JEE 2012 :: General Questions :: Chemistry

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Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct?

Answer:

Explanation:

Converting all of them into Fischer projection

Since 'M' and 'N' have --OH on the same side and opposite side respectively, they cannot be a mirror image, they are diastereomers.

'M and 'O'are identical

Note Fischer projection represents eclipse form of sawhorse projection'
For comparison purposes, similar types of eclipse conformers must be drawn i'e', both vertically up or both vertically down

'M' aod 'P' are non-superimposable mirror images, hence, enantiomers

'M' and Q' are non-identical they are distereomers.

With reference to the scheme given which of the given statement (s) about T, U, V and W is (are) correct?

Answer:

Explanation:

(a) T is an ester hydrolysis in hot aqueous alkali

(b) $LiAlH_{4}$ reduces ester to alcohol as

'U" No chiral carbon optically inactive'
(c) U on treatment with an excess of acetic anhydride forms a diester as

(d) U on treatment with CrO₃ l H⁺  undergo oxidation to diacid which gives
effervescence with $NaHCO_{3}$

For the given aqueous reaction which of the statement(s) is (are) true?

Answer:

Explanation:

$K_{3}[Fe^{3+}(CN)_{6}]+KI(excess)$  $\rightarrow$

                        $K_{4}[Fe^{+2}(CN)_{6}]+KI_{3}$ (redxox)( Brownish yellow solution)

 $K_{4}[Fe(CN)_{6}]+ZnSO_{4}$ $\rightarrow$

$K_{2}Zn_{3}[Fe(CN)_{6}]_{2}$ or $K_{2}Zn[Fe(CN)_{6}]$ (white ppt)

 $I_{3}^{-}$(Brownish yellow filtrate)+$2Na_{2}S_{2}O_{3}$ $\rightarrow$   $Na_{2}S_{4}O_{6}$(clear solution)+2NaI+$I_{2}$ (Turns strach solution blue)

 $K_{2}Zn[Fe(CN)_{6}]$ react with NaOH as

  $K_{2}Zn[Fe(CN)_{6}]+NaOH$  $\rightarrow$  $[Zn(OH)_{4}]^{2-}+[Fe(CN)_{6}]^{4-}$

The reversible expansion on an ideal gas under adiabatic and isothermal conditions is shown in the figure.
Which of the following statement(s) is (are) correct?

Answer:

Explanation:

(a) Since, change of state ($p_{1},V_{1},T_{1}$) to ($p_{2}, V_{2},T_{2}$) is isothermal therefore $T_{1}=T_{2}$

(b) Since, change of state ($p_{1},V_{1},T_{1}$) to ($p_{3}, V_{3},T_{3}$)  is an adiabatic  expansion it brings about cooling of gas , therefore $T_{3} <T_{1}$

(c)   Work-done is the area under the  curve of p-V diagram' As obvious  from the given diagram, magnitude

of area under the isothermal curve  adiabatic curve , hence

$w_{isothermal}$   > $W_{adiabatic}$

Note: Here only magnitudes of work is being considered otherwise both works have negative sign.

(d)$\triangle U=nC_{v}\triangle T$

In isothermal Process, $\triangle U=0,$ as $\triangle T=0$ in adiabatic process

 $\triangle U=nC_{v}$ ($T_{3}-T_{1})<0$ as $T_{3}<T_{1}$

$\Rightarrow$   $\triangle U_{isothermal}$ >  $\triangle U_{adiabatic}$

The given graph/data I, II, III and IV , represent general trends observed for different Physisorption and chemisorption processes under mild conditions of temperature and pressure' Which of the following choice(s) about I, II, III and IV is (are) correct?

Answer:

Explanation:

Graph-I represents physisorption as in physisorption, absorbents are bonded to adsorbate through weak van der Waals' force.
Increasing temperature increases the kinetic energy of adsorbed particles increasing the rate of desorption, hence the amount of adsorption decreases.
Graph-III also represents physical adsorption as the extent of adsorption increases with Pressure.
Graph-II represents chemisorption as it is a simple activation energy diagram of a
chemical reaction.
Graph-IV represents, chemisorption as it represents the potential energy diagram
for the formation of a typical covalent bond

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