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Octane has a vapor pressure of 40. Torr at 45.1°c and 400. Torr at 104.0°c. What is its heat of vaporization?

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Answer:

The heat of vaporization of octane is 39.01KJ/mol.

Step-by-step explanation:

No chemical reaction take place without heat,because it is necessary for the formation of breaking of chemical bond.The reaction which does not require external energy is called as spontaneous energy and that which require external energy is called as non spontaneous energy.

By using Clausius - Clapeyron equation,

log(P₂/P₁) = Δh/2.303R(1/T₁- 1/T₂) eqn 1

P₁ is vapor pressure at temperature T₁

P₂ is vapor pressure at temperature T₂

Δh is heat of vaporization to find

P₁ = 40 Torr, T₁ =45.1°c = 318.25K

P₂ = 400 Torr, T₂ = 104.0°c. = 377.15K

substituting all values in eqn 1

Δh =
(2.303*8.314)/(377.15-318.25) *(377.15*318.25

Δh = 39018.55J/mol

The heat of vaporization of octane is 39.01KJ/mol.

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