3 Marks Question

Question 13 Marks

i. Determine the units of rate constant for first and zero order reaction.
ii. Show that time required for the completion of 99% of the first order reaction is twice the 90% of completion of the reaction.

Answer

i. $K =( mol )^{1- n } L ^{ n -1} s^{-1}$
For zero order, $n =0$
So, $K=( mol )^{1-0} L^{0-1} s^{-1}= s ^{-1} mol L ^{-1}$
For first order, $n =1$
$
\begin{array}{l}
K=(mol)^{1-n} L^{n-1} s^{-1} \\
\text { So, } K=(mol)^{1-1} L^{1-1} s^{-1} \\
=s^{-1}
\end{array}
$
ii. For a first order reaction,
$
\begin{array}{l}
t=\frac{2.303}{K} \log \frac{[A]_0}{[A]} \\
{[A]_0=a,[A]=a-\frac{a \times 99}{100}=0.01 a} \\
t(99 \%)=\frac{2.303}{K} \log \frac{a}{0.01 a} \\
=\frac{2.303}{K} \log 100 \\
=\frac{2.303}{K} \times 2 \ldots \text { (i) }
\end{array}
$
For $90 \%$ completion of reaction,
$
\begin{array}{l}
{[A]=a-\frac{a \times 99}{100}=0.1 a} \\
t(90 \%)=\frac{2.303}{K} \log \frac{a}{0.1 a} \\
=\frac{2.303}{K} \times 1 \ldots .(i)
\end{array}
$
Dividing equation (i) by equation (ii), we get
$
t(99 \%)=2 \times t(90 \%)
$
Hence, the time taken to complete 9% of the first order reaction is twice the time required for the completion of 90% of the reaction.

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Question 23 Marks

Tert-Butylbromide reacts with aq. NaOH by S_N1 mechanism while n-butylbromide reacts by SN₂ mechanism. Why?

Answer

Tert-butylbromide undergoes substitution by S_N1 unimolecular substitution mechanism because it is able to form a stable carbocation in the first step after cleavage of the halide group. The carbocation then reacts with the nucleophile OH^-. On the other hand, primary halide n-butylbromide cannot form a stable carbocation so it undergoes S_N2 bimolecular substitution mechanism,which is a one-step substitution that involves the attack of OH^- and simultaneous leaving of X^- to form n-butyl alcohol.

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Question 33 Marks

Write the mechanism of acid dehydration of ethanol to yield ethane.

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Question 43 Marks

Calculate the emf of the cell
$Mg ( s )\left| Mg ^{+2}(0.1 M )\right| \mid Cu ^{+2}\left(1 \times 10^{-3} M \mid Cu ( s )\right)$
Given, $\left.E _{C u^{2+} / C u}^{\ominus}=+0.34 V, E _{M g^{+2} / M g}^{\ominus}=-2.37 V\right)$

Answer

We have
$
Mg(s)\left|Mg^{+2}(0.1 M)\right| \mid Cu^{+2}\left(1 \times 10^{-3} M \mid Cu(s)\right)
$
Half cell reactions of this cell are:
At Cathode (Reduction):
$
C u^{2+}(a q)+2 e^{-} \rightarrow C u(s)
$
At Anode (Oxidation):
$
M g(s) \rightarrow M g^{2+}(a q)+2 e^{-}
$
For this cell, we have, $n =2$ moles of electrons.
$
\begin{array}{l}
\text { (Given, } E_{Cu 2+}^{\ominus} / Cu \\
\left.E_{\text {cell }}^{\ominus}=+0.34 V, E_{Mg^{+2} / Mg}^{\ominus}=-2.37 V\right) \\
E_{\text {cell }}^{\ominus}=E_{\left(C Cu^{2+} / Cu\right)}^{\ominus}-E_{\left(Mg^{2+} / Mg\right)}^{\ominus}-E_{\text {oxidation }}^{\ominus} \\
=0.34-(-2.37) V \\
=2.71 V
\end{array}
$
According to Nernst equation
$
\begin{array}{l}
E_{\text {cell }}=E_{\text {cell }}^{\ominus}-\frac{0.059}{n} \log \frac{\left[Mg^{2+}\right]}{\left[Cu^{2+}\right]} \\
E_{\text {cell }}=E_{\text {cell }}^{\ominus}-\frac{0.059}{2} \log \left[\frac{0.1}{10^{-3}}\right] \\
=2.71-0.0295 \log 10^2 \\
=2.71-0.0295 \times 2 \\
=2.651 \\
\therefore E_{\text {cell }}=2.651 V
\end{array}
$

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Question 53 Marks

Write the Nernst equation and calculate emf of the following cell at 298 K:
$\begin{array}{l} Cr \left| Cr ^{3+}(0 \cdot 1 M ) \| Fe ^{2+}(0 \cdot 01 M )\right| Fe \\ \text { Given : } E _{ Cr ^{3+} / Cr }^{\ominus}=-0 \cdot 75 V \\ E _{ Fe ^{2+} / Fe }^{\ominus}=-0 \cdot 45 V \\ (\log 10=1)\end{array}$

Answer

$\begin{array}{l}E_{\text {cell }}=E_{\text {cell }}^0-\frac{0.059}{6} \log \frac{\left[ Cr ^{3+}\right]^2}{\left[ Fe ^{2+}\right]^3} \\ E_{\text {cell }}^0=-0.45-(-0.75) \\ =0.30 V \\ E_{\text {cell }}=0.3-\frac{0.059}{6} \log \frac{(0-1)^2}{(0.01)^3} \\ =0 \cdot 3-.00985 \log \frac{\left(10^{-1}\right)^2}{\left(10^{-2}\right)^3} \\ =0.3-.00985 \times 4 \log 10 \\ =0.3-0.0394 \\ =0.2606 V\end{array}$

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Question 63 Marks

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.
i. PhMgBr and then H₃O⁺
ii. Tollens' reagent
iii. Semicarbazide and weak acid
iv. Excess ethanol and acid
v. Zinc amalgam and dilute hydrochloric acid

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Question 73 Marks

How can phenol be converted to aspirin?

Answer

Phenol is converted into salicylic acid. The reaction is usually carried out by allowing sodium phenoxide to absorb carbon dioxide and then heating the product to 400 K and 4-7 atm pressure. The first unstable intermediate is formed which undergoes a proton shift to form sodium salicylate. The subsequent acidification of sodium salicylate gives.

Then aspirin is obtained by acetylating salicylic acid with acetic anhydride and conc. H₂SO₄

The preparation of Aspirin from salicylic acid is an example of an electrophilic substitution reaction in which carbon dioxide is an electrophile.

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Question 83 Marks

How much charge is required for the following reductions:
i. 1 mol of $Al ^{3+}$ to Al ?
ii. 1 mol of $Cu ^{2+}$ to Cu ?
iii. 1 mol of $MnO _4^{-}$to $Mn ^{2+}$ ?

Answer

i. $A l^{3+}+3 e^{-} \rightarrow A l$
Therefore, Required charge $=3 F$
$
\begin{array}{l}
=3 \times 96487 C \\
=289461 C
\end{array}
$
ii. $C u^{2+}+2 e^{-} \rightarrow C u$
Therefore, Required charge $=2 F$
$
\begin{array}{l}
=2 \times 96487 C \\
=192974 C
\end{array}
$
iii. $MnO _4^{-} \rightarrow Mn ^{2+}$
i.e., $M n^{7+}+5 e^{-} \rightarrow M n^{2+}$
Therefore, Required charge $=5 F$
$
\begin{array}{l}
=5 \times 96487 C \\
=482435 C
\end{array}
$

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Chemistry 3 Marks Question

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