Get Answers to all your Questions

header-bg qa

fG0 at 500 K for substance ‘S’ in liquid state and gaseous state are +100.7 kcal mol −1 and +103 kcal mol−1 , respectively. Vapour pressure of liquid ‘S’ at 500 K is approximately equal to :
(R=2 cal K−1 mol−1 )

  • Option 1)

    0.1 atm

  • Option 2)

    1 atm

  • Option 3)

    10 atm

  • Option 4)

    100 atm

 

Answers (1)

best_answer

As we have learnt,

 

Δ G of equilibrium -

\Delta G_{0}= -2.303 RT \log K_{c}
 

- wherein

At Equilibrum    

\Delta G= 0

and Q= K_{c}

\\*\Delta G^{o}_{Reaction} = 103 - 100.7 = 2.3 kcal = 2.3\times10^{3}\;cal \\*\Delta G^{o} = -2.3RT\log K_{p} \\* 2.3 \times 10^{3} = -2.3 \times 2\times 500 \log K_{p} \\* \log K_{p} = -1 \\*K_{p} = 10^{-1} = 0.1\;atm

 

 


Option 1)

0.1 atm

Option 2)

1 atm

Option 3)

10 atm

Option 4)

100 atm

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

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