Question

Rhombic and monoclinic sulphur are at equilibrium at the transition temperature (${95.5}^{o}C$). The enthalpy of transition at ${95.5}^{o}C$ is $353.8J{mol}^{-1}$. The entropy of transition is:

• A. $0.064J{K}^{-1}{mol}^{-1}$
• B. $0.96J{K}^{-1}{mol}^{-1}$
  
• C. $0.016J{K}^{-1}{mol}^{-1}$
  
• D. $0.008J{K}^{-1}{mol}^{-1}$
  

Answer 1

Answer: (B)

$0.96J{K}^{-1}{mol}^{-1}$

$T=95.5+273=368.5K$

$\Delta H=+353.8J{mol}^{-1}$

Now, at equilibrium,

$\Delta S=\frac{\Delta H}{T}=\frac{353.8}{368.5}$

$\Delta S=0.96J{K}^{-1}{mol}^{-1}$

Now at $T>{95.5}^{o}C,T.\Delta S>\Delta H,i.e.,\Delta G=-ve$ or the forward reaction will proceed, i.e.,$S_M$ is more stable above ${95.5}^{o}C$.