For the reaction:

X_{2}O_{4}\left ( l \right )\rightarrow 2XO_{2}(g)

\Delta U=2.1k \ cal, \Delta S=20 cal\ K^{-1}  at 300K

Hence \Delta G is :-

  • Option 1)

    2.7k cal

  • Option 2)

    -2.7 k cal

  • Option 3)

    9.3k cal

  • Option 4)

    -9.3k cal

 

Answers (1)
P Plabita

 

Molar heat capacity for isobaric process C(p) -

dH=n\: C_{P}\: dT
 

- wherein

dH=dE+d(PV)

or

dH=dE+n\, R\, dT

 

 \Delta H= \Delta U + \Delta n_{g}RT

Given, \Delta U=2.1 Kcal\;\;\;\; \Delta n_{g}=2

R- 2 \times 10^{-3}Kcal;\;\;T=300 K

\therefore \Delta H= 2.1+2 \times 2 \times 10^{-3} \times 300=3.3 Kcal

Also, \Delta G= \Delta H-T \Delta S

Given, \Delta S=20 \times 10^{-3}Kcal\;K^{-1}

On putting the value of \Delta H and \Delta S in the equation, we get - 

\Delta G = 3.3-300 \times 20 \times 10^{-3}= -2.7 Kcal

 


Option 1)

2.7k cal

This option is incorrect.

Option 2)

-2.7 k cal

This option is correct.

Option 3)

9.3k cal

This option is incorrect.

Option 4)

-9.3k cal

This option is incorrect.

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