You visited us 1 times! Enjoying our articles? Unlock Full Access!

Entropy

Enthalpy chan...

Question

Enthalpy change for the process,
$H_{2} O (s) ⇌ H_{2} O (l)$
is $6.01 k J m o l^{- 1} .$ The entropy change for 1 mole of ice at its melting point will be

12 J K^{- 1} m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

22 J K^{- 1} m o l^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

100 J K^{- 1} m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30 J K^{- 1} m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $22 J K^{- 1} m o l^{- 1}$
The entropy change in a phase transformation (fusion or melting here) is given as:
$△ S_{f u s i o n} = \frac{△ H_{f u s i o n}}{T_{f u s i o n}}$
melting point of ice = $273 K$
so
$△ S_{f u s i o n} = \frac{6.01 \times 1000}{273}$

$= 22 J K^{- 1} m o l^{- 1}$

Suggest Corrections

Similar questions

H_{2} O (i c e) ⇌ H_{2} O (w a t e r)

m o l^{- 1}

H_{2} O (s) ⇌ H_{2} O (l)

6.01 k J m o l^{- 1} .

H_{2} O (i c e) ⇌ H_{2} O (w a t e r)

6.01 K J {m o l}^{- 1}

1

6.01 k J {m o l}^{- 1}

1

m o l e^{- 1}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program