VITEEE 2016 :: Chemistry :: General questions

• Chemistry - General questions

The enthalpy change for a given reaction at 298 K is -x Jmol^-1, For the reaction to be spontaneous at 298 K , the entropy change at that temperature

Answer:

Explanation:

$\triangle G=\triangle H -T\triangle S$

For spontaneous reaction $\triangle H=-ve$

$T\triangle S > \triangle H$

$\triangle G=\frac{-q}{T}=-\frac{-X}{298}=\frac{X}{298}$

So, $\triangle G $ can be negative

Standard entropy  of X₂,Y₂  and XY₃  are 60, 40 and 50 JK^-1 mol^-1 , respectively , For the reactions

$\frac{1}{2}X_{2}+\frac{3}{2}Y_{2}\rightarrow XY_{2}, \triangle H=-30kJ, $   , to be at equilibrium , the temperature will be 

Answer:

Explanation:

For a reaction to be equilibrium   $\triangle$ G=0,   Since ,  $\triangle G=\triangle H-T\triangle S$ sop at equilibrium

 $\triangle H -T\triangle S=0$

 For the reaction

 $\frac{1}{2}X_{2}+\frac{3}{2}Y_{2}\rightarrow XY_{3}$   ;   $\triangle$ H= -30kJ

(given)

 Calculating $\triangle$ S for the above reaction , we get

 $\triangle S= 50-\left[\frac{1}{2}\times 60+\frac{3}{2}\times40\right]JK^{-1}$

 = 50-(30+60)=JK^-1 =-40 JK^-1 

At equilibrium , T $\triangle$ S= $\triangle$ H

           [$\because$   $\triangle$ G=0]

 $\therefore$     $T\times (-40)=-30\times1000[\because 1kJ=1000J]$

 or        $T= \frac{-30\times 1000}{-40} or 750 K$

 At 298 K , the conductivity of a saturated solution  , of AgCl  in water is 2.6 x 10-6S cm^-1. Its solubility product at 298 K,

 Given :  $\lambda^{\infty}(Ag^{+})=63.0 S cm^{2}mol^{-1}$

             $\lambda^{\infty}(Cl^{-})=67.0 S cm^{2}mol^{-1}$

Answer:

Explanation:

 Solubility 

$S= \frac{1000k}{\lambda_{AgCl}^{0}}=\frac{1000\times2.6\times 10^{-6}}{\lambda_{Ag+}^{0}+\lambda_{Cl-}^{0}}$

$\frac{2.6\times10^{-3}}{63+67}=2\times10^{-5}mol L^{-1}; K_{sp}=S^{2}$

For hydrogen -oxygen fuel cell, the cell reaction is 

$2H_{2}(0)+O_{2}(g)\rightarrow 2H_{2}O(l)$

 if   $\triangle G_f^0  (H_{2}O)=-237.2kJmol^{-1}$   , then emf of this cell is 

Answer:

Explanation:

 We know   $\triangle$ G⁰= -nFE⁰

$\therefore$        $E^{0}=\frac{\triangle G^{0}}{nF}$

 $= +\frac{237.2\times1000J}{2\times96500}=1.23 V[\therefore n=2]$

Which type of carbocation is/are formed when 

 is treated with an acid?

Answer:

Explanation:

• Chemistry - General questions