Question

53. The latent heat of fusion of ice is 5.99 KJ/mol at its melting point. Then
(i) delta S for fusion of 900 g ice and
(in) delta S for freezing of liquid water are respectively​

Answer 1

Answer:

For (i): The $\Delta S$ for fusion of given amount of ice is 1.097 kJ/K.

For (ii): The $\Delta S$ for freezing for given amount of water is -1.097 kJ/K.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of water = 900 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$\text{Moles of water}=\frac{900g}{18g/mol}=50mol$

We are given:

Latent heat of fusion = 5.99 kJ/mol

For 1 mole of ice, the latent heat of fusion is 5.99 kJ.

So, for 50 moles of ice, the latent heat of fusion will be $\frac{5.99}{1}\times 50=299.5kJ$

• For (i):

The equation for melting of ice follows:

$H_2O(s)\rightleftharpoons H_2O(l)$

To calculate the entropy change for melting of ice, we use the equation:

$\Delta S=\frac{\Delta H_{fusion}}{T}$

$\Delta H_{fusion}=299.5kJ$

T = melting point of ice = 0°C = 273 K     (Conversion used:  $T(K)=T(^oC)+273$  )

Putting values in above equation, we get:

$\Delta S=\frac{299.5kJ}{273K}=1.097kJ/K$

Hence, the $\Delta S$ for fusion of given amount of ice is 1.097 kJ/K.

• For (ii):

The equation for melting of ice follows:

$H_2O(l)\rightleftharpoons H_2O(s)$

To calculate the entropy change for freezing of water, we use the equation:

$\Delta S=-\frac{\Delta H_{fusion}}{T}$

$\Delta H_{fusion}=299.5kJ$

T = freezing point of water = 0°C = 273 K

Putting values in above equation, we get:

$\Delta S=-\frac{299.5kJ}{273K}=-1.097kJ/K$

Hence, the $\Delta S$ for freezing for given amount of water is -1.097 kJ/K.