JEE 2016 :: Advanced 2016 :: Chemistry 2016

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016

 On complete hydrogenation, natural  rubber produces

Answer:

Explanation:

Natural rubber is formed by polymerisation of isoprene.

This co-polymer is formed from propylene and ethylene.

The increasing of atomic radii of the following group 13 elements is

Answer:

Explanation:

Due to poor shielding of d-orbital in Ga, the atomic radius of Ga is smaller than that of Al. Thus, Ga< Al< In< TI

Among  $[Ni(CO)_{4}]$,  $[NiCl_{4}]^{2-}$ . $[Co(NH_{3})_{4}]Cl_{2}] Cl$  , $Na_{3}[CoF_{6}]$,   $Na_{2}O_{2}$  and  $CsO_{2}$  , the total number of paramagnetic compound is

Answer:

Explanation:

One mole of an ideal gas at 300K  in thermal contact with surroundings expands isothermally from 1.0 L  to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy  of surrounds ( $\triangle S_{surr}$ ) in J  $K^{-1}$  is (1L atm=101.3 J)

Answer:

Explanation:

By the first law.$\triangle  E=Q+W$

 For isothermal expansion, $\triangle  E=0$ 

  $\therefore$ Q=-W

    $-Q_{irrev}=W_{irrev}=p\triangle V$

                      =3(2-1)= 3L atm

   Also,  $\triangle S_{surr}= \frac{Q_{irrev}}{T}=\frac{(-3\times101.3)J}{300 K}$

                                    $= -\frac{303.9}{300}=-1.013 JK^{-1}$

Thermal decomposition of gaseous X₂, to gaseous X at 298 K takes place according to the following equation

$X_{2}(g)\rightleftharpoons2X(g)$

The standard reaction Gibbs energy, Δ_rG⁰, of this reaction is positive. At the start of the reaction, there is one mole of X₂ and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given, R = 0.083 L bar K^-1 mol^-1)

The incorrect  statement among the following reaction , is

Answer:

Explanation:

a)          $K_{p}= \frac{4px^{2}}{(4-x^{2})}=px^{2}(\because 4>>>x)$

$\therefore x\alpha \sqrt{\frac{1}{p}}$

If p decreases, x increases. Equilibrium is shifted in the forward  side. Thus, statement(a) is correct

(b)       At the start of the reaction , Q=0 where , Q is the reaction quotient

   $\triangle G= \triangle G^{0}+2.303RT \log Q$

   since   $\triangle G^{0}>0$  , thus  $\triangle G$ is -ve

Hence, dissociation take place spontaneously,

 Thus (b) is correct

 (c) If we use x= 0.7 and p=2 bar,

  then  $K_{p}=\frac{4\times 2(0.7)^{2}}{[4-(0.7)^{2}]}=1.16>1$

Thus, (c) is incorrect

(d) At equilibium, $\triangle G=0$

$\therefore$      $\triangle G ^{0}=-2.303RT\log K_{p}$

Since , $\triangle G ^{0}=+ve$

$Hence, K_{p}< 1,K_{C}= \frac{K_{p}}{(RT)}$

Then $ K_{C}< 1$ , Thus (d) is correct

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016