3 Marks Question

Question 13 Marks

State Kohlrausch's law of independent migration of ions. How can the degree of dissociation of acetic acid in a solution be calculated from its molar conductivity data?

Answer

Kohlrausch law of independent migration of ions: It states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. If $\lambda^{\circ} Na ^{+}$and $\lambda^{\circ} Cl ^{-}$are limiting molar conductivity for sodium chloride is given by $
\lambda_m^0(NaCl)=\lambda^o Na+\lambda^0 Cl^{-}
$
Calculation of degree of dissociation of weak electrolyte like acetic acid. The degree of dissociation $\alpha$ is given by: $
\alpha=\frac{\lambda_m}{\lambda_m^0}
$ where $\lambda_m$ be molar conductivity and $\lambda_m^0$ be the limiting molar conductivity.

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Question 23 Marks

Draw the structures of major monohalo products in each of the following reactions:
(1)

(2)

(3)

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Question 33 Marks

Calculate the emf of the following cell:
$ Mg ( s )\left| Mg ^{2+}(0.2 M ) \| Ag ^{+}\left(1 \times 10^{-3} M \right)\right| Ag ( s )$
$E ^0\left( Ag ^{+} / Ag \right)=0.80 V$
$E ^0\left( Mg ^{2+} / Mg \right)=-2.37 V$

Answer

applying nernst equation
$E_{\text {cell }}=E_{\text {cell }}^0-\frac{0.0591}{2} \log \frac{\left[Mg^{2+}\right]}{\left[Ag^{2+}\right]}$
$E^0\left(A g^{+} / Ag\right)-E^0\left(Mg^{2+} / Mg\right)-\frac{0.0591}{2} \log \frac{0.2}{\left(10^{-3}\right)^2}$
$=+0.80 V-(-2.37 V)-\frac{0.0591}{2} \log \left(2 \times 10^5\right)$
$=+3.17 V-\frac{0.0591}{2}\left[\log 2+\log 10^5\right]$
$=+3.17 V-\frac{0.0591}{2} \times 5.3010$
$=+3.17 V-0.1566 V$
$=3.0134 V$

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Question 43 Marks

How would you bring about the following conversions?
i. Propanal to butanone
ii. Benzaldehyde to benzophenone
iii. Benzoyl chloride to benzonitrile

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Question 53 Marks

Write the reactions and conditions for the following conversions:
i. 2-Propanone into 2-methyl-2-Proponal
ii. n-Propyl alcohol into hexane

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Question 63 Marks

i. Write the mechanism of the following reaction:
$2 CH _3 CH _2 OH \xrightarrow[413 K]{\stackrel{H^{+}}{\longrightarrow}} CH _3 CH _2 OCH _2 CH _3+ H _2 O$
ii. Write the preparation of phenol from cumene.

Answer

(1)

(2)

(3)

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Question 73 Marks

In a reaction between $A$ and $B,$ the initial rate of reaction was measured for different initial concentrations of $A$ and $B$ as given below$:$

$A/mol \ L^{-1}$	$0.20$	$0.20$	$0.40$
$B/mol \ L^{-1}$	$0.30$	$0.10$	$0.05$
$ro/mol \ L^{-1}s^{-1}$	$5.07 \times 10^{-5}$	$5.07 \times 10^{-5}$	$1.43 \times 10^{-4}$

What is the order of the reaction with respect to $A$ and $B \ ?$

Answer

Consider the order of the reaction with respect to $A$ is $x$ and with respect to $B$ is $y .$
Therefore, $r_0=k[A]^x[B]^y$
$5.07 \times 10^{-5}=k[0.20]^x[0.30]^y \ldots \ldots(\text { (i) }$
$5.07 \times 10^{-5}=k[0.20]^x[0.10]^y \ldots . .(\text { ii) }$
$1.43 \times 10^{-4}=k[0.40]^x[0.05]^y \ldots . .(\text { iii) }$
Dividing equation $(i)$ by $(ii),$ we obtain
$\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}}=\frac{k[0.20]^x[0.30]^y}{k[0.20]^x[0.10]^y}$
$1=\frac{[0.30]^y}{[0.10]^y}\left(\frac{0.30}{0.10}\right)^0=\left(\frac{0.30}{0.10}\right)^y$
$y=0$
Dividing equation $(iii)$ by $(ii),$ we obtain
$\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{k[0.40]^x[0.05]^y}{k[0.20]^x[0.30]^y}$
$\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{[0.40]^y}{[0.20]^y}\left[\text { Since } y=0,[0.05]^y=[0.30]^y=1\right]$
$2.821=2^x$
$\log 2.821=x \log 2(\text { Taking } \log \text { on both sides }) x=\frac{\log 2.821}{\log 2}$
$=1.496$
$= 1.5$ (approximately)
Hence, the order of the reaction with respect to $A$ is $1.5$ and with respect to $B$ is $0.$

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Question 83 Marks

What are fuel cells? Explain the electrode reactions involved in the working of $H_2 - O_2$ fuel cell.

Answer

Fuel cells: Those galvanic cells in which chemical energy of combustion of fuels like hydrogen, methane, etc. is converted into electrical energy are called fuel cells.
$H_2-O_2$ fuel cell$:$ The cell consists of three compartments separated by a porous electrode. Hydrogen gas is introduced into one compartment and oxygen into another compartment. These gases then diffuse slowly through the electrode and react with an electrolyte that is in the central part of the cell. The electrodes are made of porous carbon and electrolyte is a resin containing concentrated aqueous sodium hydroxide solution. Hydrogen is oxidized at anode and oxygen is reduced at the cathode.

The electrode reactions involved in the working of the $H_2-O_2$ fuel cell are as$:$ At cathode,
$O_2(g)+2 H_2 O(l)+4 e^{-} \longrightarrow 4 OH^{-}(a q)$
At anode,
$2 H_2(g)+4 OH^{-}(a q) \longrightarrow 4 H_2 O(l)+4 e^{-}$
Overall cell reaction$:2 H _2(g)+ O _2(g) \longrightarrow 2 H _2 O ( l )$

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