Question

The ΔH vap of a certain compound is 46.55 kJ⋅ mol−1 and its ΔS vap is 67.37 J⋅mol−1⋅K−1.

Answer 1

Answer:

The boiling point of this compound is  417.8$1^{o}$ C

Explanation:

We know that

ΔG = ΔH-TΔS

where

ΔG is the Gibb's free energy of vaporization

ΔS is the entropy of vaporization

ΔH is the enthalpy of vaporization

T is boiling point Temperature in kelvin

Assuming ΔG = 0, we get

ΔH = TΔS

T = ΔH / ΔS

  = $\frac{46.55kJmol-1}{67.37Jmol-1K-1}$

  = $\frac{46.55 x 1000 Jmol-1}{67.37Jmol-1K-1}$

  = 690.96 K

  = 690.96 - 273.15 deg Celsius

  = 417.8$1^{o}$ C

Hence the boiling point of this compound is 417.8$1^{o}$ C

Answer 2

690 Kelvin is the boiling point of this compound.

Explanation:

Enthalpy is the sum of internal energy and the product of pressure and volume that is how much energy is in the substance.

Entropy is the measurement of randomness and measure of thermal energy per unit of temperature.

ΔH vap of compound is 46.55 kJ⋅ mol− or  J.MOL-1

ΔS vap  is 67.37 J⋅mol−1⋅K−1.

The boiling point or temperature can be calculated by the formula:

T= $\frac{ΔH vap}{ΔS vap}$

T =$\frac{46550}{67.37}$      

  = 690 Kelvin

The boiling point is the temperature when atmospheric temperature gets equal to