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Question

Chemistry Question on Thermodynamics

ΔrG\Delta_{r} G^{\circ} at 500K500 \,K for substance 'S' in liquid state and gascous state are +100.7kcalmol1+100.7 \,kcal\, mol ^{-1} and +103kcalmol1+103 \,kcal \,mol ^{-1}, respectively. Vapour pressure of liquid ' S ' at 500K500\, K is approximately equal to (R=2calK1mol1)\left(R=2 \,cal K ^{-1} \,mol ^{-1}\right)

A

0.1 atm

B

1 atm

C

10 atm

D

100 atm

Answer

0.1 atm

Explanation

Solution

We have S(liquid)S (gas) S {\text{(liquid)}} \rightleftharpoons S {\text { (gas) }} ΔGreaction =ΔrG( vapor )ΔfG (liquid) \Delta G_{\text {reaction }}^{\circ} =\Delta_{r} G^{\circ}(\text { vapor })-\Delta_{f} G^{\circ} \text { (liquid) } =103100.7=103-100.7 =2.3kcalmol1=2.3 \,kcal \,mol ^{-1} We know ΔGExtion =RTlnK\Delta G_{\text {Extion }}^{\circ}=-R T \ln K 2.3×103=2×500×lnKp2.3 \times 10^{3}=-2 \times 500 \times \ln K_{p} lnKp=2.3ln \,K_{p}=-2.3 Kp=0.1 \Rightarrow K_{p}=0.1  vapour pressure =0.1atm \Rightarrow \text { vapour pressure }=0.1\, atm