JEE 2011 :: General Questions :: Chemistry

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For the first order reaction $2N_{2}O_{5}(g) \rightarrow  4NO_{2}(g)+O_{2} (g)$

Answer:

Explanation:

(a) For a first-order reaction. the concentration of reactant remaining after time t is given by $[A]=[A]_{0}e^{-kt}$. Therefore, the concentration of  reactant decreases exponentially  with time 

(b)   Rise in temperature  increase rate constant (k)  and  therefore decrease half-life $(t_{1/2})$ as

                           $t_{1/2}=\frac{ln 2}{k}$

 (d) For a first order reaction. if 100 moles of reactant is taken  initially, after  n half-lives reactant remaining is given by percentage

  $A= 100(\frac{1}{2})^{n}=100(\frac{1}{2})^{8}=0.3906$

$\Rightarrow $  A  reacted =100-0.3906=99.6%

(c) Half-life of first order reaction is independent of initial concentration

Reduction of the metal centre in aqueous permanganate ion involves

Answer:

Explanation:

 In neutral  medium ;  $MnO_{4}^{-}  \rightarrow   MnO_{2} (Mn^{7+} +3e^{-} \rightarrow Mn^{4+})$

In  alkaline medium  :  $MnO_{4}^{-} \rightarrow MnO_{2} (Mn^{7+}+3e^{-} \rightarrow Mn^{4+})$

In acidic medium:    $Mn_{4}^{-} \rightarrow Mn^{2} (Mn^{7+}+5e^{-} \rightarrow Mn^{2+})$

Among the following complexes

$(K-P),K_{3}[Fe(CN)_{6}](K),$

$[Co(NH_{3})_{6}Cl_{3}(L),Na_{3}[Co(ox)_{3}](M)$

$[Ni(H_{2}O)_{6}]Cl_{2}(N); K_{2}[Pt(CN)_{4}](O)$ and 

$[Zn(H_{2}O)_{6}](NO_{3})_{2}$(P)

 the diamagnetic complexes are

Answer:

Explanation:

For a diamagnetic complex,  there should not be any unpaired electron in the valence shell of central metral.In $K_{3}[Fe(CN)_{6}],Fe(III)$ has $d^{5}$- configration (odd electrons). Hence it is  paramagnetic. In $[Co(NH_{3})_{6}]Cl_{3},Co(III)$ has d⁶ configuration in a strong ligand field, hence  all the electrons are paired and the complex is diamagnetic  .In $Na_{3}[Co(ox)_{3}],Co(III)$ has $d^{6}$- configuration  and oxalate being a chelating  ligand, very strong ligand and all the six electrons remains paired in lower  $t_{2g}$  level. diamagnetic.In $[Ni(H_{2}O)_{6}]Cl_{2},Ni(II)$ has $3d^{8}$ -configuration and $H_{2}O$ ia a weak ligand hence 

In $K_{2}[Pt (CN)_{4}]Pt(II)$ has $d^{8}$- configuration and $CN^{-}$  is a strong ligand, hence all the eight electrons are spin paired. Therefore, complex is diamagnetic

In $[Zn(H_{2}O)_{6}](NO_{3})_{2},Zn(II)$  has ${_{3}}d^{10}$ configuration spin paired , hence diamagnetic

Passing $H_{2}S$ gas into a mixture of $Mn^{2+},Ni^{2+},Cu^{2+}$ and $Hg^{2+}$ ions in an acidified aqueous solution precipitates

Answer:

Explanation:

In an acidic medium,$H_{2}S$ is very feebly ionized giving very small concentration of sulphide ion for precipitation. Therefore, the most insoluble salts Cus and HgS are precipitated only.

Consider the following cell reaction.

  $2Fe (s)+O_{2}(g)+4H^{+}(aq) \rightarrow$

                   $2Fe^{2+}(aq)+2H_{2}O(l),$  E°=1.67 V

At       $[FE^{2+}]=10^{-3}$M, $P(O_{2})=0.1 atm$

and  pH=3 , the cell potential at 25° C is 

Answer:

Explanation:

The half reactions are

 $Fe(s) \rightarrow  Fe^{2+}(aq)+2e^{-} \times 2$

 $O_{2}(g)+4H^{+}+4e^{-} \rightarrow 2H_{2}O+2Fe(s)$

  $+O_{2}(g)+4H^{+} \rightarrow   2Fe^{2+}(aq)+2H_{2}O(l)$

  $E= E^{0}-\frac{0.059}{4} log \frac{(10^{-3})^{2}}{(10^{-3})^{4})(0.1)}=1.57V$

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