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

Formula: AH = TAS

Where, AH symbolizes Enthalpy of Fusion,

T represents Temperature in Kelvin and AS represent Entropy.

According to the question,

» Enthalpy of Fusion = AH = 1.435 Kcal/mol

Enthalpy of Fusion = AH = 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 (AS)

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

1435 cal/mol = 273 K xAS

» AS = 1435 cal/mole / 273 K

» AS = 5.256 cal/mole.K

+ AS = 5.26 cal/mole.K

Hence Option (1) is the correct answer.

Answer 2

Answer:

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 3

Answer:

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 4

Answer:

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 5

Answer:

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 6

Answer:

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.