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Q.

For the phase transformation S(rhombic) ~ S(monoclinic), $Δ_{f} H^{\circ}$ (S, monoclinic) $= 0.33 kJ {mol}^{- 1}$ , $S^{0}$ (S, monoclinic) $= 32.6 {JK}^{- 1} {mol}^{- 1}$ and $S^{\circ}$ ( S, rhombic)= $31.8 {JK}^{- 1} {mol}^{- 1}$ .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

$Δ_{r} H^{\circ} = {Δe}^{\circ}$ (S monoclinic) $- Δ_{f} H^{\circ}$ (S rhombic) $= 0.33 kJ {mol}^{- 1} - 0 = 0.33 kJ {mol}^{- 1}$

$Δ_{r} S^{\circ} = S^{\circ}$ (S, monoclinic) $- S^{\circ}$ (S, rhombic) $= (32.6 - 31.8) {JK}^{- 1} {mol}^{- 1} = 0.8 {JK}^{- 1} {mol}^{- 1}$

$T = \frac{Δ_{r} H^{\circ}}{Δ_{r} S^{\circ}} = \frac{0.33 \times 10^{3} J {mol}^{- 1}}{0.8 {JK}^{- 1} {mol}^{- 1}} = 412.5 K$

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