Question 13 Marks
How the following conversions can be carried out?
i. 2-Bromopropane to 1-bromopropane
ii. Chloroethane to butane
iii. Benzene to diphenyl
i. 2-Bromopropane to 1-bromopropane
ii. Chloroethane to butane
iii. Benzene to diphenyl
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Question 23 Marks
The first order rate constant for the decomposition of ethyl iodide by the reaction
$C_2H_5I(g) \rightarrow C_2H_4(g) + HI(g)$
at $600\ K$ is $1.60 \times 10^{-2} s^{-1}$. Its energy of activation is $209\ kJ/mol.$ Calculate the rate constant of the reaction at $700 Κ.$
$C_2H_5I(g) \rightarrow C_2H_4(g) + HI(g)$
at $600\ K$ is $1.60 \times 10^{-2} s^{-1}$. Its energy of activation is $209\ kJ/mol.$ Calculate the rate constant of the reaction at $700 Κ.$
Answer
View full question & answer→Here $T_1=600 K$
$T_2=700 K$
$E_a=209 KJ / mol$
$=209000 Jmol^{-1}$
$k_1=1.60 \times 10^{-5} s^{-1}$
$k_2=?$
Using the formula
$\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right]$
$\log k_2-\log k_1=\frac{E_a}{2.303 R}\left[\frac{700-600}{600 \times 700}\right]$
$\log k_2-\log 1.60 \times 10^{-5}=\frac{209000}{2.303 \times 8.314}\left[\frac{100}{600 \times 700}\right]$
$\log k_2=\log 1.60 \times 10^{-5}+2.599$
$\log k_2=-4.796+2.599$
$=-2.197$
$k_2=\text { anti } \log (-2.197)$
$=6.36 \times 10^{-3} s^{-1}$
$T_2=700 K$
$E_a=209 KJ / mol$
$=209000 Jmol^{-1}$
$k_1=1.60 \times 10^{-5} s^{-1}$
$k_2=?$
Using the formula
$\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right]$
$\log k_2-\log k_1=\frac{E_a}{2.303 R}\left[\frac{700-600}{600 \times 700}\right]$
$\log k_2-\log 1.60 \times 10^{-5}=\frac{209000}{2.303 \times 8.314}\left[\frac{100}{600 \times 700}\right]$
$\log k_2=\log 1.60 \times 10^{-5}+2.599$
$\log k_2=-4.796+2.599$
$=-2.197$
$k_2=\text { anti } \log (-2.197)$
$=6.36 \times 10^{-3} s^{-1}$
Question 33 Marks
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Answer
View full question & answer→A lead storage battery consists of a lead anode, a grid of lead packed with lead dioxide $(\ce{PbO2})$ as cathode and a $38\%$ solution of sulphuric acid $(\ce{H2SO4})$ as an electrolyte.
When the battery is in use, the following cell reactions take place:
At anode: $\ce{Pb (s)+ SO 4^{2-}(a q) \rightarrow PbSO 4(s)+2 e^{-}}$
At cathode: $\ce{PbSO 4(s)+ SO 4^{2-}(a q)+4 H ^{+}(a q)+2 e^{-} \rightarrow PbSO 4(s)+2 H 2 O (l)}$
The overall cell reaction is given by,
$ \ce{Pb(s)+PbO2(s)+2 H2 SO4(aq) \rightarrow 2 PbSO4(s)+2 H2 O(l)}$
When battery is charging, the reverse of all these reactions takes place.
Hence, on charging$, \ce{PbSO4(s)}$ present at the anode and cathode is converted into $\ce{Pb}$ and $\ce{PbO2},$ respectively. Sulphuric acid
$(\ce{H2SO4)}$ reconstitute in the reaction. The reaction may written as:
$\ce{2 PbO 2(s)+2 H2 O ( l ) \xrightarrow{\text { charging }} Pb ( s )+ PbO 2(s)+2 H 2 SO 4}$
When the battery is in use, the following cell reactions take place:
At anode: $\ce{Pb (s)+ SO 4^{2-}(a q) \rightarrow PbSO 4(s)+2 e^{-}}$
At cathode: $\ce{PbSO 4(s)+ SO 4^{2-}(a q)+4 H ^{+}(a q)+2 e^{-} \rightarrow PbSO 4(s)+2 H 2 O (l)}$
The overall cell reaction is given by,
$ \ce{Pb(s)+PbO2(s)+2 H2 SO4(aq) \rightarrow 2 PbSO4(s)+2 H2 O(l)}$
When battery is charging, the reverse of all these reactions takes place.
Hence, on charging$, \ce{PbSO4(s)}$ present at the anode and cathode is converted into $\ce{Pb}$ and $\ce{PbO2},$ respectively. Sulphuric acid
$(\ce{H2SO4)}$ reconstitute in the reaction. The reaction may written as:
$\ce{2 PbO 2(s)+2 H2 O ( l ) \xrightarrow{\text { charging }} Pb ( s )+ PbO 2(s)+2 H 2 SO 4}$
Question 43 Marks
Silver is deposited on a metallic vessel by passing a current of 0.2 amps. for 3 hrs. Calculate the weight of silver deposited. (At mass of silver = 108 amu, 1 F = 96500 C)?
Answer
View full question & answer→By Faraday's first law of electrolysis.
w = ZIt.
Substituting I = 0.2 A, t = 3×60×60 sec, Z= 108/(1×96500)
⇒ w = 2.417 g.
2.417 g of silver is deposited.
w = ZIt.
Substituting I = 0.2 A, t = 3×60×60 sec, Z= 108/(1×96500)
⇒ w = 2.417 g.
2.417 g of silver is deposited.
Question 53 Marks
How much copper is deposited on the cathode of an electrolytic cell if a current of $5$ ampere is passed through a solution of copper sulphate for $45$ minutes?
Answer
View full question & answer→We know that
${\left[Cu=63.5 g\ \text{mol}^{-1}, 1 F=96500\ C\ \text{mol}^{-1}\right]}$
$\ce{Cu^{2+}(a q)+2 e^{-} \rightarrow Cu(s)}$
$m=Z \times I \times t$
$=\frac{63.5}{2 \times 96500} \times 5 \text{ amp} \times 45 \times 60$
$=\frac{857250}{193000}=4.44 g$
${\left[Cu=63.5 g\ \text{mol}^{-1}, 1 F=96500\ C\ \text{mol}^{-1}\right]}$
$\ce{Cu^{2+}(a q)+2 e^{-} \rightarrow Cu(s)}$
$m=Z \times I \times t$
$=\frac{63.5}{2 \times 96500} \times 5 \text{ amp} \times 45 \times 60$
$=\frac{857250}{193000}=4.44 g$
Question 63 Marks
An organic compound $A,$ which has a characteristic odour, on treatment with con. $NaOH$ forms two compounds $B$ and $C.$ Compound $B$ has molecular formula $C_7H_8O$ which on oxidation gives back $A.$ Compound $C$ is the sodium salt of an acid. $C,$ when heated with soda lime yields an aromatic hydrocarbon $D.$ Deduce the structures of $A, B, C$ and $D.$
Answer
View full question & answer→This is Cannizzaro Reaction
The molecular formula of $(B)$ and characteristic odour of $(A)$ suggests that $(A)$ is an aromatic aldehyde, $C_6H_5CHO$ and $(B)$ is alcohol, $C_6H_5CH_2OH.$ As $(C)$ is a sodium salt of an acid gives hydrocarbon $(D)$ on heating with soda lime, $(C)$ is sodium benzoate and $(D)$ is benzene. In this reaction, Benzaldehyde undergoes self oxidation and reduction$($disproportionation$).$ Therefore:-
The molecular formula of $(B)$ and characteristic odour of $(A)$ suggests that $(A)$ is an aromatic aldehyde, $C_6H_5CHO$ and $(B)$ is alcohol, $C_6H_5CH_2OH.$ As $(C)$ is a sodium salt of an acid gives hydrocarbon $(D)$ on heating with soda lime, $(C)$ is sodium benzoate and $(D)$ is benzene. In this reaction, Benzaldehyde undergoes self oxidation and reduction$($disproportionation$).$ Therefore:-
Question 73 Marks
Give the major products that are formed by heating each of the following ethers with HI.
Question 83 Marks
Write the mechanism of hydration of ethene to yield ethanol.
Answer
View full question & answer→The mechanism of hydration of ethene to form ethanol involves three steps.
