Respuesta :
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= [tex]\frac{ΔH vap}{ΔS vap}[/tex]
T =[tex]\frac{46550}{67.37}[/tex]
= 690 Kelvin
The boiling point is the temperature when atmospheric temperature gets equal to
Answer:
The boiling point of this compound is 417.8[tex]1^{o}[/tex] 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
= [tex]\frac{46.55kJmol-1}{67.37Jmol-1K-1}[/tex]
= [tex]\frac{46.55 x 1000 Jmol-1}{67.37Jmol-1K-1}[/tex]
= 690.96 K
= 690.96 - 273.15 deg Celsius
= 417.8[tex]1^{o}[/tex] C
Hence the boiling point of this compound is 417.8[tex]1^{o}[/tex] C
