Question

The entropy change for the conversion of 1 mole of $\alpha$ tin (at 13${}^0$C, 1 atm) to 1 mole of $\beta$ tin (13${}^0$C, 1 atm), if the enthalpy of transition is 2.095 KJ mol${}^{-1}$, is :

• A. 7.32 J mol${}^{-1}$ K${}^{-1}$
• B. 14.62 J K mol${}^{-1}$
• C. 56 J mol${}^{-1}$K${}^{-1}$
• D. 0

Answer 1

Answer: (A)

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

$△S=\frac{△H}{T}$ where T is in K.

So, $△S=\frac{2095}{273+13}=$ 7.32 J mol${}^{-1}$ K${}^{-1}$