse no ion is involved in the reason whose concentration may change and electrolyte is a paste of \(KOH\) and \(ZnO\) in mercury cell.

Distribution of molecular velocities at two different temperatures is shown below.

Note: At higher temperature more molecules have higher velocities and less molecules have lower velocities. As evident from fig., it is clear that with the increase in temperature the most probable velocity increases but the fraction of such molecules decreases.

\(\begin{aligned} & \mathrm{FeO}(\mathrm{s})+\mathrm{C}_{\text {(graphite) }} \rightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \\ & \Delta \mathrm{H}_{\text {reaction }}^{\circ}=\Delta \mathrm{H}_{\text {f(product })}^{\circ}-\Delta \mathrm{H}_{\mathrm{f}(\text { reactant })} \\ & \end{aligned}\)

\(=\left[\Delta \mathrm{H}_{\mathrm{f}(\mathrm{Fe})}^{\circ}+\Delta \mathrm{H}_{\mathrm{f}(\mathrm{CO})}^{\circ}\right]-\left[\Delta \mathrm{H}_{\mathrm{f}(\mathrm{Fe})}^{\circ}-\Delta \mathrm{H}_{\mathrm{f}(\mathrm{C})}^{\circ}\right] \)

\(=[0+(-110.5)]-[-266.3-0] \)

\(=156 \mathrm{~kJ}^{-1} \mathrm{~mol}^{-1} \)

\(\Delta \mathrm{S}_{\text {reaction }}^{\circ}=\Delta \mathrm{S}_{\text {product }}^{\circ}-\Delta \mathrm{S}_{\text {reactant }}^{\circ} \)

\(=\left[\Delta \mathrm{S}^{\circ}{ }_{(\mathrm{Fe})}+\Delta \mathrm{S}^{\circ}{ }_{(\mathrm{CO})}\right]-\left[\Delta \mathrm{S}^{\circ}{ }_{(\mathrm{FeO})}-\Delta \mathrm{S}^{\circ}{ }_{(\mathrm{O})}\right] \)

\( =[27.28+197.6]-[57.49+5.79] \)

\(=161 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \)

According to Gibb's equation,

\(\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}\)

The reaction becomes spontaneous when \(\Delta \mathrm{G}^{\circ}\) is atleast zero or negative.

\(0=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} \)

\(\mathrm{T} \Delta \mathrm{S}^{\circ}=\Delta \mathrm{H}^{\circ} \)

\(\Rightarrow \mathrm{T}=\frac{\Delta \mathrm{H}^{\circ}}{\Delta \mathrm{S}^{\circ}}=\frac{156 \mathrm{~kJ} \mathrm{~mol}^{-1}}{161 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}} \)

\(=\frac{156000 \mathrm{~mol}^{-1}}{161 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}}=964 \mathrm{~K}\)

The temperature at which reaction becomes spontaneous is 964 K.

\(\begin{aligned} & \mathrm{H}_{2(\mathrm{~g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HBr}_{(\mathrm{g})}, \Delta \mathrm{H}_{\mathrm{f}}^0=\text { ? } \\ & \Delta \mathrm{H}^{\circ}{ }_{\mathrm{f}}=\Sigma(\text { B.E. })_{\text {reactants }}-\Sigma(\text { B.E. })_{\text {products }} \\ & =(\text { B.E. })_{\mathrm{H}-\mathrm{H}}+(\mathrm{B} . \mathrm{E} .)_{\mathrm{Br}-\mathrm{Br}}-2(\mathrm{~B} . \mathrm{E})_{\mathrm{H}-\mathrm{Br}} \\ & =[433+192]-2(364) \mathrm{kJ} \mathrm{mol}^{-1} \\ & =(625-728) \mathrm{kJ} \mathrm{mol}^{-1}=-103 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \end{aligned}\)

Disproportionation reaction is a redox reaction in which a compound of intermediate oxidation state converts to two different compounds, one of higher and one of lower oxidation states.

(A) \(\stackrel{+4}{\mathrm{CH}_{4}}+2 \stackrel{0}{\mathrm{O}_{2}} \rightarrow \stackrel{+4-2}{\mathrm{CO}}_{2}+2 \mathrm{H}_{2} \mathrm{O}\): Not a disproportionate reaction.

(B) \(\stackrel{4+}{\mathrm{CH}_{4}}\stackrel{0}{+4 \mathrm{Cl}_{2}} \rightarrow \stackrel{+4-1}{\mathrm{CCl}_{4}}+4 \mathrm{HCl}\): Not a disproportionate reaction.

(C) \(\stackrel{0}{2 \mathrm{~F}_{2}}\stackrel{-2}{+2 \mathrm{OH}^{-}} \rightarrow \stackrel{-1}{2 \mathrm{~F^-}}+\stackrel{+2}{\mathrm{O}} \stackrel{-1}{~ \mathrm{F}_{2}} +\mathrm{H}_{2} \mathrm{O}\): Redox reaction but not disproportionate reaction.

(D) \(\stackrel{+4}{2 \mathrm{NO}_{2}}+2 \mathrm{OH}^{-} \rightarrow \stackrel{+3}{\mathrm{NO}_{2}^{-}}+\stackrel{+5}{\mathrm{NO}_{3}^{-}}+\mathrm{H}_{2} \mathrm{O}\): \(\mathrm{N}\) changes from \(+4\) to \(+3\) and \(+5\). Thus its a disproportionate reaction.