Question

The enthalpy of fusion of water is 1.435 kcal/mol.
The molar entropy change for the melting of ice at
0°C is
[AIPMT (Prelims)-2012]
(1) 5.260 cal/mol K)
(2) 0.526 cal/(mol K)
(3) 10.52 cal/mol K)
(4) 21.04 cal/(mol K)​

Answer 1

Answer:

Formula: ΔH = TΔS

Where, ΔH symbolizes Enthalpy of Fusion, T represents Temperature in Kelvin and ΔS represent Entropy.

According to the question,

→ Enthalpy of Fusion = ΔH = 1.435 Kcal/mol

→ Enthalpy of Fusion = ΔH = 1435 cal/mol

It is given melting point of ice is 0°C. Converting it to Kelvin we get:

→ Melting Point of Ice = 0°C + 273 = 273 K

Now we are required to find the value of Molar Entropy ( ΔS )

Since ΔH is in terms of Calories per mole, we can directly substitute it in the formula.

→ 1435 cal/mol = 273 K ×ΔS

→ ΔS = 1435 cal/mole / 273 K

→ ΔS = 5.256 cal/mole.K

→ ΔS ≈ 5.26 cal/mole.K

Hence Option (1) is the correct answer.

Answer 2

Given :

• The enthalpy of fusion of water is 1.435 kcal/mol = ΔH
• Melting point of ice → 0°C

To Find :

• The molar entropy change for melting ice at 0° C

Solution :

Convert the units of enthalpy of fusion of water to cal/mol.

We know, 1 kcal → 1000 cal.

•°• 1.435 kcal/mol = $\sf{\dfrac{1.435}{1000}}$

1.435 kcal/mol = $\sf{\red{1435\:cal\:^{-mol}}}$

Now, we have melting point of ice given as 0° C, let's change it into, Kelvin.

To convert it to Kelvin, we have the formula :

• $\sf{\green{C\:=\:K\:-\:273.15}}$

Where,

• C = 0° C
• K = ?

$\longrightarrow$$\sf{0\:=\:K\:-\:273.15}$

$\longrightarrow$$\sf{0+273.15=K}$

$\longrightarrow$ $\sf{K\:=\:273.15}$

Now, we have the required quantities. Let's use the formula $\sf{\red{Enthalpy\:of\:fusion}}$

Formula :

$\large{\boxed{\sf{\purple{\triangle\:H\:=\:T\:\triangle\:S}}}}$

Block in the data,

$\longrightarrow$ $\sf{1435\:cal\:^{-mol}\:=\:273.15\:\times\:\:\triangle\:S}$

$\longrightarrow$ $\sf{\dfrac{1435\:cal\:^{-mol}}{273.15}\:=\:\triangle\:S}$

$\longrightarrow$ $\sf{5.2535\:=\:\triangle\:S}$

The answer after the calculation is approximate form of option (1).

•°• Answer is Options (1)

• 5.260 cal/mol K

$\large{\boxed{\sf{\blue{Molar\:entropy\:of\:ice\:at\:0\:\degree\:celcius\:=\:5.260\:cal^{-mol}\:K}}}}$