Respuesta :
Answer: The change in entropy for the system is 2.35 J/K
Explanation:
The chemical equation follows:
[tex]C_7H_{16}(l)\rightleftharpoons C_7H_{16}(g)[/tex]
- To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of heptane = 2.8 g
Molar mass of heptane = 100.2 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of heptane}=\frac{2.8g}{100.2g/mol}=0.028mol[/tex]
To calculate the entropy change for vaporization of heptane, we use the equation:
[tex]\Delta S=\frac{n\Delta H_{vap}}{T}[/tex]
where,
[tex]\Delta S[/tex] = entropy change of the reaction
n = number of moles = 0.028 mole
[tex]\Delta H_{vap}[/tex] = heat of vaporization = 31.2 kJ/mol = 31200 J/mol (Conversion factor: 1 kJ = 1000 J)
T = temperature of the system = [tex]98.4^oC=[98.4+273]=371.4K[/tex]
Putting values in above equation, we get:
[tex]\Delta S=\frac{0.028mol\times 31200J/mol}{371.4K}\\\\\Delta S=2.35J/K[/tex]
Hence, the change in entropy for the system is 2.35 J/K
