Vapour pressure of liquid ‘S’ at 500 K is approximately equal to

Answers (1)

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

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