Question

Latent heat of fusion of ice is 0.333 kilojoules per gram the increase in entropy when 1 mole water melts at zero degree celsius will be ?

Answer 1

Answer : The change in enthalpy is 0.02198 kJ/mol K or 21.98 J/mol K

Explanation :

At melting point, the solid phase and liquid phase are in equilibrium with each other. Therefore free energy change, ΔG is zero at melting point.

The equation that relates ΔG, ΔH and ΔS is given below.

$\bigtriangleup G = \bigtriangleup H - T \bigtriangleup S$

Since ΔG = 0, the equation modifies as ,

$\bigtriangleup H = T \bigtriangleup S$

We have been given latent heat of fusion which is also the enthalpy of fusion.

Therefore, ΔH = 0.333 kJ/g

Let us find ΔH for 1 mol of ice.

The equation can be set up as follows.

$1 mol H_{2}O \times \frac{18.02g}{1mol} \times \frac{0.333kJ}{g} = 6.0 kJ/mol$

ΔH = 6.0 kJ/mol

T = 273 K

Let us rearrange the equation to solve for ΔS.

$\bigtriangleup S= \frac{\bigtriangleup H}{T}$

$\bigtriangleup S = \frac{6.0kJ/mol}{273K} = 0.02198 kJ/mol.K$

The change in enthalpy is 0.02198 kJ/mol K or 21.98 J/mol K

Answer 2

Answer:

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