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Q.

For the phase transformation     S(rhombic) ~ S(monoclinic),   Δf H (S, monoclinic) =0.33 kJ mol1,S0 (S, monoclinic) =32.6 JK1 mol1 and S  ( S, rhombic)=31.8 JK1 mol1.The temperature at which the above transformation is in equilibrium is

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a

412.5 K

b

352.5 K

c

315.2 K

d

452.5 K

answer is C.

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Detailed Solution

ΔrH=Δe(S monoclinic) ΔfH(S rhombic) =0.33 kJ mol10=0.33 kJ mol1

ΔrS=S (S, monoclinic) S(S, rhombic) =(32.631.8) JK1 mol1=0.8 JK1 mol1

T=ΔrHΔrS=0.33×103 J mol10.8 JK1 mol1=412.5 K

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