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Confused! kindly explain, - Equilibrium - JEE Main

At 320 K, a gas A_{2} is 20% dissociated to A(g). The standard free energy change at
320 K and 1 atm in J mol ^{-1}  is approximately : (R=8.314 J K ^{-1}  mol ^{-1} ;
ln 2=0.693; ln 3=1.098)

  • Option 1)

    4763

  • Option 2)

    2068

  • Option 3)

    1844

  • Option 4)

    4281

 
Answers (1)
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S subam

As we have learned

Relation between Gibbs energy and reaction Quotient -

\bigtriangleup G=\bigtriangleup G^{\circ}+RTl_{n}Q

 

 

- wherein

\bigtriangleup G^{\circ} is standard Gibbs energy.

Q\rightarrow Reaction\:Quotient

 

A_{2}(g)\rightleftharpoons 2A(g)

1-0.2 , 2\times 0.2

0.8 , 0.4

K_{p}= \frac{(P_{A}^{2})}{P_{A}^{2}}=\frac{0.4*0.4}{0.8}=0.2

 \Delta G \degree= -2.303*8.314*320 log 0.2= 4281 J/mole

 

 

 

 

 

 


Option 1)

4763

This is incorrect

Option 2)

2068

This is incorrect

Option 3)

1844

This is incorrect

Option 4)

4281

This is correct

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