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Chemistry Test - 42

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Chemistry Test - 42
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  • Question 1
    1 / -0

    The reaction of cyanamide, \(\mathrm{NH}_{2} \mathrm{CN}(\mathrm{s})\) with dioxygen was carried out in a bomb calorimeter and \(\Delta \mathrm{U}\) was found to be \(-742.7 \mathrm{KJ} \mathrm{mol}^{-1}\) at \(298 \mathrm{~K}\). Calculate the enthalpy change for the reaction at \(298 \mathrm{~K}\).

    \(\mathrm{N H}_{4} \mathrm{CN}_{(\mathrm{g})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{N}_{2(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)

    Solution

    Enthalpy change for a reaction \((\Delta \mathrm{H})\) is given by the expression,

    \(\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}\)

    Where,

    \(\Delta \mathrm{U}=\) change in internal energy

    \(\Delta \mathrm{n}_{\mathrm{g}}=\) change in number of moles

    For the given reaction,

    \(\Delta \mathrm{n}_{\mathrm{g}}=\sum \mathrm{n}_{\mathrm{g}}\) (products) - \(\sum \mathrm{n}_{\mathrm{g}}\) (reactants)

    \(\Delta \mathrm{n}_{\mathrm{g}}=(2-1.5)\) moles

    \(\Delta \mathrm{n}_{\mathrm{g}}=+0.5\) moles

    And, \(\Delta \mathrm{U}=-742.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

    \(\mathrm{T}=298 \mathrm{~K}\)

    \(\mathrm{R}=8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)

    Substituting the values in the expression of \(\Delta \mathrm{H}\)

    \(\Delta \mathrm{H}=\left(-742.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)+(+0.5 \mathrm{~mol})(298 \mathrm{~K}) 8.314 \times 10^{-3} \mathrm{kJmol}^{-1} \mathrm{~K}^{-1}\)

    \(\Delta \mathrm{H}=-742.7+1.2\)

    \(\Delta \mathrm{H}=-741.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

  • Question 2
    1 / -0

    Ethylidene chloride is a/an ___________.

    Solution

    Ethylidene chloride is a gem-dihalide. 

    • Ethylidene chloride is also known as 1,1 Dichloroethane. 
    • It has 2 chlorine atoms on the same carbon atom, thus, has 1,1 relationship. 
    • Therefore, it is geminal-dihalide.
    • Gem-dihalide is a compound that have two halogen atoms on the same carbon atom. 

    Below given the structure of Ethylidene chloride:

     

  • Question 3
    1 / -0

    Match List-I with List-II

       List-I    List-II
     A.  Sodium carbonate  (i)  Deacon
     B.  Titanium  (ii)  Castner-Kellner
     C.  Chlorine  (iii)  van-Arkel
     D.  Sodium hydroxide  (iv)  Solvay

    Choose the correct answer form the options given below.

    Solution

    (A) Sodium carbonate is manufactured by Solvay process \(\Rightarrow\) (iv) of List-II.

    Solvay process's reactions are as follows

    In Solvay's process, first ammonia \(\left(\mathrm{NH}_3\right.\) ) reacts with \(\mathrm{CO}_2\) and \(\mathrm{H}_2 \mathrm{O}\) to give ammonium bicarbonate \(\left(\mathrm{NH}_4 \mathrm{H} \mathrm{CO}_3\right)\).

    This bicarbonate react with \(\mathrm{N} \mathrm{aCl}\) to give sodium bicarbonate \((\mathrm{NaHCO})_3\) along with ammonium chloride \(\left(\mathrm{NH}_4 \mathrm{Cl}\right)\).

    At last, \(\mathrm{NaH} \mathrm{CO}_3\) is heated to form sodium carbonate as major product.

    \(\mathrm{NH}_3+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{NH}_4 \mathrm{HCO}_3 \)

    \(\mathrm{NH}_4 \mathrm{HCO}_3+\underset{\text { (Brine) }}{\mathrm{NaCl}} \rightarrow \mathrm{NaHCO}_3 \downarrow+\mathrm{NH}_4 \mathrm{Cl} \)

    \(2 \mathrm{NaHCO}_3 \rightarrow \Delta_2 \rightarrow \underset{\text { Sodium carbonate }}{\mathrm{Na}_2 \mathrm{CO}_3}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

    (B) Titanium is purified by van-Arkel method \(\Rightarrow\) (iii) of List-II

    van-Arkel method

    Crude titanium \(\xrightarrow[\mathrm{I}_2]{\mathrm{780 \textrm {K }}} \mathrm{Til}_4(\mathrm{~g}) \xrightarrow{1800 \mathrm{~K}} \underset{\text { (Pure) }}{\mathrm{Ti}}+2 \mathrm{I}_2\)

    (C) Chlorine is manufactured by Deacon process ⇒ (i) of List-II.

    Deacon process

    \(4 \mathrm{HCl}(\mathrm{g})+\underset{\text {[Air]}~~~~~~~~~~~~~\text{ 723k }~~~~~~~~~}{\mathrm{O}_2(\mathrm{~g}) \xrightarrow{\mathrm{CuCl}_2 \text { (catalyst) }}} \underset{\text { Chlorine gas }}{2 \mathrm{Cl}_2}+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

    (D) Sodium hydroxide is manufactured by Castner-Kellner cell \(\Rightarrow\)

    (ii) of List-II.

    Castner-Kellner cell reaction

    Electrolyte used is brine, saturated \(\mathrm{NaCl}\) solution. Cathode \((\mathrm{Hg})\) :

    \(\mathrm{Na}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Na} \xrightarrow{\mathrm{Hg}} \mathrm{Na} / \mathrm{Hg}\)

    Anode (carbon): \(\mathrm{Cl}^{-}+\mathrm{e}^{-} \rightarrow \frac{1}{2} \mathrm{Cl}_2 \uparrow\)

    \(2 \mathrm{Na} / \mathrm{Hg}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \underset{\text { Sodium hydroxide }}{2 \mathrm{NaOH}}+2 \mathrm{Hg}+\mathrm{H}_2 \uparrow\)

  • Question 4
    1 / -0

    Which of the following is not a heteroatom?

    Solution

    A heteroatom is any atom other than carbon or hydrogen atom. All the others apart from carbon mentioned above are heteroatoms. In other words, a heteroatom is a non-carbon atom present in a carbon structure.

    Carbon or hydrogen (termed a heteroatom) is bonded to carbon. All heteroatoms have a greater or lesser attraction for electrons than does carbon. Thus, each bond between a carbon and a heteroatom is polar, and the degree of polarity depends on the difference between the electron-attracting properties of the two.

  • Question 5
    1 / -0

    In a hexagonal closed packing lattice, coordination number of an atom in a unit cell is:

    Solution

    In the hexagonal closest packed (hcp) each ion has 12 neighboring ions, therefore, it has a coordination number of 12 and contains 6 atoms per unit cell.

    The centre atom in layer B of HCP structure is touched with 12 other atoms of the same cell. 

      

  • Question 6
    1 / -0

    Calculate (a) wavenumber and (b) frequency of yellow radiation having wavelength \(5800 Å\).

    Solution

    Given,

    Wavelength \(=5800 Å=5800 \times 10^{-10} \mathrm{~m}\)

    As we know,

    The speed of light,

    \(c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\)

    (a) Wavenumber is given as,

    \(\bar{\nu}=\frac{1}{\lambda}\)

    \(=\frac{1}{5800 \times 10^{-10} }\)

    \(=1.724 \times 10^{6} \mathrm{~m}^{-1}\)

    \(=1.724 \times 10^{4} \mathrm{~cm}^{-1}\)

    (b) The frequency is given as,

    \(\nu=\frac{c}{\lambda}\)

    \(=\frac{3 \times 10^{8} }{5800 \times 10^{-10}}\)

    \(=5.172 \times 10^{14}\) Hz

  • Question 7
    1 / -0

    In the complex \(K _{2} Fe \left[ Fe ( CN )_{6}\right]\):

    Solution

    In the complex \(K _{2} Fe \left[ Fe ( CN )_{6}\right]\), both Fe atoms are in the same oxidation state.

    Oxidation state of Fe in \(K _{2} Fe \left[ Fe ( CN )_{6}\right]\),

    \(0=2+2 x -6\)

    \(x =2\)

    In the complex \(K _{2} Fe \left[ Fe ( CN )_{6}\right]\), both the iron atoms have \(+2\) oxidation state.Since six cyanide ions surround the iron atom, its coordination number is \(6 \).

    The \(Fe ^{2+}\) ion in \(\left[ Fe ( CN )_{6}\right]^{4-}\) has outer electronic configuration of \(3 d ^{6}\). The complex is a low spin complex. It contains 0 unpaired electrons with a magnetic moment of 0 BM and is colourless.

  • Question 8
    1 / -0

    Which of the following compounds give(s) positive test with Tollens' reagent?

    A. Carboxylic acid

    B. Alcohol

    C. Alpha hydroxy ketones

    D. Aldehydes

    Solution

    Alpha hydroxy ketones and Aldehydes give a positive test with Tollens' reagent.

    Ketones generally don’t give positive test with Tollen’s reagent, but alpha hydroxyl ketones are an exception. Iodoform test, Ester test, etc., are used for alcohols. The litmus test, Sodium bicarbonate test and Ester test are used for carboxylic acid.

  • Question 9
    1 / -0

    A rain in 50% completed om shown and 75% in 4 hours. Find the order of reaction:

    Solution

    \(50 \%\) reaction completes in 2 hours. That is, half life of reaction is 2 hours

    \(t _{\left(\frac{1}{2}\right)}=2\) hours. Also given that, \(t _{\left(\frac{3}{4}\right)}=4\) hours

    We can observe that,

    \(t _{\left(\frac{3}{4}\right)}=2 \times t _{\left(\frac{1}{2}\right)}\)

    Thus, relation holds true only in first order

  • Question 10
    1 / -0

    Bonds that do not exist in tertiary structure of proteins:

    Solution

    The tertiary structure of a protein consists of the way a polypeptide is formed of a complex molecular shape. This is caused by R-group interactions such as ionic and hydrogen bonds, disulphide bridges, and hydrophobic & hydrophilic interactions.Bonds that do not exist in tertiary structure of proteins arePhosphodiester bonds.A phosphodiester bond occurs when exactly two of the hydroxyl groups in phosphoric acid react with hydroxyl groups on other molecules to form two ester bonds.

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