Get Answers to all your Questions

header-bg qa

Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below:

The energy involved in the conversion of  1/2\, Cl_{2(g)}\; to\; Cl^{-}_{(aq)}  (using data , = 240 kJ mol-1  , = = -349 kJ mol-1,   = -381 kJ mol-1)   will be

  • Option 1)

    +120 kJ mol-1

  • Option 2)

    +152 kJ mol-1

  • Option 3)

    -610 kJ mol-1

  • Option 4)

    -850 kJ mol-1

 

Answers (1)

best_answer

As we learnt in

Bond dissociation enthalpy -

It is the average of enthalpy required to dissociate the said bond present in different gaseous compound in to free atoms in gaseous state.

- wherein

N_{2}+Bond\, Energy\rightarrow 2N

 

 

Enthalpy of Hydration -

Amount of enthalpy evolved during hydration of 1 mole anhydrous compound.

- wherein

CuSO_{4(s)}+5H_{2}O\rightarrow CuSO_{4}\cdot 5H_{2}O+ \Delta H

 

 The energy involved in the conversion of \frac{1}{2}Cl_{2}(g) to Cl^{-1}(aq) is given

\Delta H=\frac{1}{2}\Delta_{diss}H_{Cl_{2}}^{(-1)}+\Delta _{eg}H_{Cl}^{(-1)}+\Delta _{hyd}H_{Cl}^{(-1)}

Substituting various values from given data

\Delta H=(\frac{1}{2}\times240)+(-349)+(-381)\:kJmol^{-1}\\=(120-349-381)=\:-610kJmol^{-1}


Option 1)

+120 kJ mol-1

This option is incorrect.

Option 2)

+152 kJ mol-1

This option is incorrect.

Option 3)

-610 kJ mol-1

This option is correct.

Option 4)

-850 kJ mol-1

This option is incorrect.

Posted by

perimeter

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE